RBSE Solutions Class 10 Science Chapter 9 Light

I. Multiple Choice Questions

Question 1. Which of the following mirrors (RBSESolutions.com) has a wide field of view?
(a) Plain mirror
(b) Convex mirror
(c) Concave mirror
(d) Parabolic mirror
Answer: (b) Convex mirror
In simple words: A convex mirror is best for seeing a wide area, like the rear-view mirror in a car. This is because its curved shape makes objects appear smaller, allowing more of the surroundings to be seen.

🎯 Exam Tip: Remember that convex mirrors diverge light, which creates a wide field of view and forms virtual, erect, and diminished images.

Question 2. Speed of light is maximum in which of the following?
(a) Water
(b) Glass
(c) Vacuum
(d) Glycerine
Answer: (c) Vacuum
In simple words: Light travels fastest in a vacuum, which is empty space, because there are no particles to slow it down. It slows down when it passes through materials like water or glass.

🎯 Exam Tip: The speed of light is a fundamental constant in a vacuum, represented by 'c'. Its speed decreases in any medium due to interactions with the medium's particles.

Question 3. Which of the following is responsible for the coin (RBSESolutions.com) appearing shallower when it is kept at the bottom of a water tank?
(a) Refraction
(b) Reflection
(c) Total internal reflection
(d) None of these
Answer: (a) Refraction
In simple words: When you look at something underwater, like a coin, it looks closer to the surface than it really is. This happens because light bends as it moves from water to air, a process called refraction.

🎯 Exam Tip: Refraction is the bending of light as it passes from one medium to another, causing optical illusions like objects appearing shifted or distorted.

Question 4. The focal length of a mirror is +60 cm. Which kind of mirror it is?
(d) Convex mirror
Answer: (d) Convex mirror
In simple words: When a mirror has a positive focal length, it means it is a convex mirror. Convex mirrors always make images that are smaller and stand upright, which is helpful in many applications.

🎯 Exam Tip: A positive focal length indicates a convex mirror or a converging lens, while a negative focal length indicates a concave mirror or a diverging lens.

Question 5. What is the focal length of a plain mirror?
(a) 0
(b) 1
(c) Infinity
(d) None of these
Answer: (c) Infinity
In simple words: A flat mirror doesn't curve light rays, so its focal point is infinitely far away. This is why it always produces an image that looks the same distance behind the mirror as the object is in front.

🎯 Exam Tip: A plane mirror is considered a special case where the radius of curvature is infinite, leading to an infinite focal length.

Question 6. Which type of image is always formed by a convex mirror?
(a) Real and erect
(b) Real and inverted
(c) Virtual and inverted
(d) Virtual and erect
Answer: (d) Virtual and erect
In simple words: A convex mirror always makes an image that appears behind the mirror, cannot be projected onto a screen, and is always upright. This kind of image is called virtual and erect.

🎯 Exam Tip: Convex mirrors are known for always producing virtual, erect, and diminished images, making them useful for wide-angle views.

Question 7. The power of a lens is +2 Dioptre. What is the focal length of this mirror?
(a) 2 m
(b) 1 m
(c) 0.5 m
(d) 0.2 m
Answer: (c) 0.5 m
In simple words: The power of a lens tells us how much it bends light, and it's the inverse of its focal length in meters. If the power is +2 Dioptres, the focal length is \(1 \div 2\), which is 0.5 meters.

🎯 Exam Tip: Remember the formula \( P = 1/f \) where \( f \) is in meters. A positive power means a converging (convex) lens, and a negative power means a diverging (concave) lens.

Question 8. Which of the following is true for a person with long sightedness?
(a) He will clearly see a nearby object
(b) He will clearly see a distant object
(c) ee nearby and distant objects
Answer: (b) He will clearly see a distant object
In simple words: A person with long sightedness, also called hypermetropia, can see far-away things clearly but struggles to focus on objects that are close by.

🎯 Exam Tip: Long sightedness means the eye converges light too weakly or the eyeball is too short, causing the image of nearby objects to form behind the retina.

Question 9. The focal length of a convex lens is 15 cm. What should be the object distance to get a real image equal to the size of object?
(a) 30 cm
(b) 15 cm
(c) 60 cm
(d) None of these
Answer: (a) 30 cm
In simple words: For a convex lens, to get a real image that is the same size as the object, you need to place the object at twice the focal length. So, if the focal length is 15 cm, the object distance should be 30 cm.

🎯 Exam Tip: This condition (object at 2F) is a key characteristic of convex lenses, resulting in a real, inverted image of the same size at 2F on the other side.

Question 10. An object is kept at infinity from a concave lens of focal length 20 cm. What is the image distance?
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) Infinity
Answer: (c) 20 cm
In simple words: When an object is placed very far away (at infinity) from a concave lens, the lens forms a virtual image at its focal point. So, the image distance will be equal to the focal length.

🎯 Exam Tip: For a concave lens, an object at infinity always forms a virtual, erect, and highly diminished image at its principal focus on the same side as the object.

Light Very Short Answer Type Questions

Question 1. An object absorbs all the colours of light falling on it. This object would appear of which colour?
Answer: If an object absorbs all colors of light, it does not reflect any light back to our eyes. Therefore, the object will appear black. Black is the absence of reflected light.
In simple words: An object that soaks up all light colors looks black because no light bounces off it for us to see.

🎯 Exam Tip: Remember that the color of an opaque object is determined by the colors of light it reflects. If all colors are absorbed, it appears black.

Question 2. What should be the minimum length of a plain mirror if you want to see your whole image in it?
Answer: The minimum length of a plane mirror required to see your full image is half of your height. This allows light rays from your head and feet to reflect to your eyes. So, if you are 160 cm tall, you need a mirror at least 80 cm long.
In simple words: You need a flat mirror that is at least half your height to see your entire body in it.

🎯 Exam Tip: This principle applies regardless of how far you stand from the mirror; the minimum required length remains half your height.

Question 4. Write any two uses of (RBSESolutions.com) convex mirror.
Answer: Convex mirrors are used in various practical applications because they always form virtual, erect, and diminished images with a wide field of view. Two common uses are: 1. **As rear-view mirrors in vehicles:** They allow the driver to see a larger area of traffic behind the car, even though objects appear smaller and farther away. 2. **As security mirrors in shops:** Their wide field of view helps monitor a large area, deterring theft and improving surveillance.
In simple words: Convex mirrors are great for car side mirrors and security cameras in stores because they let you see a lot of space at once.

🎯 Exam Tip: When explaining uses, always link the function to the specific properties of the mirror (e.g., wide field of view for convex mirrors).

Question 5. Write any two uses of concave mirror.
Answer: Concave mirrors are versatile because they can form both real and virtual images, and can converge light. Two common uses are: 1. **As shaving mirrors:** When held close to the face, they produce an enlarged, erect, and virtual image, making it easier to shave. 2. **In solar cookers:** They collect and focus sunlight onto a small area, generating high temperatures for cooking.
In simple words: Concave mirrors are used for shaving to make your face look bigger and in solar cookers to collect sunlight and make it hot.

🎯 Exam Tip: For concave mirrors, emphasize their converging properties and ability to produce magnified images when the object is within the focal length.

Question 6. Write the mirror formula.
Answer: The mirror formula mathematically relates the focal length of a spherical mirror to the object distance and image distance. This formula is crucial for calculating image positions.
\( \frac { 1 }{ u } + \frac { 1 }{ v } = \frac { 1 }{ f } \)
Where:
\( u \) = object distance
\( v \) = image distance
\( f \) = focal length
In simple words: The mirror formula is a math rule that connects how far an object is, how far its image is, and the mirror's focal length.

🎯 Exam Tip: Always remember to use the correct sign conventions (Cartesian sign convention) when applying the mirror formula to avoid errors in calculations.

Question 7. What is the relation between radius of (RBSESolutions.com) curvature and focal length of a spherical mirror?
Answer: For spherical mirrors, the focal length (\( f \)) is exactly half of the radius of curvature (\( R \)). This means the focal point lies midway between the pole and the center of curvature.
\( f = \frac { R }{ 2 } \)
Where \( f \) is the focal length and \( R \) is the radius of curvature.
In simple words: The focal length of a round mirror is always half of its radius of curvature.

🎯 Exam Tip: This relationship is fundamental for both concave and convex spherical mirrors, simplifying calculations involving their properties.

Question 8. Write magnification formula.
Answer: The magnification formula describes how much larger or smaller an image is compared to the object, and whether it's inverted or erect. It can be expressed in terms of image/object heights or distances.
\( m = \frac { \text{height of image } h' }{ \text{object height } h } = \frac { -v }{ u } \)
Where:
\( m \) = magnification
\( h' \) = height of the image
\( h \) = height of the object
\( v \) = image distance
\( u \) = object distance
In simple words: Magnification tells you if the image is bigger or smaller than the real object, and it's calculated using the heights or distances of the image and object.

🎯 Exam Tip: A negative magnification indicates a real and inverted image, while a positive magnification indicates a virtual and erect image.

Question 9. Write Snell's Law.
Answer: Snell's Law describes the relationship between the angles of incidence and refraction for light passing through an interface between two different media. It is a key law in optics explaining how light bends.
Snell's law states that:
\( \frac { \sin i }{ \sin r } = \text{constant} \)
Where:
\( i \) = angle of incidence
\( r \) = angle of refraction
The constant is the refractive index of the second medium with respect to the first.
In simple words: Snell's Law explains how light bends when it moves from one material to another, linking the angles of light and the properties of the materials.

🎯 Exam Tip: The constant in Snell's Law is specifically the relative refractive index of the second medium concerning the first, indicating how much light's speed changes.

Short Answer Type Questions

Question 11. Parallel ray from an object are incident on a (RBSESolutions.com) convex lens. What is the location of image in this case?
Answer: When parallel rays of light from an object fall on a convex lens, they converge after passing through the lens. The image is formed at the principal focus (\( F_2 \)) on the opposite side of the lens. This is why convex lenses are used to focus light.
In simple words: If light rays hit a convex lens in parallel, they will all meet at the lens's focal point on the other side, forming an image there.

🎯 Exam Tip: For a convex lens, parallel rays always converge at \( F_2 \), forming a real, inverted, and highly diminished image.

Question 12. What is the unit of power of lens?
Answer: The unit of power of a lens is the dioptre (D). One dioptre is defined as the power of a lens whose focal length is 1 meter. This unit helps eye doctors prescribe corrective lenses accurately.
\( P = \frac { 1 }{ f (\text{in meters}) } \)
So, if \( f = 1 \) meter, then \( P = 1 \) Dioptre.
In simple words: The unit for measuring how strong a lens is, is called a dioptre.

🎯 Exam Tip: Always ensure the focal length is expressed in meters when calculating lens power in dioptres; otherwise, the result will be incorrect.

Question 13. A person suffering from short sightedness is unable to see objects at which position?
Answer: A person with short sightedness, also known as myopia, is unable to see distant objects clearly. Their vision is sharp for nearby objects, but objects far away appear blurry. This happens because the light focuses in front of the retina.
In simple words: Someone with short sightedness can't see things far away very well.

🎯 Exam Tip: Myopia occurs when the eyeball is too long or the lens is too powerful, causing light to converge before reaching the retina.

Question 14. A convex lens of suitable focal length can rectify which defect of vision?
Answer: A convex lens of suitable focal length can rectify long sightedness, which is also known as hypermetropia. In this condition, the eye's lens cannot converge light enough, causing distant objects to appear blurry. The convex lens helps converge the light rays before they enter the eye.
In simple words: A convex lens can fix long sightedness, helping people see far away more clearly.

🎯 Exam Tip: Convex lenses are converging lenses, which are used to increase the converging power of the eye in cases of hypermetropia.

Question 15. What is cataract?
Answer: Cataract is an eye condition where the natural lens of the eye becomes cloudy or opaque. This causes vision to become blurry, hazy, or less colorful, similar to looking through a frosted window. It can be treated with surgery.
In simple words: Cataract is when the eye's natural lens gets cloudy, making it hard to see clearly.

🎯 Exam Tip: Cataracts commonly develop with age and are a leading cause of blindness globally, but they are treatable with a surgical procedure to replace the cloudy lens.

Question 16. Which type of image if formed by a shaving mirror?
Answer: A shaving mirror is a concave mirror. When an object (the face) is placed very close to a concave mirror (within its focal length), it forms a virtual, enlarged, and erect image. This magnified image helps in precise shaving.
In simple words: A shaving mirror makes an image that looks bigger and upright, but it's not a real image you can catch on a screen.

🎯 Exam Tip: Concave mirrors produce enlarged images when the object is placed between the pole and the principal focus.

Question 2. What do you understand by lateral inversion?
Answer: Lateral inversion is the phenomenon where the right side of an object appears as the left side in its image formed by a plane mirror, and vice-versa. For example, if you raise your right hand, your image in a mirror will appear to raise its left hand. This effect is commonly observed in everyday mirrors.
In simple words: Lateral inversion means that when you look in a flat mirror, your right side looks like your image's left side, and your left side looks like its right.

🎯 Exam Tip: Understand that lateral inversion is a reversal of sides, not an up-down inversion, which is why ambulance text is written in a mirrored way.

Question 3. With the help of a ray diagram, show the location (RBSESolutions.com) of image when an object is between centre of curvature and focus of a concave mirror.
Answer: When an object is placed between the centre of curvature (C) and the principal focus (F) of a concave mirror, a real, inverted, and enlarged image is formed beyond the centre of curvature (C). The light rays meet to form this image.

In simple words: When an object is placed between C and F for a concave mirror, the image forms beyond C, is bigger, real, and upside down.

🎯 Exam Tip: Always draw at least two principal rays to accurately locate the image and describe its characteristics for a ray diagram.

Question 4. Explain Cartesian sign convention for spherical mirror.
Answer: The Cartesian sign convention is a set of rules used to determine the signs of distances and heights when working with spherical mirrors. These rules ensure consistent calculations.
Following are the Cartesian sign conventions for spherical mirror:

• All distances are measured from the pole (P) of the mirror.
• Incident light is shown coming from the Left Hand Side (LHS) of the mirror.
• Distances measured towards the LHS of the mirror are taken as negative, and those measured towards the RHS are taken as positive.
• The height measured perpendicular to and above the principal axis is taken as positive.
• The height measured perpendicular to and below the principal axis is taken as negative.

These conventions help to standardize calculations and interpret results for image formation by mirrors.
In simple words: Cartesian sign convention gives rules for positive and negative directions for measurements around a mirror, like distances and heights, to make calculations easy.

🎯 Exam Tip: Mastering the sign conventions is crucial for solving numerical problems in optics accurately; a wrong sign can lead to incorrect results.

Question 5. Explain refraction of (RBSESolutions.com) light and write the laws of refraction.
Answer: Refraction of light is the bending of light as it passes from one transparent medium to another, such as from air to water. This bending occurs because light changes its speed when it enters a new medium. This change in speed causes the light path to deviate.
Following are the laws of refraction:

• The incident ray, the refracted ray, and the normal to the interface at the point of incidence all lie in the same plane.
• The ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)) is constant for a given pair of media and a given color of light. This is known as Snell's Law.
  \( \frac { \sin i }{ \sin r } = \text{constant} \)

In simple words: Refraction is when light bends as it moves from one material to another because its speed changes. There are two main rules for how this bending happens.

🎯 Exam Tip: Always remember that the angle of incidence and refraction are measured with respect to the normal, an imaginary line perpendicular to the surface.

Question 6. What are the different types of convex and concave lenses?
Answer: Lenses come in various types, depending on their curved surfaces, which determine how they bend light.
**Types of convex lenses:** Convex lenses are thicker in the middle and thinner at the edges, causing parallel light rays to converge.

• **Biconvex lens:** Both surfaces are convex.
• **Plano-convex lens:** One surface is flat, and the other is convex.
• **Concavo-convex lens:** One surface is concave and the other is convex, but the convex surface has a greater curvature, resulting in overall converging power.

**Types of concave lenses:** Concave lenses are thinner in the middle and thicker at the edges, causing parallel light rays to diverge.

• **Biconcave lens:** Both surfaces are concave.
• **Plano-concave lens:** One surface is flat, and the other is concave.
• **Convexo-concave lens:** One surface is convex and the other is concave, but the concave surface has a greater curvature, resulting in overall diverging power.

In simple words: Convex lenses are those that make light come together, and they have shapes like being curved on both sides or flat on one. Concave lenses spread light out, and they also have different shapes like being curved inwards on both sides.

🎯 Exam Tip: Remember that convex lenses are converging lenses, and concave lenses are diverging lenses; their names indicate their primary effect on light.

Question 7. Define main focus and optical (RBSESolutions.com) centre of spherical lens.
Answer: The main focus and optical centre are important points used to describe how a spherical lens behaves with light.
**Main Focus (Principal Focus):** When rays of light parallel to the principal axis pass through a spherical lens, they either converge to a point on the principal axis (for a convex lens) or appear to diverge from a point on the principal axis (for a concave lens). This point is called the main focus or principal focus of the spherical lens. A lens has two principal foci.
**Optical Centre (O):** The optical centre is the central point of a lens. A ray of light passing through the optical centre of a lens goes undeviated, meaning it does not bend. This point acts as a reference for measuring distances in lens equations.
In simple words: The main focus is where parallel light rays meet or seem to come from after passing through a lens. The optical centre is the middle point of the lens, where light rays pass straight through without bending.

🎯 Exam Tip: For ray diagrams, always remember that rays passing through the optical centre are not refracted, making it a reliable reference point.

Question 8. What do you understand by radius of curvature and centre of curvature of spherical lens?
Answer: Spherical lenses are made from parts of spheres, and their properties are defined by these spheres.
**Centre of Curvature (C):** The curved surfaces of a lens are parts of two spheres. The centre of these imaginary spheres is called the centre of curvature. A lens has two centres of curvature, usually denoted as \( C_1 \) and \( C_2 \), or \( 2F_1 \) and \( 2F_2 \).
**Radius of Curvature (R):** The radius of curvature is the radius of the imaginary spheres from which the lens surfaces are taken. It is the distance between the optical centre of the lens and its centre of curvature. Lenses can have different radii of curvature for each surface.
In simple words: The centre of curvature is the center of the big imaginary ball that each curved part of the lens comes from. The radius of curvature is just how big that imaginary ball is, measured from its center to the lens surface.

🎯 Exam Tip: Remember that each curved surface of a lens has its own centre and radius of curvature, which can be the same or different depending on the lens design.

Question 9. Write the laws of refraction through spherical lens.
Answer: The laws of refraction for spherical lenses are based on how light bends when passing through curved surfaces.
Following are the laws of refraction through spherical lens:

• A ray of light parallel to the principal axis passes through the main focus (\( F_2 \)) after refraction through a convex lens. For a concave lens, it appears to diverge from the main focus (\( F_1 \)).
• A ray of light passing through the principal focus (\( F_1 \)) of a convex lens becomes parallel to the principal axis after refraction. For a concave lens, a ray directed towards \( F_2 \) becomes parallel to the principal axis after refraction.
• A ray of light passing through the optical centre (O) of the lens goes undeviated (without bending).

In simple words: For spherical lenses, light rays that are straight before the lens will go through a special point called the focus. Rays going through the center of the lens won't bend at all.

🎯 Exam Tip: These three rules are essential for drawing accurate ray diagrams to determine the position and nature of images formed by lenses.

Image Formation by Concave Lens

Object at infinity: When an object is at infinity, the image is formed at \( F_1 \). The image is smaller, virtual, and erect.
Object between infinity and optical centre: When an object is placed anywhere between infinity and the optical centre, the image is formed between \( F_1 \) and O. The image is smaller, virtual, and erect.
In simple words: For a concave lens, an object very far away makes a tiny, upright image at the first focal point. If the object is anywhere else, the image is still small, upright, and forms between the focal point and the lens's center.

🎯 Exam Tip: Concave lenses always produce virtual, erect, and diminished images, regardless of the object's position, and are used for correcting short-sightedness.

Question 11. What do you understand by power of lens?
Answer: The power of a lens is a measure of its ability to converge (bring together) or diverge (spread out) light rays. A lens with a shorter focal length has greater power because it bends light more strongly. It is defined as the reciprocal of the focal length, measured in dioptres (D).
\( P = \frac { 1 }{ f } \)
The SI unit of power of lens is Diopter (D).
In simple words: Lens power tells you how much a lens bends light. A stronger lens with a shorter focal length has more power.

🎯 Exam Tip: A positive power indicates a converging (convex) lens, while a negative power indicates a diverging (concave) lens.

Question 12. What do you understand by near (RBSESolutions.com) sightedness? How is it rectified?
Answer: Near sightedness, also called myopia, is a vision defect where a person can see nearby objects clearly but finds it difficult to see distant objects. This occurs because the image of a distant object forms in front of the retina, rather than directly on it. This can happen due to an increase in the curvature of the eye lens or an elongation of the eyeball.
This defect is rectified by using a concave lens of suitable focal length. The concave lens diverges the incoming light rays slightly before they enter the eye, allowing the image to form correctly on the retina.
In simple words: Near sightedness means distant objects are blurry because the eye focuses light too early. It's fixed by wearing concave lenses.

🎯 Exam Tip: When correcting myopia, the concave lens used should have a focal length equal to the far point of the myopic eye.

Question 13. What do you understand by far sightedness? How is it rectified?
Answer: Far sightedness, also known as hypermetropia, is a vision defect where a person finds it difficult to see nearby objects clearly, but distant objects are seen well. This happens because the image of nearby objects forms behind the retina. This can occur if the focal length of the eye lens is too long or if the eyeball is too short.
This defect is rectified by using a convex lens of suitable focal length. The convex lens helps to converge the light rays further before they enter the eye, ensuring the image forms precisely on the retina.
In simple words: Far sightedness means close objects are blurry because the eye focuses light too late. It's fixed by wearing convex lenses.

🎯 Exam Tip: Hypermetropia correction requires a converging (convex) lens to increase the overall converging power of the eye system.

Question 15. What do you understand by power of accommodation and field of vision of eye?
Answer: The human eye has amazing abilities that allow us to see the world around us.
**Power of Accommodation:** This is the ability of the human eye to adjust its focal length to clearly see objects at different distances, from nearby to far away. This adjustment is done by the ciliary muscles, which change the curvature of the eye lens.
**Field of Vision:** This refers to the total area that can be seen by one or both eyes at any given moment without moving the head. The normal human binocular (two-eye) field of vision is slightly more than 180 degrees horizontally, providing a wide panorama. This wide field of vision helps with spatial awareness and peripheral detection.
In simple words: Power of accommodation is how your eye changes focus to see things far or near. Field of vision is how wide an area your eyes can see without moving your head.

🎯 Exam Tip: Emphasize that power of accommodation is due to the flexibility of the eye lens and the action of ciliary muscles.

Question 16. An object is kept between infinity and \(2F_1\) of a convex lens. Explain (RBSESolutions.com) the location of image with the help of a suitable ray diagram.
Answer: When an object is placed between infinity and \(2F_1\) of a convex lens, the image is formed between \(F_2\) and \(2F_2\) on the other side of the lens. This image is always smaller than the object, real, and inverted. The convex lens converges the light rays to form this image.

In simple words: When an object is placed between very far away and twice the focal length of a convex lens, its image forms between the focal length and twice the focal length on the other side, and it's smaller and upside down.

🎯 Exam Tip: Always draw parallel rays converging to F2 and rays through the optical center remaining undeviated to locate the image for convex lenses.

Question 5. With the help of suitable ray diagrams, explain image formation by convex lens for following:
(a) Object is between focus and optical centre
(b) Object is at focus
(c) Object is between \(F_1\) and \(2F_1\)
(d) Object is at \(2F_1\)
(e) Object is between infinity and \(2F_1\)
Answer: The nature, position, and size of the image formed by a convex lens depend on the object's position relative to the lens. Here's a summary of the image formation for different object positions:

**(a) Object is between focus and optical centre:**
When the object is placed between the principal focus (\(F_1\)) and the optical centre (O), the image is formed on the same side of the lens as the object. This image is enlarged, virtual, and erect.

**(b) Object is at focus (\(F_1\)):**
When the object is placed at the principal focus (\(F_1\)), the image is formed at infinity. It is highly enlarged, real, and inverted. The light rays become parallel after refraction, never meeting at a finite point.

**(c) Object is between \(F_1\) and \(2F_1\)):**
When the object is placed between \(F_1\) and \(2F_1\), the image is formed beyond \(2F_2\). This image is enlarged, real, and inverted. This setup is often used for projectors.

**(d) Object is at \(2F_1\)):**
When the object is placed at \(2F_1\), the image is formed at \(2F_2\) on the other side of the lens. This image is real, inverted, and exactly the same size as the object. This is a symmetrical case for convex lenses.

**(e) Object is between infinity and \(2F_1\)):**
When the object is placed anywhere between infinity and \(2F_1\), the image is formed between \(F_2\) and \(2F_2\) on the other side of the lens. This image is diminished (smaller), real, and inverted.

In simple words: How a convex lens forms an image depends on where the object is. It can make images that are bigger, smaller, or the same size, and they can be real (inverted) or virtual (upright), depending on how far the object is from the lens.

🎯 Exam Tip: Always use at least two standard rays (parallel to axis, through focus, or through optical center) to accurately draw ray diagrams for lenses.

Question 6. Explain different types of vision defects and their rectification.
Answer: There are different types of vision defects that can affect how a person sees. Each defect has a specific cause and can be corrected using certain lenses.

• Myopia (Near-sightedness): A person with myopia finds it hard to see distant objects clearly, but nearby objects appear sharp. This happens because the eye's lens has too much curvature, or the eyeball is too long. As a result, the image of distant objects forms in front of the retina. To fix this, a concave lens is used, which diverges light rays before they enter the eye, allowing the image to form correctly on the retina.
• Presbyopia: This defect occurs with age when the ciliary muscles of the eye become weak. These muscles help change the shape of the eye's lens to focus on objects at different distances. When they weaken, a person struggles to see both near and far objects clearly. Bifocal lenses, which have two different focusing powers, are used for correction.
• Astigmatism: This condition occurs due to an irregular curve in the cornea, which is the clear front part of the eye. People with astigmatism have trouble seeing vertical and horizontal lines clearly at the same time. This defect is corrected using cylindrical lenses. Understanding these defects helps in choosing the right corrective measures to improve vision.

In simple words: Vision defects like myopia (near-sightedness), presbyopia (age-related focus issues), and astigmatism (blurred vision from uneven eye surface) make it hard to see clearly. Each can be fixed with specific types of corrective lenses.

🎯 Exam Tip: When explaining vision defects, always mention the cause of the defect, its symptoms (what the person struggles to see), and the type of lens used for its correction.

Light Numerical Questions

Question 1. Focal length of a concave mirror is 30 cm. If object is at a distance of 40 cm then find the image distance. Find the magnification as well.
Answer: For a concave mirror, the focal length \( f \) is negative, so \( f = -30 \) cm. The object distance \( u \) is always taken as negative, so \( u = -40 \) cm. We use the mirror formula to find the image distance \( v \).

The mirror formula is:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Now, substitute the values:
\[ \frac{1}{v} + \frac{1}{-40} = \frac{1}{-30} \]
\[ \frac{1}{v} = \frac{1}{-30} - \frac{1}{-40} \]
\[ \frac{1}{v} = \frac{-1}{30} + \frac{1}{40} \]
To add these fractions, find a common denominator, which is 120:
\[ \frac{1}{v} = \frac{-4}{120} + \frac{3}{120} \]
\[ \frac{1}{v} = \frac{-4+3}{120} \]
\[ \frac{1}{v} = \frac{-1}{120} \]
\[ v = -120 \text{ cm} \]

The image distance is \( -120 \) cm. The negative sign indicates that the image is formed on the same side as the object, which means it is a real image.

Next, we calculate the magnification \( M \). For a mirror, the magnification formula is:
\[ M = \frac{-v}{u} \]
Substitute the values of \( v \) and \( u \):
\[ M = \frac{-(-120)}{-40} \]
\[ M = \frac{120}{-40} \]
\[ M = -3 \]

The magnification is \( -3 \). The negative sign confirms that the image is real and inverted. The value \( 3 \) shows that the image is three times larger than the object. This magnification is consistent with a real, inverted, and magnified image formed by a concave mirror when the object is placed between the center of curvature and the principal focus.
In simple words: The mirror forms an image 120 cm from it, on the same side as the object. This image is real and inverted, and it is 3 times bigger than the object.

🎯 Exam Tip: Remember to use the correct sign conventions for focal length and object distance. For concave mirrors, \( f \) is negative. For real objects, \( u \) is always negative. A negative image distance \( v \) means a real image, while a positive \( v \) means a virtual image. A negative magnification \( M \) means a real and inverted image, and a positive \( M \) means a virtual and erect image.

Question 3. An object is at a distance of 60 cm from a convex lens of focal length 30 cm. If height of object is 3 cm then find the location and nature of image.
Answer: For a convex lens, the focal length \( f \) is positive, so \( f = 30 \) cm. The object distance \( u \) is always taken as negative, so \( u = -60 \) cm. The height of the object \( h \) is \( 3 \) cm. We use the lens formula to find the image distance \( v \).

The lens formula is:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Now, substitute the values:
\[ \frac{1}{v} - \frac{1}{-60} = \frac{1}{30} \]
\[ \frac{1}{v} + \frac{1}{60} = \frac{1}{30} \]
\[ \frac{1}{v} = \frac{1}{30} - \frac{1}{60} \]
To subtract these fractions, find a common denominator, which is 60:
\[ \frac{1}{v} = \frac{2}{60} - \frac{1}{60} \]
\[ \frac{1}{v} = \frac{2-1}{60} \]
\[ \frac{1}{v} = \frac{1}{60} \]
\[ v = 60 \text{ cm} \]

The image distance is \( 60 \) cm. The positive sign indicates that the image is formed on the opposite side of the lens from the object, which means it is a real image. Since the object is placed at \( 2f \) (which is \( 2 \times 30 = 60 \) cm) for a convex lens, the image is formed at \( 2f \) on the other side. This is a special case where object at \( 2f \) produces image at \( 2f \) of same size, real and inverted.

Next, we calculate the height of the image \( h' \) using the magnification formula for lenses:
\[ M = \frac{h'}{h} = \frac{v}{u} \]
Substitute the values:
\[ \frac{h'}{3} = \frac{60}{-60} \]
\[ \frac{h'}{3} = -1 \]
\[ h' = -3 \text{ cm} \]

The height of the image is \( -3 \) cm. The negative sign means the image is inverted. The magnitude \( 3 \) cm indicates that the image is the same size as the object. So, the image is formed at 60 cm on the other side of the lens, is real, inverted, and the same size as the object.
In simple words: The lens forms an image 60 cm away on the other side. The image is real, upside down, and the same size as the object (3 cm tall).

🎯 Exam Tip: Always remember that for convex lenses, \( f \) is positive. A positive \( v \) means a real image on the opposite side of the lens. The sign of magnification \( M \) tells you the image's orientation: negative for inverted (real) and positive for erect (virtual).

Question 5. Focal length of a concave mirror is 30 cm. If object is at a distance of 20 cm then find the location and nature of image.
Answer: For a concave mirror, the focal length \( f \) is negative, so \( f = -30 \) cm. The object distance \( u \) is always taken as negative, so \( u = -20 \) cm. We use the mirror formula to find the image distance \( v \).

The mirror formula is:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Now, substitute the values:
\[ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-30} \]
\[ \frac{1}{v} = \frac{1}{-30} - \frac{1}{-20} \]
\[ \frac{1}{v} = \frac{-1}{30} + \frac{1}{20} \]
To add these fractions, find a common denominator, which is 60:
\[ \frac{1}{v} = \frac{-2}{60} + \frac{3}{60} \]
\[ \frac{1}{v} = \frac{-2+3}{60} \]
\[ \frac{1}{v} = \frac{1}{60} \]
\[ v = 60 \text{ cm} \]

The image distance is \( 60 \) cm. The positive sign indicates that the image is formed behind the mirror. This means the image is virtual. When an object is placed between the pole and the focal point of a concave mirror, a virtual, erect, and magnified image is formed behind the mirror.
In simple words: The concave mirror forms an image 60 cm behind it. This image is virtual, meaning light rays do not actually meet there, and it will be upright.

🎯 Exam Tip: Pay close attention to the object's position relative to the mirror's focal length. If \( u \) is less than \( |f| \) for a concave mirror, the image is always virtual, erect, and magnified, formed behind the mirror.

Question 7. Find the magnification by a convex lens of focal length 10 cm if erect image of object is formed at minimum distance of clear vision.
Answer: For a convex lens, the focal length \( f \) is positive, so \( f = 10 \) cm. The minimum distance of clear vision (often denoted as D or near point) for a normal eye is 25 cm. Since an erect image is formed, it must be virtual, and by convention, virtual images formed by lenses are on the same side as the object and have a negative image distance, so \( v = -25 \) cm.

We use the lens formula to find the object distance \( u \):
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Now, substitute the values:
\[ \frac{1}{-25} - \frac{1}{u} = \frac{1}{10} \]
\[ -\frac{1}{u} = \frac{1}{10} + \frac{1}{25} \]
To add these fractions, find a common denominator, which is 50:
\[ -\frac{1}{u} = \frac{5}{50} + \frac{2}{50} \]
\[ -\frac{1}{u} = \frac{5+2}{50} \]
\[ -\frac{1}{u} = \frac{7}{50} \]
\[ u = -\frac{50}{7} \text{ cm} \]

The object distance is \( -\frac{50}{7} \) cm. The negative sign confirms that the object is placed in front of the lens. This means the object is placed within the focal length of the convex lens, creating a virtual and erect image.

Next, we calculate the magnification \( M \) using the formula for lenses:
\[ M = \frac{v}{u} \]
Substitute the values of \( v \) and \( u \):
\[ M = \frac{-25}{-\frac{50}{7}} \]
\[ M = -25 \times \frac{-7}{50} \]
\[ M = \frac{175}{50} \]
\[ M = 3.5 \]

The magnification is \( 3.5 \). The positive sign confirms that the image is erect, which matches the problem statement. The value \( 3.5 \) indicates that the image is 3.5 times larger than the object. This is a common setup for a simple magnifying glass.
In simple words: With a convex lens, if an upright image is formed at the closest distance a normal eye can see clearly, the image appears 3.5 times larger than the actual object.

🎯 Exam Tip: For problems involving minimum distance of clear vision, remember that it's usually 25 cm and that for an erect image formed by a convex lens, the image is virtual, so \( v \) is negative.

Question 2. Image formed by plane mirror is
(a) real and erect
(b) real and inverted
(c) virtual and erect
(d) virtual and inverted
Answer: (c) virtual and erect
In simple words: A flat mirror always makes an image that looks like it's behind the mirror (virtual) and stands upright (erect).

🎯 Exam Tip: Remember that plane mirrors always produce images that are virtual, erect, and laterally inverted, and the size of the image is the same as the object.

Question 3. A concave mirror gives, real, inverted and same size image if the object is placed
(a) at F
(b) at infinity
(c) at C
(d) beyond C
Answer: (c) at C
In simple words: For a curved mirror that makes light come together (concave mirror), if you put an object at its center of curvature (C), the image will be real, upside down, and the exact same size.

🎯 Exam Tip: Know the standard image formation cases for concave mirrors. Placing an object at the center of curvature (C) is a key scenario where the image properties are specific: real, inverted, and same size, also at C.

Question 4. Power of a lens is -40, its focal length is
(a) 4 m
(b) - 40 cm
(c) - 0.25 m
(d) - 25 m.
Answer: (c) - 0.25 m
In simple words: If a lens has a power of -40, it means its focal length is -0.025 meters, or -2.5 centimeters. We find this by dividing 1 by the power.

🎯 Exam Tip: Remember the relationship between power (P) and focal length (f): \( P = 1/f \), where \( f \) is in meters. A negative power indicates a concave lens, which has a negative focal length.

Question 6. The mirror that always gives virtual and erect image of the object but image of smaller size than the size of the object is
(a) Plane mirror
(b) Concave mirror
(c) Convex mirror
(d) None of the options
Answer: (c) Convex mirror
In simple words: A convex mirror always creates an image that appears behind the mirror (virtual), stands upright (erect), and is smaller than the actual object.

🎯 Exam Tip: Convex mirrors are commonly used as rear-view mirrors because they provide a wide field of view and always form virtual, erect, and diminished images, regardless of the object's position.

Question 7. All the distances in case of spherical mirror are measured in relation to
(a) object to image
(b) the pole of the mirror
(c) the focus of the mirror
(d) the image to the object.
Answer: (b) the pole of the mirror
In simple words: When measuring distances for a curved mirror, you always start from the middle point on the mirror's surface, which is called the pole.

🎯 Exam Tip: The pole is the central reference point for all measurements in spherical mirrors, as per the Cartesian sign convention. Distances are measured along the principal axis from this point.

Question 8. The radius of curvature and focal length of a concave mirror are
(a) positive
(b) negative
(c) both
(d) none of these
Answer: (b) negative
In simple words: For a concave mirror, both its radius of curvature and focal length are considered negative because they are located in front of the mirror, in the direction opposite to incident light.

🎯 Exam Tip: Remember the sign conventions: for concave mirrors and lenses, both focal length and radius of curvature are typically negative because they lie on the left side (in front) of the optical element.

Question 10. The ratio of the speed of light in vacuum to that in a medium is known as
(a) magnification
(b) refraction
(c) refractive index
(d) Snell's law
Answer: (c) refractive index
In simple words: The refractive index tells us how much slower light travels in a material compared to how fast it travels in empty space.

🎯 Exam Tip: The refractive index is a fundamental property of a medium that quantifies how light propagates through it, and it's always greater than or equal to 1.0 (for vacuum).

Question 11. In optics, an object which has higher refractive index is called
(a) optically rarer
(b) optically denser
(c) optical density
(d) refractive index
Answer: (b) optically denser
In simple words: A material that slows light down more than another material has a higher refractive index and is called optically denser.

🎯 Exam Tip: Optically denser materials cause light to bend more towards the normal when it enters from a rarer medium, due to the decrease in light's speed.

Question 12. Convex lens forms a real, point sized image at focus, the object is placed
(a) at focus
(b) between F and 2F
(c) at infinity
(d) at 2F
Answer: (c) at infinity
In simple words: If an object is extremely far away (at infinity), a convex lens will bring its light rays together to form a very small, real image exactly at its focal point.

🎯 Exam Tip: Remember that parallel rays from a distant object converge at the focal point after passing through a convex lens, forming a real, inverted, and highly diminished image.

Question 14. The radius of curvature of a mirror is 20 cm the focal length is
(a) 20 cm
(b) 10 cm
(c) 40 cm
(d) 5 cm
Answer: (b) 10 cm
In simple words: For any spherical mirror, the focal length is always half of its radius of curvature. So, if the radius is 20 cm, the focal length is 10 cm.

🎯 Exam Tip: The relationship \( f = R/2 \) is crucial for spherical mirrors. This formula means the focal point (F) is exactly halfway between the pole (P) and the center of curvature (C).

Question 15. The refractive indices of some media are given below: In which of these is the speed of light minimum and maximum, respectively.

(a) X-minimum, W-maximum
(b) Z-minimum, W-maximum
(c) W-minimum, X-maximum
(d) X-minimum, Z-maximum
Answer: (c) W-minimum, X-maximum
In simple words: Light moves slowest in the material with the highest refractive index (W), and fastest in the material with the lowest refractive index (X).

🎯 Exam Tip: Remember that the speed of light in a medium is inversely proportional to its refractive index. A higher refractive index means light travels slower, and a lower refractive index means light travels faster.

Question 17. A mirror that has very wide field view is
(a) concave
(b) convex
(c) plane
(d) none of these
Answer: (b) convex
In simple words: A convex mirror spreads out the reflected light, making it useful for seeing a larger area, which is why it has a wide field of view.

🎯 Exam Tip: Convex mirrors are ideal for applications requiring a wide field of view, such as security mirrors in shops or rear-view mirrors in vehicles, because they always form diminished and erect images.

Question 18. If the object is placed at focus of a concave mirror, the image is formed at
(a) infinity
(b) focus
(c) centre of curvature
(d) between F and O.
Answer: (a) infinity
In simple words: When an object is placed at the focal point of a concave mirror, the reflected light rays become parallel and never meet, meaning the image appears to be formed infinitely far away.

🎯 Exam Tip: This is a crucial case for concave mirrors: object at focus produces image at infinity (parallel rays), and conversely, parallel rays from infinity converge at the focus.

Question 19. The image formed on the retina of human eye is
(a) virtual and erect
(b) real and inverted
(c) virtual an inverted
(d) real and erect
Answer: (b) real and inverted
In simple words: The lens in our eye forms an actual image on the retina, but it is upside down and real. Our brain then flips this image so we see things correctly.

🎯 Exam Tip: Remember that the human eye's lens is a convex lens, which forms a real, inverted, and diminished image on the retina. The brain then processes this inverted image to perceive it as erect.

Question 21. The least distance of distinct vision for a young adult with normal vision is
(a) 25 m
(b) 20 m
(c) 25 cm
(d) 20 cm
Answer: (c) 25 cm
In simple words: A normal young adult can see things clearly when they are at least 25 centimeters away from their eyes.

🎯 Exam Tip: This value, 25 cm, is a standard benchmark for the near point of distinct vision and is important in understanding eye defects like hypermetropia.

Question 22. The persistence of vision for normal eye is
(a) \( \frac{1}{10} \)th of a second
(b) \( \frac{1}{16} \)th of a second
(c) \( \frac{1}{6} \)th of a second
(d) \( \frac{1}{18} \)th of a second
Answer: (b) \( \frac{1}{16} \) th of a second
In simple words: Our eyes hold onto an image for a very short time after we see it, about one-sixteenth of a second. This makes moving pictures look smooth.

🎯 Exam Tip: Persistence of vision is the phenomenon that makes us perceive a series of still images as continuous motion, forming the basis of cinematography.

Question 23. The phenomenon of light responsible for the working of the human eye is
(a) reflection
(b) refraction
(c) power of accommodation
(d) persistence of vision.
Answer: (b) refraction
In simple words: Our eyes work by bending light as it enters, focusing it onto the retina so we can see. This bending of light is called refraction.

🎯 Exam Tip: While other phenomena like accommodation are involved, the primary principle by which the eye's lens and cornea form an image is refraction, bending light to focus it on the retina.

Question 25. The colored light that refracts most while passing through a prism is
(a) Yellow
(b) Violet
(c) Blue
(d) Red
Answer: (b) Violet
In simple words: When white light passes through a prism, violet light bends the most because it has the shortest wavelength.

🎯 Exam Tip: Remember the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red) for the colors of the spectrum. Violet light has the shortest wavelength and thus deviates the most, while red light has the longest wavelength and deviates the least.

Question 26. The amount of light entering the human eye is controlled by
(a) Ciliary muscles
(b) Pupil
(c) Cornea
(d) Iris
Answer: (b) Pupil
In simple words: The pupil is the black hole in the center of your eye that gets bigger or smaller to let in the right amount of light, just like a camera's aperture.

🎯 Exam Tip: The iris, which is the colored part of the eye, controls the size of the pupil. So, while the iris adjusts, it is the pupil that directly regulates the light entering the eye.

Question 27. Which part of the eye mostly refracts light entering the eye from external objects?
(a) Lens
(b) Cornea
(c) Iris
(d) Pupil
Answer: (b) Cornea
In simple words: The cornea, the clear outer layer of the eye, does most of the light bending to help focus images onto the retina.

🎯 Exam Tip: While the crystalline lens also refracts light, the cornea is responsible for the largest amount of refraction due to the significant change in refractive index between air and the cornea.

Question 29. Long-sightedness or hypermetropia can be corrected by
(a) Planar lens
(b) Concave lens
(c) Convex lens
(d) Bifocal lens
Answer: (c) Convex lens
In simple words: Long-sightedness, where far-away objects are clear but close-up ones are blurry, is fixed by using a convex lens, which helps to focus light rays correctly onto the retina.

🎯 Exam Tip: Hypermetropia occurs when light focuses behind the retina; a convex (converging) lens is needed to bring the focal point forward onto the retina.

Question 30. A student of class 10, is not able to see clearly the black board question when seated at a distance of 5 m from the board, the defect he is suffering from is
(a) Myopia
(b) Hypermetropia
(c) Presbyopia
(d) Astigmatism
Answer: (a) Myopia
In simple words: If someone cannot see far-off things clearly, like a blackboard from a distance, they have myopia, which is also known as near-sightedness.

🎯 Exam Tip: Myopia is characterized by the inability to see distant objects clearly, while near objects remain clear. This is due to the eye's focal length being too short or the eyeball being too long.

Question 31. The part of eye that determines the colour of the eye of a person is
(a) Pupil
(b) Cornea
(c) Retina
(d) Iris
Answer: (d) Iris
In simple words: The iris is the colorful ring around the pupil of your eye, and its natural pigments give your eyes their unique color.

🎯 Exam Tip: Beyond color, the iris also plays a crucial role in controlling the size of the pupil, thereby regulating the amount of light that enters the eye.

Question 33. The colour of sky is blue during day time, red during sunset and black at night. This is due to
(a) Scattering of light
(b) Small particles present in atmosphere
(c) Atmospheric refraction
(d) All of the options
Answer: (d) All of the options
In simple words: The sky changes colors because of how sunlight interacts with tiny bits in the air and how light bends through the atmosphere. All these things together make the sky look blue, red, or black at different times.

🎯 Exam Tip: Remember that the blue color of the sky is primarily due to Rayleigh scattering, while the red hues at sunrise/sunset are also due to scattering (blue light is scattered away, leaving red light), and atmospheric refraction plays a role in how we perceive the sun's position and light intensity.

Question 34. The eye defect represented by the figure is
(a) Myopia
(b) Hypermetropia
(c) Cataract
(d) Presbyopia
Answer: (a) Myopia

In simple words: The diagram shows light focusing in front of the retina, which is a common problem for people with myopia, or near-sightedness.

🎯 Exam Tip: Be able to identify diagrams for different eye defects. Myopia is characterized by light rays converging before the retina, while hypermetropia shows them converging behind it.

Light Very Short Answer Type Questions

Question 1. What is light?
Answer: Light is a type of electromagnetic radiation that allows us to see things. It travels as waves and does not need any material to pass through, meaning it can travel even in empty space. We perceive the world around us because of how light interacts with objects and our eyes.
In simple words: Light is a form of energy that helps us see. It travels like a wave and does not need air or water to move.

🎯 Exam Tip: When defining light, key terms to include are "electromagnetic radiation" and "sensation of sight." Mentioning that it doesn't require a medium emphasizes its wave nature.

Question 3. Define reflection of light.
Answer: Reflection of light is the phenomenon where light rays bounce back into the same medium after hitting a smooth and polished surface. This principle is why we can see objects when light from a source hits them and then reflects into our eyes.
In simple words: When light hits a shiny surface and bounces back, it is called reflection.

🎯 Exam Tip: Remember that reflection always occurs in the same medium and follows the laws of reflection, where the angle of incidence equals the angle of reflection.

Question 4. Define incident ray.
Answer: An incident ray is the ray of light that falls upon or strikes a surface, such as a mirror or any polished surface. It represents the path of light before it interacts with the surface.
In simple words: An incident ray is the beam of light that first hits a mirror or any shiny surface.

🎯 Exam Tip: Always draw an arrowhead on incident rays in diagrams to show the direction of light. It's crucial for understanding the path of light.

Question 5. Define reflected ray.
Answer: A reflected ray is the ray of light that bounces back from a surface into the same medium after the incident ray has struck it. This ray carries the visual information that allows us to perceive reflected images.
In simple words: A reflected ray is the light beam that bounces off a surface after hitting it.

🎯 Exam Tip: In diagrams, the reflected ray also needs an arrowhead to indicate its direction after bouncing off the surface.

Question 6. Define normal.
Answer: The normal is an imaginary line drawn perpendicular (at 90 degrees) to the reflecting surface at the point where the incident ray strikes it. It serves as a reference line for measuring the angles of incidence and reflection.
In simple words: The normal is an imaginary straight line that stands perfectly upright (90 degrees) on a reflecting surface at the point where light hits it.

🎯 Exam Tip: Always draw the normal as a dashed line in ray diagrams to show it is imaginary, and make sure it's perpendicular to the surface.

Question 7. Define angle of incidence.
Answer: The angle of incidence is the angle formed between the incident ray and the normal at the point of incidence. This angle is crucial for applying the laws of reflection and refraction.
In simple words: The angle of incidence is the space between the incoming light ray and the normal line.

🎯 Exam Tip: Remember that the angle of incidence is always measured from the normal, not from the surface itself.

Question 8. Define angle of reflection.
Answer: The angle of reflection is the angle formed between the reflected ray and the normal at the point of incidence. According to the law of reflection, this angle is always equal to the angle of incidence.
In simple words: The angle of reflection is the space between the light ray bouncing back and the normal line.

🎯 Exam Tip: In problems, always ensure that the angle of reflection equals the angle of incidence when dealing with reflection from a smooth surface.

Question 10. What are the properties of image formed by a plane mirror?
Answer: The image formed by a plane mirror has several distinct properties:

• It is virtual and erect, meaning it appears to be behind the mirror and stands upright.
• The size of the image is exactly equal to the size of the object.
• The image is laterally inverted, which means the left side of the object appears as the right side in the image, and vice-versa.
• The image is formed as far behind the mirror as the object is in front of it.

A plane mirror creates a perfect, undistorted, virtual replica of the object.
In simple words: A plane mirror makes an image that looks like it's behind the mirror, stands upright, is the same size as the actual object, and is flipped sideways (left becomes right). The image appears as far behind the mirror as the object is in front.

🎯 Exam Tip: Key terms like "virtual", "erect", "same size", "laterally inverted", and "equidistant" are essential when describing properties of images formed by plane mirrors.

Question 11. What are spherical mirrors?
Answer: Spherical mirrors are mirrors whose reflecting surface is a part of a hollow sphere. They can be of two types: concave mirrors, which curve inwards like the inside of a spoon, and convex mirrors, which curve outwards like the back of a spoon. These mirrors are fundamental in understanding optics due to their ability to converge or diverge light rays.
In simple words: Spherical mirrors are curved mirrors, like parts cut from a big ball. They can either curve inwards or outwards.

🎯 Exam Tip: Clearly differentiate between concave (converging) and convex (diverging) mirrors, as their properties and uses are opposite.

Question 12. State mirror formula and write it mathematically.
Answer: The mirror formula is a mathematical relationship that connects the focal length (f) of a spherical mirror with the object distance (u) and the image distance (v). It is an essential tool for solving problems related to image formation by mirrors. It is given by:
\( \frac { 1 }{ v } + \frac { 1 }{ u } = \frac { 1 }{ f } \)
Where:
\( u \) = Object distance (distance of the object from the pole of the mirror)
\( v \) = Image distance (distance of the image from the pole of the mirror)
\( f \) = Focal length (distance of the principal focus from the pole of the mirror)
This formula applies to both concave and convex mirrors, provided the correct sign conventions are used.
In simple words: The mirror formula is a math rule that connects how far an object is from a mirror, how far its image is formed, and the mirror's focal length. It helps us calculate any of these if we know the other two.

🎯 Exam Tip: Always use the Cartesian sign conventions consistently when applying the mirror formula to avoid errors in calculations.

Question 13. Give the relation between focal length and radius of curvature.
Answer: For a spherical mirror (both concave and convex), the focal length (f) is directly related to its radius of curvature (R). The focal length is exactly half of the radius of curvature. This means the principal focus lies exactly midway between the pole and the centre of curvature.
Mathematically, this relation is:
\( f = \frac { R }{ 2 } \)
In simple words: For a curved mirror, the focal length is exactly half of its radius of curvature. This means the point where light focuses is halfway between the mirror and its center of the sphere it came from.

🎯 Exam Tip: Remember that this relation holds true for spherical mirrors with small apertures relative to their radius of curvature.

Question 15. Define refraction of light.
Answer: Refraction of light is the phenomenon where a light ray changes its direction as it passes from one transparent medium into another. This change in direction occurs because the speed of light is different in different media, causing the light wave to bend. A common example is a straw appearing bent in a glass of water.
In simple words: Refraction is when light bends as it moves from one material, like air, into another material, like water, causing its path to change.

🎯 Exam Tip: Always associate refraction with the bending of light due to a change in speed when moving between different transparent media.

Question 16. State laws of refraction.
Answer: The laws of refraction govern how light behaves when it passes from one medium to another. There are two main laws:

• The incident ray, the refracted ray, and the normal to the interface of the two transparent media at the point of incidence, all lie in the same plane.
• Snell's Law: For a given pair of media and a specific colour of light, the ratio of the sine of the angle of incidence (\( \sin i \)) to the sine of the angle of refraction (\( \sin r \)) is constant. This constant is called the refractive index of the second medium with respect to the first.
  \( \frac { \sin i }{ \sin r } = \text{Constant} \)

Snell's law is crucial for calculating how much light bends.
In simple words: The first law says the incoming light, the bent light, and the normal line all stay in the same flat area. The second law, called Snell's Law, says that for any two materials and a specific color of light, the ratio of the sine of the incident angle to the sine of the refracted angle is always the same number

🎯 Exam Tip: Remember that Snell's Law applies only to monochromatic light and a specific pair of media; the constant value changes for different colours or media.

Question 17. What do you observe when light ray passes through rectangular slab?
Answer: When a light ray passes through a rectangular glass slab, several observations can be made:

• The angle of incidence at the first surface is equal to the angle of emergence at the second surface.
• The incident ray is parallel to the emergent ray, but they are not along the same line.
• A phenomenon called lateral displacement occurs, where the emergent ray is shifted sideways from the path of the incident ray.
• The amount of lateral displacement is directly proportional to the thickness of the glass slab and the angle of incidence.

This lateral shift is noticeable for thicker slabs and larger angles.
In simple words: When light goes through a glass slab, the light that comes out is parallel to the light that went in, even though it shifts a bit. The amount of shift depends on how thick the glass is and at what angle the light entered.

🎯 Exam Tip: In diagrams, ensure that the incident and emergent rays are drawn parallel and show the lateral displacement clearly.

Question 18. Define lateral displacement.
Answer: Lateral displacement refers to the perpendicular distance between the original path of an incident ray and the path of the emergent ray when light passes through a transparent material with parallel faces, such as a glass slab. It quantifies how much the light ray has shifted sideways from its initial line of travel. This displacement occurs due to the two refractions at the parallel surfaces.
In simple words: Lateral displacement is how much the light ray shifts sideways when it passes through something like a glass slab, measured as the straight distance between its original path and its final path.

🎯 Exam Tip: Remember that lateral displacement is zero if the light ray strikes the slab normally (at 90 degrees), as there is no bending.

Question 20. Define absolute refractive index.
Answer: The absolute refractive index of a medium is defined as the ratio of the speed of light in a vacuum to the speed of light in that specific medium. It indicates how much the light ray will bend, or slow down, when it enters that medium from a vacuum. Since a vacuum is the fastest medium for light, the absolute refractive index is always greater than or equal to one.
In simple words: The absolute refractive index tells us how much light bends when it goes from empty space (vacuum) into a specific material. It's a way to measure how much a material slows down light.

🎯 Exam Tip: Remember that the absolute refractive index of air is very close to 1, often taken as 1 for practical calculations, similar to vacuum.

Question 21. What is the unit of refractive index?
Answer: The refractive index is a dimensionless quantity, which means it has no unit. It is a ratio of two similar physical quantities (speeds of light), so their units cancel out. Therefore, it is represented simply as a numerical value.
In simple words: The refractive index is just a number; it does not have any units because it's a ratio of two speeds.

🎯 Exam Tip: Always write the refractive index as a pure number without any unit, as it indicates a ratio.

Question 22. What is the relation between optical density, refractive index and speed of light?
Answer: Optical density, refractive index, and the speed of light are closely related. A medium with a higher refractive index is considered optically denser. In an optically denser medium, the speed of light is less, causing the light rays to bend more towards the normal when entering from a rarer medium. Conversely, a medium with a lower refractive index is optically rarer, and light travels faster through it. This relationship helps explain phenomena like why diamonds sparkle so much, as light slows down significantly inside them.
In simple words: If a material has a high refractive index, light moves slower through it, and we call it optically denser. If it has a low refractive index, light moves faster, and we call it optically rarer.

🎯 Exam Tip: Remember: Higher refractive index = Higher optical density = Lower speed of light = More bending.

Question 23. State lens formula and write it mathematically.
Answer: The lens formula is a mathematical equation that relates the focal length (f) of a spherical lens to the object distance (u) and the image distance (v). It is a critical formula used to predict the location and nature of images formed by lenses. It is given by:
\( \frac { 1 }{ v } - \frac { 1 }{ u } = \frac { 1 }{ f } \)
This formula is applicable to both convex and concave lenses, provided that the Cartesian sign conventions are correctly applied to the values of u, v, and f.
In simple words: The lens formula is a math rule that links the distance of an object from a lens, the distance of the image formed, and the lens's focal length. It helps to calculate any of these values if the other two are known.

🎯 Exam Tip: Pay close attention to the minus sign between \( \frac{1}{v} \) and \( \frac{1}{u} \) in the lens formula, as it differentiates it from the mirror formula.

Question 24. Define magnification of lens.
Answer: The magnification (m) produced by a lens is a measure of how much larger or smaller an image is compared to the object, and whether it is inverted or erect. It is defined as the ratio of the height of the image (h') to the height of the object (h). It can also be expressed in terms of image distance (v) and object distance (u).
\( \text{Magnification (m)} = \frac { \text{Height of image (h')} }{ \text{Height of object (h)} } = \frac { v }{ u } \)
A positive magnification indicates an erect (virtual) image, while a negative magnification indicates an inverted (real) image. If \( |m| > 1 \), the image is enlarged; if \( |m| < 1 \), it is diminished; and if \( |m| = 1 \), it is the same size.
In simple words: Magnification tells us how much bigger or smaller an image is compared to the actual object. It's a ratio of the image height to the object height, or the ratio of image distance to object distance.

🎯 Exam Tip: The sign of magnification is crucial: positive for erect/virtual images, negative for inverted/real images. The magnitude indicates size change.

Question 26. What is the magnification of a plane mirror?
Answer: The magnification of a plane mirror is always \( m = +1 \). The positive sign indicates that the image formed is virtual and erect, meaning it is upright relative to the object. The magnitude of 1 signifies that the image is the same size as the object. This is a consistent property of all images formed by plane mirrors.
In simple words: For a plane mirror, the magnification is +1. This means the image is the same size as the object and stands upright.

🎯 Exam Tip: A magnification of +1 uniquely identifies the properties of an image formed by a plane mirror: virtual, erect, and same size.

Question 27. Which lens bends a light ray more the one having shorter or longer focal length?
Answer: The lens with the shorter focal length bends light rays more significantly. This is because a shorter focal length indicates a more powerful lens, which has a greater ability to converge (for convex lenses) or diverge (for concave lenses) light rays. Therefore, a lens that bends light more strongly will have a focal point closer to the lens.
In simple words: A lens that has a shorter focal length will bend light rays more strongly than a lens with a longer focal length.

🎯 Exam Tip: Remember that power of a lens is inversely proportional to its focal length (\( P = 1/f \)), so a shorter focal length means higher power and more bending.

Question 28. If a convex lens is used to focus sunlight on a paper, where should the paper be placed so that it catches fire.
Answer: To make a paper catch fire using a convex lens and sunlight, the paper should be placed precisely at the principal focus of the lens. At this point, all the parallel rays of sunlight converge, concentrating a large amount of light energy into a very small area, which generates enough heat to ignite the paper. This demonstrates the converging property of a convex lens.
In simple words: To make paper catch fire using a convex lens and sunlight, you need to hold the paper exactly at the lens's main focus point where all the light rays come together.

🎯 Exam Tip: The principal focus is the point of maximum heat concentration for a convex lens when parallel rays (like sunlight) fall on it.

Question 29. What happens if a light falls on a glass slab making 90° at its surface?
Answer: If a light ray falls on a glass slab making a 90° angle with its surface (i.e., normally or perpendicularly), it undergoes normal refraction. In this specific case, the light ray passes straight through the glass slab without any deviation or bending. The angle of incidence is 0°, and therefore, the angle of refraction is also 0°, meaning the ray travels along the normal.
In simple words: If light hits a glass slab straight on, at a 90-degree angle to the surface, it will pass right through without bending or changing direction.

🎯 Exam Tip: Light only bends when it hits a new medium at an angle other than 90 degrees to the surface. Perpendicular incidence means no deviation.

Question 30. Where should be an object placed in front of convex lens so as to use it as a
Answer: To use a convex lens as a magnifying glass, the object should be placed between its optical center (O) and its principal focus (F). When the object is in this position, the convex lens forms a virtual, erect, and enlarged image on the same side as the object, making it appear magnified. This is a common application of convex lenses in simple magnifying glasses.
In simple words: To use a convex lens like a magnifying glass, you need to place the object very close to the lens, between its center and its focus point. This makes the object look bigger and upright.

🎯 Exam Tip: For a convex lens to act as a magnifying glass, the object must always be placed within its focal length.

Question 32. What is the function of iris?
Answer: The iris is the colored part of the eye that functions to control the size of the pupil. By adjusting the pupil's diameter, the iris regulates the amount of light entering the eye. In bright light, the iris contracts to make the pupil smaller, reducing light entry, while in dim light, it dilates to make the pupil larger, allowing more light to enter, thus helping to optimize vision in varying light conditions.
In simple words: The iris is the colored part of your eye, and its job is to make the pupil bigger or smaller to let in the right amount of light

🎯 Exam Tip: The iris acts like a diaphragm in a camera, controlling the aperture (pupil) to manage light intake for clear vision.

Question 33. Why is inverted image formed on the retina of human eye?
Answer: An inverted (upside-down) image is formed on the retina of the human eye because the eye's crystalline lens is a convex lens. Convex lenses converge light rays, causing them to cross over before reaching the retina, which naturally forms a real and inverted image. The brain then processes this inverted image and interprets it as upright, allowing us to see the world correctly.
In simple words: Our eye lens is shaped like a convex lens, which naturally creates an upside-down (inverted) image on the retina. Our brain then flips this image so we see things the right way up.

🎯 Exam Tip: Understand that the brain's interpretation is crucial; the physical image on the retina is always inverted.

Question 34. What type of signals are generated and sent to the brain by light sensitive cells of retina?
Answer: The light-sensitive cells (rods and cones) in the retina generate electrical signals when stimulated by light. These electrical signals are then transmitted via the optic nerve to the brain. The brain interprets these electrical impulses as visual information, allowing us to perceive images, colours, and movements. This conversion of light energy into electrical signals is a fundamental process of vision.
In simple words: The light-sensitive cells in your retina turn the light they receive into tiny electrical messages, which are then sent to your brain for processing.

🎯 Exam Tip: Remember that vision is a two-step process: light conversion to electrical signals in the retina, followed by brain interpretation.

Question 35. What is the function of crystalline lens of human eye?
Answer: The primary function of the crystalline lens in the human eye is to provide the necessary adjustment in focal length so that light rays from objects at different distances are precisely focused onto the retina. This ability to change focal length, known as accommodation, allows the eye to form clear images of both near and distant objects, ensuring sharp vision across various depths. It is a vital component for sharp, adaptable vision.
In simple words: The eye's crystalline lens helps to focus light rays from objects at various distances, making sure a clear image lands right on the retina. This is how we can see both near and far things.

🎯 Exam Tip: The crystalline lens's flexibility and ability to change focal length is key to the eye's power of accommodation.

Question 36. What holds the crystalline lens in the human eye?
Answer: The crystalline lens in the human eye is held in position by suspensory ligaments, which are attached to the ciliary muscles. These ciliary muscles contract and relax to change the tension in the suspensory ligaments, thereby altering the shape and focal length of the crystalline lens. This mechanism is essential for the eye's ability to focus on objects at different distances.
In simple words: Small muscles called ciliary muscles hold the eye's lens in place and change its shape to help us focus

🎯 Exam Tip: Ciliary muscles and suspensory ligaments work together to enable the eye to accommodate, changing the lens's shape.

Question 37. N. Processing math: 42% hich crystalline lens of human eye becomes opaque.
Answer: The crystalline lens of the human eye becomes opaque in a condition known as cataract. Cataract is a common eye condition, often related to aging, where the clear lens of the eye becomes cloudy or foggy. This opaqueness scatters light, leading to blurry vision, faded colours, and difficulty seeing at night. Without treatment, it can lead to blindness.
In simple words: When the eye's natural lens becomes cloudy and blocks light, it is called a cataract. This makes it hard to see clearly.

🎯 Exam Tip: Cataracts primarily affect the transparency of the eye's lens, leading to impaired vision, and are often treated with surgery to replace the cloudy lens.

Question 39. In which type of eye defect far point of the eye gets reduced?
Answer: The far point of the eye gets reduced in Myopia, also commonly known as nearsightedness. In myopia, a person can see nearby objects clearly, but distant objects appear blurry. This occurs because the image of distant objects is formed in front of the retina, rather than directly on it, moving the farthest clear viewing distance closer to the eye.
In simple words: Myopia, also known as nearsightedness, is an eye problem where a person cannot see distant objects clearly because the far point of their vision moves closer to the eye.

🎯 Exam Tip: Myopia is corrected using a concave lens, which diverges light rays before they enter the eye, pushing the image back onto the retina.

Question 40. In which type of eye defect near point of the eye becomes more than 25 cm?
Answer: The near point of the eye becomes more than 25 cm in Hypermetropia, which is also called farsightedness. In this condition, a person can see distant objects clearly, but finds it difficult to focus on nearby objects. This happens because the image of close objects is formed behind the retina, making the closest point for clear vision farther than the normal 25 cm. It's often due to a shorter eyeball or a lens that is too weak.
In simple words: In hypermetropia, or farsightedness, a person's near point moves further away than 25 cm, making it difficult to see close-up objects clearly.

🎯 Exam Tip: Hypermetropia is corrected using a convex lens, which converges light rays before they enter the eye, bringing the image forward onto the retina.

Question 41. What is dispersion of light?
Answer: Dispersion of light is the phenomenon where white light splits into its constituent colours (spectrum) when it passes through a transparent medium like a prism. This happens because different colours of light travel at slightly different speeds within the medium, causing them to bend by different amounts. The result is a beautiful band of seven colours, often remembered by the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).
In simple words: Dispersion of light is when white light spreads out into all the colors of the rainbow, like when it passes through a prism.

🎯 Exam Tip: Remember that violet light bends the most, and red light bends the least, during dispersion through a prism.

Question 42. What is spectrum?
Answer: A spectrum is the band of seven distinct colours (Violet, Indigo, Blue, Green, Yellow, Orange, Red) that is obtained when white light is dispersed, typically by passing through a prism. Each colour in the spectrum corresponds to a different wavelength and frequency of light. Rainbows are a natural example of a visible light spectrum.
In simple words: A spectrum is the band of colors, like red, orange, yellow, green, blue, indigo, and violet, that we see when white light is split apart.

🎯 Exam Tip: The order of colours in a spectrum (VIBGYOR) is important and should be remembered correctly.

Question 43. Give one main difference between the lens of human eye and lens of camera.
Answer: The main difference between the lens of a human eye and a camera lens lies in their ability to change focal length. The human eye lens is flexible and can change its shape, allowing its focal length to be adjusted automatically by ciliary muscles to focus on objects at various distances (accommodation). In contrast, a typical camera lens has a fixed focal length, meaning it cannot change its shape to refocus. Instead, the focus in a camera is adjusted by moving the lens closer or further from the sensor.
In simple words: The human eye's lens can change its shape and focal length to focus on things at different distances, but a camera lens usually has a fixed focal length.

🎯 Exam Tip: Focus on the concept of "accommodation" for the human eye, which is the key difference from a fixed camera lens

Question 2. Define centre of curvature, principal axis, optical centre, aperture, focus and focal length for a lens.
Answer: The key terms related to spherical lenses are:

• (a) Centre of curvature: It is the centre of the imaginary spheres from which each of the two curved surfaces of the lens forms a part. A lens has two centres of curvature, often denoted as \( C_1 \) and \( C_2 \) or \( 2F_1 \) and \( 2F_2 \).
• (b) Principal axis: This is an imaginary straight line that passes symmetrically through the optical centre of the lens and connects its two centres of curvature. It is normal to both spherical surfaces of the lens.
• (c) Optical centre: This is the central point of the lens, usually located on the principal axis. A ray of light passing through the optical centre of a lens usually goes undeviated.
• (d) Focus: A lens has two principal foci. The first principal focus (\( F_1 \)) is the point on the principal axis where rays appear to diverge from (concave lens) or converge to (convex lens) after refraction, if they were initially parallel to the principal axis. The second principal focus (\( F_2 \)) is the point where rays parallel to the principal axis converge (convex lens) or appear to diverge from (concave lens) after passing through the lens.
• (e) Focal length: This is the distance between the optical centre of the lens and its principal focus (either \( F_1 \) or \( F_2 \)). It determines the power of the lens.

Understanding these terms is fundamental for drawing ray diagrams and applying lens formulas.
In simple words: The center of curvature is the center of the sphere from which the lens is made. The principal axis is a line through the lens's center. The optical center is the lens's middle point where light passes straight. The focus is where light rays meet or seem to come from after going through the lens. Focal length is the distance from the lens's center to its focus

🎯 Exam Tip: Note that a lens has two foci and two centres of curvature, unlike mirrors which have only one of each.

Question 3. Write nature, position and relative size of image formed by convex lens.
Answer: The nature, position, and relative size of the image formed by a convex lens depend on the position of the object, as summarized in the table below:

A convex lens is highly versatile, forming both real and virtual images depending on object placement, which makes it useful in many optical instruments.
In simple words: For a convex lens, the image changes depending on where the object is placed. It can be smaller, same size, or larger, and can be real and upside-down, or virtual and upright, depending on the object's distance from the lens

🎯 Exam Tip: Memorizing this table and understanding the ray diagrams for each position is crucial for mastering image formation by convex lenses.

Question 4. Give the sign conventions for spherical mirrors.
Answer: The Cartesian sign conventions are a set of rules used to assign positive or negative signs to various distances and heights when dealing with spherical mirrors. These conventions ensure consistency in calculations using the mirror formula. Here are the key rules:

• All distances are measured from the pole (P) of the mirror.
• The incident light is always shown coming from the left-hand side (LHS) of the mirror.
• Distances measured towards the left of the mirror (in the direction opposite to incident light) are taken as negative, while distances measured towards the right (in the direction of incident light) are taken as positive.
• Heights measured perpendicular to and above the principal axis are taken as positive.
• Heights measured perpendicular to and below the principal axis are taken as negative.

These conventions are critical for accurately determining image properties.

In simple words: The sign conventions for spherical mirrors help us know whether distances and heights are positive or negative, depending on where they are measured from the mirror and in what direction. This helps in using mirror formulas correctly

🎯 Exam Tip: Always imagine light coming from the left when applying sign conventions. This simplifies understanding and consistency.

Question 5. Write nature, position and relative size of image formed by cancave lens.
Answer: A concave lens always forms a virtual, erect, and diminished (smaller than the object) image, regardless of the object's position. The image is always located between the optical center and the focal point on the same side of the lens as the object. This consistent image formation makes concave lenses useful for various applications like reducing image size or correcting certain vision defects.
In simple words: A concave lens always creates an image that appears upright, smaller than the actual object, and looks like it's in front of the lens on the same side as the object.

🎯 Exam Tip: Remember that for a concave lens, images are always virtual, erect, and diminished, simplifying ray diagrams and problem-solving.

Question 6. Give sign conventions for spherical lenses.
Answer: The Cartesian sign conventions for spherical lenses are similar to those for mirrors, designed to provide a consistent system for calculations. These conventions are crucial for correctly using the lens formula and understanding image characteristics. The main rules include:

• All distances are measured from the optical centre (O) of the lens.
• Incident light is assumed to come from the left-hand side of the lens.
• Distances measured in the direction of incident light (right of the optical centre) are positive, while those measured against it (left of the optical centre) are negative.
• Heights above the principal axis are positive, and heights below the principal axis are negative.

These conventions help in accurately determining the image characteristics.

In simple words: Sign conventions for spherical lenses help us use lens formulas correctly. They tell us whether distances (like object or image distance) and heights should be treated as positive or negative, depending on their direction from the lens.

🎯 Exam Tip: A key difference from mirrors is the sign of focal length: convex lenses have positive focal length, while concave lenses have negative focal length.

Question 7. Two medium with refractive index 1.31 and 1.50 are given. In which case (i) bending of light is more? (ii) speed of light is more?
Answer: We compare the refractive indices to determine the bending and speed of light:

• (i) Bending of light is more in the medium with the higher refractive index, which is 1.50. This is because a higher refractive index indicates a greater change in the speed of light, leading to more pronounced bending.
• (ii) Speed of light is more in the medium with the lower refractive index, which is 1.31. Light travels faster in optically rarer media (lower refractive index) and slows down in denser media.

This relationship highlights how different materials affect light's behaviour.
In simple words: Light bends more in the material with a higher refractive index (1.50). Light travels faster in the material with a lower refractive index (1.31).

🎯 Exam Tip: Remember the inverse relationship: higher refractive index means lower speed of light and greater bending.

Question 8. Re osene oil is 1.44 and that of water is 1.33. A ray of light enters
Answer: The question is incomplete. However, assuming it asks for the refractive index of water with respect to kerosene oil, and how light behaves when entering water from kerosene:
Given: Refractive index of kerosene oil \( n_k = 1.44 \)
Refractive index of water \( n_w = 1.33 \)
To find the refractive index of water with respect to kerosene oil (\( n_{kw} \)):
\( n_{kw} = \frac { n_w }{ n_k } = \frac { 1.33 }{ 1.44 } \approx 0.9236 \)
Since \( n_k > n_w \), kerosene oil is optically denser than water. When a ray of light enters from an optically denser medium (kerosene) to an optically rarer medium (water), it will bend away from the normal. This is a fundamental principle of refraction.
In simple words: The question is cut off, but if we need to compare how light moves from kerosene to water, we use their refractive index values. Kerosene is optically denser than water, so light would bend away from the normal line when moving from kerosene into water.

🎯 Exam Tip: When light goes from a denser to a rarer medium, it bends away from the normal; from rarer to denser, it bends towards the normal.

Question 9. Which is optically denser out of the two medium M₁= 1.71 (refractive index) and M2 = 1.36 (refractive index). How does speed of light change when it travels from optically rarer to denser medium.
Answer: Out of the two given media, medium M₁ with a refractive index of 1.71 is optically denser compared to medium M2 with a refractive index of 1.36. This is because a higher refractive index implies a greater optical density. When light travels from an optically rarer medium (like M2) to an optically denser medium (like M₁), its speed decreases, causing the light ray to bend towards the normal. This change in speed is responsible for the phenomenon of refraction. For example, light travels slower in glass than in air.
In simple words: Medium M₁ is optically denser because it has a higher refractive index (1.71). When light moves from a less dense material to a more dense material, it slows down.

🎯 Exam Tip: Remember that a larger refractive index indicates a slower speed of light and a greater optical density.

Question 10. Comment on the size, position of the image formed by a concave mirror of focal length 18 cm when an object is placed: (i) at 22 cm (ii) 14 cm (iii)40 cm. in front of mirror without calculations.
Answer: For a concave mirror with a focal length (\( f \)) of 18 cm, the centre of curvature (\( C \)) would be at \( 2f = 36 \) cm. Here's how the image is formed for different object positions:

• (i) Object at 22 cm: Since 22 cm is between \( F \) (18 cm) and \( C \) (36 cm), the image will be formed beyond \( C \). The image will be real, inverted, and magnified.
• (ii) Object at 14 cm: Since 14 cm is between \( F \) (18 cm) and the pole \( P \) (0 cm), the image will be formed behind the mirror. The image will be virtual, erect, and magnified.
• (iii) Object at 40 cm: Since 40 cm is beyond \( C \) (36 cm), the image will be formed between \( F \) and \( C \). The image will be real, inverted, and diminished.

These different image characteristics are crucial for understanding the diverse applications of concave mirrors, from shaving mirrors to astronomical telescopes.
In simple words: For a concave mirror, where the object is placed makes a big difference. If the object is between the focal point and the center of curvature (22 cm), the image is magnified and real. If it's inside the focal point (14 cm), the image is virtual and magnified. If it's beyond the center of curvature (40 cm), the image is smaller and real.

🎯 Exam Tip: Knowing the relationship between object position (relative to F and C) and image characteristics is key for concave mirrors. Practice with ray diagrams for each case.

Question 11. Complete the following ray diagrams:
Answer: The completed ray diagrams illustrating the path of light rays for spherical mirrors are shown below:

Ray diagrams are essential for visually understanding the path of light and the formation of images in spherical mirrors. Each rule of ray tracing helps predict the image's characteristics.
In simple words: Ray diagrams show how light travels. For a mirror, rays hitting parallel to the axis go through the focus, rays going towards the focus reflect parallel, and rays going towards the center of curvature bounce back along the same path.

🎯 Exam Tip: Always draw at least two principal rays to locate the image accurately in ray diagrams.

Question 12. With the help of a ray diagram show how a pencil appears when dipped in water.
Answer: When a pencil is dipped partially in water, it appears bent or broken at the water-air interface due to the phenomenon of refraction. Light rays traveling from the part of the pencil in the water bend away from the normal as they pass from water (a denser medium) into air (a rarer medium). Our brain then traces these bent rays back in a straight line, making the submerged part of the pencil appear higher and shifted from its actual position. This creates the illusion of a bent pencil.

In simple words: When a pencil is dipped in water, it looks bent or broken at the water's surface. This happens because light rays from the pencil in the water bend as they move from water (denser) into air (rarer) before reaching our eyes, making the pencil appear to be in a different position.

🎯 Exam Tip: When drawing, show the actual path of light as solid lines and the apparent path (where the brain perceives it) as dashed lines.

Question 13. Define power of lens. What is the S. I. unit of power of a lens? If power of lens is +2D what is the nature and focal length of the lens?
Answer:
Power of lens: The power of a lens is a measure of its ability to converge (bring together) or diverge (spread out) light rays. A more powerful lens bends light more strongly. It is defined as the reciprocal of its focal length (f) in meters. Thus, \( P = \frac { 1 }{ f } \).
S.I. unit of power of a lens: The S.I. unit of power of a lens is 'dioptre' (D). One dioptre is the power of a lens whose focal length is 1 meter.
If the power of a lens is +2 D:
The positive sign indicates that it is a convex lens (converging lens).
The focal length (\( f \)) can be calculated as \( f = \frac { 1 }{ P } = \frac { 1 }{ 2 } = 0.5 \) meters. This lens would be useful for correcting farsightedness.
In simple words: The power of a lens tells us how much it can bend light. A stronger lens (higher power) bends light more. Its unit is dioptre. If a lens has a power of +2D, it's a convex lens, and its focal length is 0.5 meters.

🎯 Exam Tip: Remember that focal length must always be in meters when calculating the power of a lens in dioptres.

Question 15. The refractive index of water is 1.33 and kerosene is 1.44. Calculate the refractive index of kerosene with respect to water.
Answer: To calculate the refractive index of kerosene with respect to water, we use the formula for relative refractive index.
Given:
Refractive index of water (\( n_w \)) = 1.33
Refractive index of kerosene (\( n_k \)) = 1.44
The refractive index of kerosene with respect to water (\( n_{wk} \)) is calculated as:
\( n_{wk} = \frac { \text{Absolute refractive index of kerosene} }{ \text{Absolute refractive index of water} } = \frac { n_k }{ n_w } \)
\( n_{wk} = \frac { 1.44 }{ 1.33 } \approx 1.0827 \)
This value indicates how much light would bend if it were to pass from water into kerosene. A value greater than 1 means kerosene is optically denser than water.
In simple words: To find the refractive index of kerosene compared to water, we divide kerosene's refractive index by water's refractive index. This number (about 1.082) tells us how much light will bend when it moves from water into kerosene.

🎯 Exam Tip: When calculating relative refractive index, the refractive index of the 'with respect to' medium goes in the denominator.

Question 16. Draw ray diagrams to show the image formed by a concave lens for the object placed at (i) infinity (ii) Between f and 2f of the lens.
Answer: Here are the ray diagrams for image formation by a concave lens:
(i) Object at infinity:
When an object is placed at infinity, the concave lens forms a virtual, erect, and highly diminished image at its principal focus (\( F_1 \)) on the same side as the object. This is why a concave lens is used in applications where a small field of view needs to be presented in a larger, virtual form.
(ii) Object between \( F \) and \( 2F \) of the lens:
When an object is placed anywhere between the focal point (\( F \)) and twice the focal length (\( 2F \)) of a concave lens, the image formed is always virtual, erect, and diminished, located between the optical center and the focal point on the same side of the lens as the object. Concave lenses consistently produce this type of image.
In simple words: For a concave lens, light rays from any object will always form an image that is smaller, upright, and appears on the same side as the object, between the lens and its focal point. This is true whether the object is very far away or closer to the lens.

🎯 Exam Tip: Always remember the three principal rays for concave lenses: parallel ray appears to diverge from F, ray through O goes undeviated, and ray directed towards F emerges parallel.

Question 17. Draw a ray diagram to show the path of light when it travels through glass slab.
Answer: When a ray of light passes through a rectangular glass slab, it undergoes refraction twice: first upon entering the slab from air, and then upon exiting the slab back into air. The ray bends towards the normal when entering the denser glass medium and bends away from the normal when exiting into the rarer air medium. The emergent ray is parallel to the incident ray but is laterally displaced, meaning it is shifted sideways from its original path.

In simple words: When a light ray enters a flat piece of glass, it bends a little bit. Then, when it leaves the glass, it bends again, but it comes out moving in the same direction as it started, just shifted a little to the side.

🎯 Exam Tip: Ensure that the incident ray, emergent ray, and the two normals are correctly drawn, and highlight the lateral displacement clearly.

Question 18. Draw the following diagram in your answer book and show the formation of image of the object AB with the help of suitable rays.
Answer:
In simple words: When an object is placed between the center of curvature (C) and the focal point (F) of a concave mirror, the rays of light from the object reflect and meet beyond C. This forms an image that is larger, real, and upside down.

🎯 Exam Tip: Remember the principal rays for mirrors: parallel rays pass through the focus, rays through the focus become parallel, and rays through the center of curvature reflect back along the same path. Always use at least two rays to locate the image.

Question 20. An object of 2 cm high is placed at a distance of 64 cm from a white screen, on placing a convex lens at a distance of 32 cm from the object it is found that a distant image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.
Answer: Since the object-screen distance is double the object-lens separation, the object is at a distance of 2f from the lens, and the image should be of the same size as the object.
So \( 2f = 32 \implies f = 16 \text{ cm} \)
Height of image = Height of object = 2 cm
In simple words: When a convex lens is placed at a distance from an object such that the object is at twice its focal length (2F₁), a real, inverted image of the same size is formed at twice the focal length on the other side (2F₂). This setup ensures the light rays cross perfectly to make an image exactly the same size as the object.

🎯 Exam Tip: Remember that for a convex lens, when the object is placed at 2F₁, the image is formed at 2F₂ and is real, inverted, and of the same size. This is a key scenario often tested in exams.

Question 21. Redraw the given diagram and show the path of refracted ray.
Answer:
In simple words: When light enters a glass slab, it bends towards the normal because glass is denser than air. When it leaves the slab and enters the air again, it bends away from the normal. The emergent ray will be parallel to the original incident ray, but shifted slightly sideways.

🎯 Exam Tip: Remember that for a rectangular glass slab, the emergent ray is always parallel to the incident ray, but with a lateral shift. The angle of incidence equals the angle of emergence.

Question 22. A convex lens has a focal length of 10 cm. At what distance from the lens should the object be placed so that it gives a real and inverted image 20 cm away from the lens? What would be the size of the image formed if the object is 2 cm high? With the help of a ray diagram show the formation of the image in this case.
Answer: Here, focal length \( f = +10 \text{ cm} \). The image is real and inverted, so the image distance \( v = +20 \text{ cm} \). The height of the object is 2 cm (say +ve).
Using the lens formula \( \frac {1}{v} - \frac {1}{u} = \frac {1}{f} \), we get:
\( \frac {1}{20} - \frac {1}{u} = \frac {1}{10} \)
\( \frac {1}{u} = \frac {1}{20} - \frac {1}{10} \)
\( \frac {1}{u} = \frac {1 - 2}{20} = \frac {-1}{20} \)
\( u = -20 \text{ cm} \)
This means the object should be placed 20 cm in front of the lens. Since the object distance (20 cm) is equal to twice the focal length (2 x 10 cm), the image will be formed at 2F₂ (20 cm) on the other side and will be of the same size.
The height of the image will be 2 cm (same as the object, but inverted).

In simple words: To get a real and inverted image 20 cm away from a convex lens with a focal length of 10 cm, the object must be placed 20 cm in front of the lens. The image formed will be exactly the same size as the object. This happens when the object is placed at a distance equal to twice the focal length.

🎯 Exam Tip: For a convex lens, if the object distance is 2f, the image distance will also be 2f on the other side, and the image size will be equal to the object size, real, and inverted. This is a critical configuration to remember for ray diagrams and calculations.

Question 24. Why does a ray of light bend when it travels from one medium into another?
Answer: A ray of light bends when it travels from one medium to another because of a change in its speed. Light travels at different speeds in different materials. When light moves from a medium where it travels fast (like air) to a medium where it travels slower (like glass), it changes direction. This bending is called refraction. It happens to minimize the time taken for light to travel from one point to another in different media. This is a principle known as Fermat's Principle.
In simple words: Light bends when it goes from one material to another because its speed changes. Like a car slowing down when it goes from a road onto mud, it changes direction.

🎯 Exam Tip: The key reason for refraction is the change in the speed of light as it passes from one medium to another. Mentioning this change in speed is crucial for full marks.

Question 25. What are the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of virtual image by a concave mirror.
Answer: A minimum of two rays are required for locating the image formed by a concave mirror. To form a virtual image using a concave mirror, the object must be placed between the pole (P) and the principal focus (F) of the mirror.
In simple words: You need at least two light rays to find where an image forms. For a concave mirror to make a virtual image, the object must be placed very close to the mirror, between its pole and focal point. The image will appear magnified and behind the mirror, making it virtual and erect.

🎯 Exam Tip: Always specify the minimum number of rays needed (two) and clearly label the object, mirror, principal axis, focal point (F), center of curvature (C), pole (P), and the formed image in your ray diagrams. For virtual images, show extensions of reflected rays as dashed lines.

Question 27. How does eye control the amount of light entering it?
Answer: The iris in our eye controls how much light enters. The iris is a colored part that changes the size of the pupil, which is the black hole in the center of the eye. In bright light, the iris makes the pupil smaller to let in less light, protecting the eye. In dim light, it makes the pupil larger to let in more light, helping us see better. This adjustment happens automatically and quickly.
In simple words: The iris of our eye changes the pupil's size. It makes the pupil smaller when it's bright and larger when it's dark, controlling how much light gets in.

🎯 Exam Tip: When describing the eye's light control, remember to name the iris as the part responsible for adjusting the pupil's size, and explain how it widens or constricts in response to light intensity.

Question 28. Why danger signals are red?
Answer: Danger signals are red because red light scatters the least when it travels through the atmosphere. This means red light can travel the farthest distance without much loss of intensity due to scattering by dust or fog particles. Because of its low scattering, red light can be seen clearly from a maximum distance, making it ideal for warnings. This principle helps ensure that important signals are visible even in challenging weather conditions.
In simple words: Red light scatters less than other colors, so it can travel a long way without fading. This makes red danger signals easy to see from far away.

🎯 Exam Tip: The key concept here is "least scattering" for red light. Mentioning that red light scatters the least and can travel the maximum distance is essential for a complete answer

Question 29. Why do you take time to see objects when you enter a dim lighted room from outside in the sun?
Answer: When you are outside in bright sunlight, your iris makes your pupils very small to protect your eyes from too much light. When you suddenly enter a dimly lit room, your pupils are still small, so very little light enters your eyes, making it hard to see. It takes a few moments for the iris to expand the pupils, allowing more light in. At the same time, the light-sensitive cells in your retina (rods) also need some time to adjust to the low light conditions, which is called dark adaptation. Both these processes help you to see clearly in the dim room after a short delay.
In simple words: When you go from bright sun to a dark room, your eye's pupil is small and needs time to get bigger to let in more light. Your eye cells also need a moment to adjust to the darkness, so you can't see clearly right away.

🎯 Exam Tip: Focus on two main points: the iris adjusting the pupil size (dilation) and the light-sensitive cells (rods) in the retina adapting to low light. Both contribute to the delayed vision in dim conditions.

Question 30. Why are two eyes more helpful for us to see as compared to one?
Answer: Having two eyes is much more helpful than having just one for several reasons. Each eye sees a slightly different view, and our brain combines these two images to create a single, three-dimensional view. This process, called stereoscopic vision, helps us judge distances and perceive depth more accurately. With one eye, our field of vision is about 150 degrees, but with two eyes, it increases to about 180 degrees, allowing us to see more of our surroundings. Also, two eyes help us see better in dim light or darkness because they can collect more light together and provide a clearer image.
In simple words: Two eyes give us a wider view and help us see depth and judge distances much better than one eye can. They also help us see more clearly in low light.

🎯 Exam Tip: Highlight the benefits of having two eyes: increased field of vision, better depth perception (stereoscopic vision), and improved ability to see in low light conditions.

Question 32. Explain the phenomenon which causes twinkling of stars.
Answer: The twinkling of stars is caused by atmospheric refraction. Stars are very far away and act like tiny point sources of light. As light from a star travels through Earth's atmosphere, it passes through different layers of air with varying temperatures and densities. These layers constantly move and change, causing the light rays to bend (refract) unevenly. This continuous bending means that the amount of star light reaching our eyes changes every moment. Sometimes more light reaches, sometimes less, making the star appear to twinkle. Planets, being closer, act as extended sources, so the light variations from different parts average out, and they do not twinkle.
In simple words: Stars twinkle because their light bends and shifts as it passes through Earth's moving and changing air. This makes the amount of light we see from them change all the time.

🎯 Exam Tip: The key phrase is "atmospheric refraction" caused by varying densities of air layers. Emphasize that stars are point sources, which makes the twinkling noticeable, unlike planets which are extended sources.

Question 33. Why does a ray of light splits into different colours on passing through a glass prism?
Answer: A ray of white light splits into different colors when it passes through a glass prism because different colors of light travel at slightly different speeds inside the prism. When light enters a new medium, it refracts (bends). The amount of bending depends on the wavelength (color) of the light and its speed in that medium. Violet light travels the slowest and bends the most, while red light travels the fastest and bends the least. Because each color bends by a different amount, the white light separates into its seven constituent colors, forming a spectrum. This phenomenon is called dispersion of light.
In simple words: White light splits into colors in a prism because each color travels at a slightly different speed through the glass, causing them to bend by different amounts.

🎯 Exam Tip: Explain that different colors (wavelengths) of light have different speeds in the prism material, leading to different angles of refraction and thus separating the colors. Mention "dispersion" as the phenomenon.

Question 34. The sun appears to be red at the time of sunset and sunrise. Give the reason.
Answer: During sunrise and sunset, the sun appears red because the sunlight has to travel a much longer distance through the Earth's atmosphere to reach our eyes. As the light travels this longer path, most of the shorter wavelength colors, like blue and green, are scattered away by the tiny particles in the atmosphere. Red light, having the longest wavelength, scatters the least. Therefore, the light that finally reaches our eyes is predominantly red, making the sun and the surrounding sky appear reddish. This effect is a result of Rayleigh scattering.

In simple words: When the sun is low in the sky, its light travels through a lot of air. This air scatters away most of the blue and green colors. So, only the red and orange light reach our eyes, making the sun look red.

🎯 Exam Tip: The key points are "longer path through atmosphere" and "scattering of shorter wavelengths (blue/green)". Emphasize that red light scatters the least, making it visible. Referencing Rayleigh scattering can also earn extra points.

Question 35. Give reason for early sunrise and delayed sunset.
Answer: Early sunrise and delayed sunset are due to atmospheric refraction. When the sun is slightly below the horizon, light from the sun enters the Earth's atmosphere. The atmosphere acts like a lens, bending the light rays towards the Earth's surface. This bending causes the sun to appear higher than its actual position. Because of this, we can see the sun about two minutes before it actually rises above the horizon and for about two minutes after it has actually set. This phenomenon makes our days seem a bit longer.
In simple words: The Earth's air bends sunlight. So, we see the sun a little early in the morning and a little late in the evening, even when it is still below the horizon. This makes our day feel longer.

🎯 Exam Tip: The core concept is "atmospheric refraction" causing the sun's light to bend, making it appear above the horizon when it's still below it. Remember to state that it adds about two minutes to both sunrise and sunset.

Question 36. What is the direction of rainbow formation? What is the position of red colour in rainbow?
Answer: A rainbow is always formed in the direction opposite to the sun. This means if the sun is in the west, the rainbow will appear in the east. The position of red color in a rainbow is at the top (outermost arc). The violet color is always at the bottom (innermost arc), and all other colors are arranged in between, following the VIBGYOR sequence.
In simple words: A rainbow always appears in the sky opposite to where the sun is. The red color is always on the outside, at the top of the rainbow arc.

🎯 Exam Tip: Remember two key facts: the rainbow forms opposite the sun, and red is always on the outside/top, while violet is on the inside/bottom.

Question 37. What is internal reflection?
Answer: Internal reflection, specifically Total Internal Reflection (TIR), occurs when a ray of light travels from a denser medium (like water) to a rarer medium (like air). If the angle at which the light hits the boundary (angle of incidence) is greater than a specific value called the critical angle, the light ray does not pass into the rarer medium. Instead, it reflects entirely back into the denser medium. This phenomenon makes light behave like it's hitting a mirror from the inside of the denser material. This is why a diamond sparkles so much; light gets trapped inside due to total internal reflection.

In simple words: When light tries to go from a thick material (like water) to a thinner one (like air) but hits the surface at a very sharp angle, it bounces back completely into the thick material. It's like a mirror from the inside.

🎯 Exam Tip: To define Total Internal Reflection, state the two conditions: light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle. Also, mention the consequences, such as the light reflecting entirely.

Question 38. A short-sighted person cannot see clearly beyond 5m. Calculate the power of lens required to correct his vision to normal?
Answer: For a short-sighted (myopic) person, the far point is not at infinity but closer, in this case, 5m. To correct this, a concave lens is used which forms a virtual image of a distant object (at infinity) at the person's far point. So, the image distance \( v = -5 \text{ m} \). The object is at infinity, so \( u = -\infty \).
Using the lens formula \( \frac {1}{v} - \frac {1}{u} = \frac {1}{f} \):
\( \frac {1}{-5} - \frac {1}{-\infty} = \frac {1}{f} \)
\( \frac {1}{-5} - 0 = \frac {1}{f} \)
\( f = -5 \text{ m} \)
The power of the lens \( P = \frac {1}{f} \)
\( P = \frac {1}{-5} = -0.20 \text{ Dioptre} \)
So, the power of the required lens is \( -0.20 \text{ D} \). The negative sign indicates it is a concave lens.
In simple words: A person with short-sightedness who can only see up to 5 meters needs a special lens. This lens has a focal length of -5 meters, which means it has a power of -0.20 Diopters. This negative power indicates it is a diverging (concave) lens.

🎯 Exam Tip: For myopia correction, the image of an object at infinity must be formed at the person's far point. Always use the lens formula with \( u = -\infty \) and \( v = -\text{far point} \), and remember that concave lenses have negative power.

Question 39. Why can't we see object very close to our eye?
Answer: We cannot see objects clearly when they are very close to our eyes because our eye lens has a limit to how much it can change its focal length. For an object to be seen clearly, its image must form precisely on the retina. The ciliary muscles adjust the curvature of the eye lens to change its focal length, a process called accommodation. However, these muscles can only contract to a certain extent. If an object is brought too close (closer than the near point, typically 25 cm for a normal eye), the ciliary muscles cannot contract enough to make the lens converge light sufficiently to form a sharp image on the retina. The image forms behind the retina, resulting in a blurred vision.
In simple words: Our eyes cannot focus on objects that are too close because the eye lens cannot bend light enough to make a clear image on the retina. There's a limit to how much our eye muscles can adjust.

🎯 Exam Tip: The key concept is the "power of accommodation" of the eye lens and its limit. Emphasize that beyond the near point (usually 25 cm), the ciliary muscles cannot adjust the focal length enough to form a sharp image on the retina.

Question 41. Give the difference between myopia and hypermetropia.
Answer:

In simple words: Myopia means you see nearby things clearly but distant things blurry, often because your eyeball is too long. Hypermetropia means you see distant things clearly but nearby things blurry, often because your eyeball is too short.

🎯 Exam Tip: When differentiating between eye defects, always mention what a person can and cannot see, the location of the image on the retina, the associated change in eyeball length or focal length, and the type of corrective lens used

Question 42. Distinguish between presbyopia and hypermetropia.
Answer:

In simple words: Hypermetropia is when you can't see close objects well, usually because your eyeball is too short. Presbyopia is an age-related issue where the eye's muscles get weak and the lens becomes stiff, making it hard to focus on both near and far objects.

🎯 Exam Tip: The key difference is the cause: Hypermetropia relates to eyeball shape or lens power, while Presbyopia is an age-related loss of accommodation due to weak ciliary muscles and reduced lens flexibility. Also, remember their respective corrective lenses.

Question 44. What is meant by dispersion of white light? Draw a ray diagram to show the dispersion of white light by a glass prism. Give reason why do we get different colours of light?
Answer: Dispersion of white light is the phenomenon where white light splits into its seven constituent colors (Violet, Indigo, Blue, Green, Yellow, Orange, Red - VIBGYOR) when it passes through a transparent medium like a glass prism.
Different colors of light have different wavelengths, and because of this, they travel at slightly different speeds when passing through a medium like glass. This difference in speed causes each color to bend (refract) by a different amount when it enters and leaves the prism. Violet light bends the most because it slows down more, while red light bends the least because it slows down less. This variation in bending angles causes the colors to separate, forming a spectrum. The different colors of light we see are simply different wavelengths of electromagnetic radiation that our eyes can detect.
In simple words: White light splits into colors when it goes through a prism because each color's speed changes differently in the glass. This makes them bend at different angles, separating them into a rainbow.

🎯 Exam Tip: When drawing the ray diagram, ensure the white light enters the prism and then disperses into the VIBGYOR spectrum, with violet bending the most and red bending the least. Clearly label the incident ray, prism, and dispersed colors.

Question 45. When we see any object through the hot air over the fire, it appears to be wavy, moving slightly. Explain.
Answer: When we look at objects through the hot air above a fire, they appear to be wavy or shimmer. This phenomenon is caused by the varying refractive index of the air. Hot air is less dense than the cooler air surrounding it, making its refractive index slightly different. As light from the object travels through these layers of hot, unevenly dense air, it undergoes continuous and irregular refraction (bending). The path of light changes rapidly and randomly as the hot air moves, causing the apparent position of the object to shift slightly. This gives the object a wavy or flickering appearance. It's similar to how stars twinkle due to atmospheric refraction.
In simple words: Hot air above a fire moves around and has a different density than cool air. Light passing through this uneven hot air bends in different ways, making objects behind it look blurry and wavy.

🎯 Exam Tip: The key concept is that hot air has a different (lower) refractive index due to lower density. Emphasize that the continuous, irregular refraction of light through these moving hot air currents causes the wavy appearance.

Question 46. Study the diagram given below and answer the questions that it follows: (a) Name the defect and give reason. (b) Give 2 causes for this defect. (c) Give the correction – draw diagram for the same.
Answer:
(a) The defect is Myopia, also known as short-sightedness. This is because the parallel rays of light from a distant object converge and form an image in front of the retina, instead of directly on it. This results in distant objects appearing blurred.
(b) Two common causes for Myopia are:
(i) The eyeball becomes elongated, meaning its length from front to back increases. This causes the retina to be too far from the eye lens.
(ii) The focal length of the eye lens decreases, meaning the lens becomes too convergent (too powerful), causing light rays to focus too strongly.
(c) Myopia is corrected by using a concave (diverging) lens of suitable focal length. This lens diverges the incoming light rays slightly before they enter the eye, allowing the eye's natural lens to focus the image correctly onto the retina.
In simple words: Myopia, or near-sightedness, happens when the eye focuses light in front of the retina, making distant things blurry. This can be because the eyeball is too long or the lens is too powerful. It is fixed by wearing a concave lens that spreads out the light a bit before it enters the eye, allowing it to focus correctly on the retina.

🎯 Exam Tip: For eye defects, remember to include (a) the name of the defect, (b) a brief description of what vision is affected, (c) the two primary causes (eyeball length or lens curvature/focal length), and (d) the type of corrective lens used, along with a ray diagram showing both the defect and its correction.

Light Long Answer Type Questions

Question 1. (a) It is desired to obtain an erect image of an object using a concave mirror of focal length 20 cm.
(i) What should be the range of distance of the object from the mirror?
(ii) Will the image be bigger or smaller than the object?
(iii) Draw a ray diagram to show the image formation in this case.
(b) One-half of a convex lens of focal length 20 cm is covered with a black paper.
(i) Will the lens produce a complete image of the object?
(ii) Show the formation of image of an object placed at 2Fa of such covered lens with the help of a ray diagram.
(iii) How will the intensity of the image formed by half covered lens compare with non-covered lens?
Answer:
(a) (i) To get an erect image from a concave mirror, the object must be placed between the pole (P) and the focus (F) of the mirror. Since the focal length is 20 cm, the range of object distance should be between 0 cm and 20 cm from the mirror's pole.
(ii) In this case, the image formed will always be virtual, erect, and larger than the object.
(iii) Here is the ray diagram for image formation when the object is between P and F of a concave mirror:

(b) (i) Yes, a complete image will be formed even if half of the lens is covered. Each part of the lens can form a full image. The image will be less bright because fewer light rays pass through.
(ii) Here is the ray diagram for an object placed at \(2F_1\) of a convex lens:

(iii) The intensity of the image formed by a half-covered lens will be less than that formed by a full, uncovered lens. This is because fewer light rays pass through the lens to form the image, making it dimmer.
In simple words: For a concave mirror, to see an upright (erect) image, the object must be very close, within 20 cm. The image will look bigger. If you cover half a convex lens, you still see the whole image, but it will be less bright.

🎯 Exam Tip: Remember that for an erect image in a concave mirror, the object must always be placed within its focal length. For a convex lens, even a partial lens forms a complete image, just with reduced brightness.

Question 2. (i) Rear view mirror in vehicles
(ii) Solar furnace
(iii) Torch
(iv) Solar cooker
(v) To get the full length image of tall building.
Answer:
(i) A convex mirror is used as a rear-view mirror in vehicles. It provides a virtual, diminished image and offers a wider field of view, helping the driver see more of the road behind.
(ii) A concave mirror is used in a solar furnace. It collects and concentrates parallel beams of sunlight to a single point, generating high heat.
(iii) A concave mirror is used in a torch. It reflects light rays as a parallel beam, which helps in focusing the light over a long distance.
(iv) A concave mirror is also used in a solar cooker. Similar to a solar furnace, it gathers sunlight and focuses it to a point to cook food.
(v) A convex mirror is used to view the full-length image of a tall building. Its wide field of view allows a large area to be seen, though the image will be smaller.
In simple words: Convex mirrors are good for seeing a wide area, like in car mirrors. Concave mirrors are good for focusing light, like in torches or solar cookers.

🎯 Exam Tip: Understand the key property of each mirror type: convex mirrors diverge light and give a wide field of view with virtual, diminished images, while concave mirrors converge light and can form real or virtual, magnified or diminished images depending on object position.

Question 3. A convex lens has a focal length of 15 cm. At what distance from the lens should the object be placed so that it forms on its other side a real and inverted image 30 cm away from the lens? What would be the size of image formed if the object is 5 cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
Answer:
Given:
Focal length of convex lens, \( f = +15 \text{ cm} \)
Image distance (real and inverted image formed on the other side), \( v = +30 \text{ cm} \)
Object height, \( h = 5 \text{ cm} \)

First, we will find the object distance (\( u \)) using the lens formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{30} - \frac{1}{u} = \frac{1}{15} \)
\( \frac{1}{u} = \frac{1}{30} - \frac{1}{15} \)
To subtract, find a common denominator (30):
\( \frac{1}{u} = \frac{1}{30} - \frac{2}{30} \)
\( \frac{1}{u} = \frac{1 - 2}{30} \)
\( \frac{1}{u} = \frac{-1}{30} \)
\( \implies u = -30 \text{ cm} \)
The negative sign indicates that the object is placed on the left side of the lens, as expected. The object is placed at a distance of 30 cm from the lens, which is at \(2f\) (since \(2 \times 15 = 30\) cm).

Next, we will find the height of the image (\( h' \)) using the magnification formula:
\( M = \frac{h'}{h} = \frac{v}{u} \)
\( \frac{h'}{5} = \frac{+30}{-30} \)
\( \frac{h'}{5} = -1 \)
\( \implies h' = -5 \text{ cm} \)
The negative sign indicates that the image is inverted, and the size of the image is 5 cm, meaning it is the same size as the object.

Here is the ray diagram for image formation by a convex lens when the object is placed at \(2F_1\):

In simple words: When an object is placed at two times the focal length of a convex lens, its image forms at the same distance on the other side. The image will be real, upside down, and the same size as the object.

🎯 Exam Tip: Always remember the sign conventions for lenses and mirrors. For a convex lens, if the object is at \(2f\), the image is also at \(2f\) on the opposite side, real, inverted, and the same size as the object.

Question 4. Draw a neat labelled diagram of human eye and explain the working of each part of it.
Answer:The human eye is a complex organ that works like a camera to help us see the world. It consists of several parts, each with a specific function:

(a) **Cornea:** This is the clear, transparent outer layer at the front of the eye. Its main job is to refract (bend) most of the light entering the eye.
(b) **Iris:** This is the colored part of the eye, which is a muscular diaphragm. It controls the size of the pupil, thus regulating the amount of light that enters the eye. When it's very bright, the iris makes the pupil smaller.
(c) **Pupil:** This is the small, black opening in the center of the iris. It acts like an aperture, allowing light to pass through to the lens and retina.
(d) **Lens:** The eye lens is a convex lens made of transparent, flexible material. It focuses the light rays onto the retina, adjusting its focal length to see objects at different distances. This adjustment helps us see both near and far objects clearly.
(e) **Ciliary muscles:** These muscles are attached to the eye lens. They change the shape of the lens, and thereby its focal length, to help the eye focus on objects at various distances. When these muscles relax, the lens becomes thinner, increasing its focal length for distant vision. When they contract, the lens becomes thicker, decreasing its focal length for near vision.
(f) **Retina:** This is the light-sensitive screen at the back of the eye. It contains millions of light-sensitive cells (rods and cones) that convert light into electrical signals. These cells are like the film in a camera, where the image is formed.
(g) **Optic nerve:** This nerve carries the electrical signals generated by the retina to the brain. The brain then interprets these signals, forming the image we perceive.
In simple words: The human eye has many parts that work together to help us see. The cornea and lens bend light, the pupil controls how much light enters, the iris changes the pupil's size, and the retina acts like a screen where images are formed. The optic nerve sends these image signals to the brain.

🎯 Exam Tip: When describing the parts of the eye, focus on the function of each component and how they work together to achieve vision. A clear, well-labelled diagram can earn you extra marks.

Question 5. Describe with the help of diagram, how the refraction of light takes place through a glass prism.
Answer:When a ray of light passes through a glass prism, it undergoes refraction (bending of light) twice. Here's how it happens:

1. **First Refraction (Air to Glass):** An incident ray (I) travels from a rarer medium (air) into a denser medium (glass) at the first face of the prism. As it enters the glass, it bends towards the normal. The angle formed between the incident ray and the normal is the angle of incidence (\( \angle i \)), and the angle formed between the refracted ray and the normal inside the prism is the angle of refraction (\( \angle r \)).
2. **Second Refraction (Glass to Air):** The refracted ray then travels through the glass and strikes the second face of the prism. At this point, it travels from a denser medium (glass) back into a rarer medium (air). As it exits the prism, it bends away from the normal. This emerging ray is called the emergent ray (E). The angle between the emergent ray and the normal is the angle of emergence (\( \angle e \)).

Due to these two refractions, the emergent ray is always deviated from the original path of the incident ray. The angle between the incident ray's direction and the emergent ray's direction is called the angle of deviation (\( \angle D \)). A prism spreads out white light into its constituent colors (dispersion) because each color bends by a slightly different amount.

Here is a diagram illustrating refraction through a glass prism:

In simple words: When light goes into a glass prism and comes out, it bends twice. The first bend is as it enters, and the second is as it leaves. This makes the light change its path, and if it's white light, it splits into different colors.

🎯 Exam Tip: Clearly label the incident ray, refracted ray, emergent ray, normals, and the angles of incidence, refraction, emergence, and deviation in your prism diagram. Emphasize that light bends towards the normal when entering a denser medium and away when entering a rarer medium.

Question 6. Name three refractive defects of vision with the help of diagram. Explain the reasons and their rectification.
Answer:The three common refractive defects of vision are Myopia, Hypermetropia, and Presbyopia, which occur when the eye cannot focus light correctly on the retina.

**1. Myopia (Nearsightedness):**
**Description:** A person with myopia can see nearby objects clearly but has difficulty seeing distant objects distinctly.
**Reasons:** This defect occurs due to:
(i) The eyeball becoming too long (elongation of the eyeball).
(ii) The focal length of the eye lens becoming too short (excessive curvature of the cornea or eye lens).
In this condition, the image of a distant object is formed in front of the retina, not directly on it.
**Rectification:** Myopia can be corrected by using a concave lens of appropriate focal length. This lens diverges the light rays before they enter the eye, pushing the image back onto the retina.

**Correction of Myopia:**
**2. Hypermetropia (Farsightedness):**
**Description:** A person with hypermetropia can see distant objects clearly but finds it difficult to see nearby objects distinctly.
**Reasons:** This defect occurs due to:
(i) The eyeball being too short (shortening of the eyeball).
(ii) The focal length of the eye lens being too long (decreased converging power of the eye lens).
In this condition, the image of a nearby object is formed behind the retina.
**Rectification:** Hypermetropia can be corrected by using a convex lens of appropriate focal length. This lens converges the light rays before they enter the eye, bringing the image forward onto the retina.

**Correction of Hypermetropia:**
**3. Presbyopia:**
**Description:** This is an age-related vision defect where a person finds it difficult to see nearby objects clearly. Sometimes, they may also have difficulty seeing distant objects. This is like a combination of both myopia and hypermetropia, which is why it often requires a special lens.
**Reasons:** It is caused by:
(i) The weakening of the ciliary muscles that hold and adjust the eye lens.
(ii) The reduced flexibility of the eye lens itself, making it harder to change its shape and focal length.
**Rectification:** Presbyopia is corrected by using bi-focal lenses. These lenses have both a concave part (for distant vision) and a convex part (for near vision), typically with the reading (convex) portion at the bottom.
In simple words: Myopia means you can't see far things clearly, and it's fixed with a concave lens. Hypermetropia means you can't see near things clearly, and it's fixed with a convex lens. Presbyopia happens when you get older and can't see clearly both near and far, so you need special lenses that have both types.

🎯 Exam Tip: When explaining vision defects, clearly state the specific symptoms, the underlying cause (e.g., eyeball length, lens flexibility), and the type of corrective lens used. Diagrams for Myopia and Hypermetropia are crucial to show how light rays are focused incorrectly and then corrected.