Samacheer Kalvi Class 10 Science Solutions Chapter 2 Optics

Column - I	Column - II
A Speed of light	(i) \( \frac{\sin i}{\sin r} \)
B Scattering of light by colloids	(ii) Smoke and pollen
C Refractive index	(iii) Rayleigh scattering
D Mie scattering	(iv) Tyndall scattering
	(v) \( 3 \times 10^8 \) m/s

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)
In simple words: Speed of light matches with its value in vacuum. Scattering by colloids is Tyndall scattering. Refractive index is related to Snell's law. Mie scattering is seen in things like smoke and pollen.

🎯 Exam Tip: For matching questions, eliminate options you are sure of first. Understand the core concept of each term to make accurate matches, especially in physics.

V. Assertion and reasoning type.

 

Question 1. Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small. Reason: Refractive index of the medium is inversely proportional to the velocity of the light.
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.
Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Both statements are correct. The reason explains why the assertion is true: when a material has a high refractive index, light slows down in it because the refractive index and light's speed are inversely linked. This link explains why light is slower in denser materials.

🎯 Exam Tip: In assertion-reason questions, first check if both statements are individually true. Then, check if the reason logically explains the assertion by inserting "because" or "therefore" between them.

 

Question 2. Assertion: Myopia is due to the increase in the converging power of eye lens. Reason: Myopia can be corrected with the help of concave lens.
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.
Answer: (a) If both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Myopia (nearsightedness) occurs because the eye lens converges light too much, causing the image to form in front of the retina. This defect can be fixed by using a concave lens, which spreads out light rays before they enter the eye, making the image fall correctly on the retina. The correction directly addresses the problem.

🎯 Exam Tip: Associate eye defects with the type of lens used for correction: concave for myopia (over-convergence) and convex for hypermetropia (under-convergence).

VI. Answer Briefly.

 

Question 1. What is refractive index?
Answer: The refractive index is a measure that tells us how much the speed of light changes when it passes from one medium to another. It helps us understand if light travels fast or slow in a particular material. This value shows how much light will bend when entering that material. In simple terms, it's a number that describes how light behaves when it enters a new substance.
In simple words: Refractive index tells us how fast or slow light travels in a material compared to a vacuum, and how much it bends when entering that material.

🎯 Exam Tip: Mentioning both the change in speed and the bending of light (refraction) is key when defining refractive index.

 

Question 2. State Snell's law.
Answer: Snell's law describes how light bends when it passes from one medium to another. It states that the ratio of the sine of the angle at which light hits the surface (angle of incidence) to the sine of the angle at which it bends within the new medium (angle of refraction) is always a constant value. This constant is equal to the ratio of the refractive indices of the two media. This law is also written as:
\( \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \)
In simple words: Snell's law explains how light bends when it crosses from one material to another. It connects the angles of the light rays and the refractive properties of the two materials.

🎯 Exam Tip: Clearly state the ratio of sines and the equality with the ratio of refractive indices. Including the mathematical formula \( \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \) is essential for full marks.

 

Question 3. Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer: When an object is placed between the principal focus (F) and twice the focal length (2F) of a convex lens, a real, inverted, and magnified image is formed beyond 2F on the other side of the lens. The diagram below illustrates this.

In simple words: When an object is placed between the focal point and twice the focal point of a convex lens, the image it creates will be real, larger than the object, and upside down, appearing further away from the lens on the opposite side.

🎯 Exam Tip: Always use at least two principal rays to accurately locate the image formed by a lens. Ensure your ray diagram includes the object, lens, principal axis, focal points (F and 2F), and the image with proper arrowheads on rays.

 

Question 4. Define dispersion of light.
Answer: Dispersion of light is the phenomenon where a beam of white light, or composite light, splits into its different component colours when it passes through a transparent medium like a glass prism or water. This happens because each color of light has a different wavelength and therefore travels at a slightly different speed within the medium, causing them to bend at different angles. A rainbow is a natural example of light dispersion.
In simple words: Dispersion of light means white light separating into its different colors, like a rainbow, when it goes through a transparent material.

🎯 Exam Tip: Remember that dispersion occurs due to the different speeds (and thus different refractive indices) of different wavelengths of light in a medium.

 

Question 5. State Rayleigh's law of scattering.
Answer: Rayleigh's scattering law explains why the sky appears blue. It states that the amount of light scattered by particles is inversely proportional to the fourth power of its wavelength. This means that shorter wavelengths of light, like blue and violet, are scattered much more strongly than longer wavelengths, such as red and orange light.
Amount of scattering 'S' \( \propto \frac{1}{\lambda^4} \)
In simple words: Rayleigh's law says that tiny particles scatter blue light much more than red light because blue light has a shorter wavelength. This is why the sky looks blue.

🎯 Exam Tip: For Rayleigh scattering, the key is the inverse fourth power relationship with wavelength. Make sure to specify that this law applies when the scatterer particle size is much smaller than the wavelength of light.

 

Question 6. Differentiate convex lens and concave lens.

Answer: The main differences between convex and concave lenses lie in their shape, how they affect light, and their uses for vision correction. Convex lenses bulge outwards, bring light together (converge), form mostly real images, and help with farsightedness (hypermetropia). Concave lenses curve inwards, spread light out (diverge), always form virtual images, and fix nearsightedness (myopia). Their distinct shapes lead to opposite optical properties.
In simple words: Convex lenses are thicker in the middle, converge light, make real images, and fix farsightedness. Concave lenses are thinner in the middle, spread light out, make virtual images, and fix nearsightedness.

🎯 Exam Tip: When differentiating, ensure you provide clear contrasting points for each characteristic (shape, action on light, image type, and common application) to score full marks.

 

Question 7. What is the power of accommodation of the eye?
Answer: The power of accommodation of the eye is its remarkable ability to adjust the focal length of its own lens. This adjustment allows the eye to clearly focus on objects that are at different distances, whether they are very close or far away. This dynamic change in focal length is primarily achieved with the help of the ciliary muscles, which modify the shape of the eye lens. Without accommodation, we would only be able to see objects clearly at one specific distance.
In simple words: The eye's power of accommodation is its ability to change the focal length of its lens to see objects clearly at different distances, both near and far.

🎯 Exam Tip: Highlight that accommodation involves changing focal length and that the ciliary muscles are responsible for this adjustment. This flexibility ensures a sharp image on the retina at varying distances.

 

Question 8. What are the causes of 'Myopia'?
Answer: Myopia, also known as nearsightedness, occurs due to several reasons related to the structure and focusing ability of the eye. The main causes include:
1. The lengthening of the eye ball, making it too long from front to back.
2. The focal length of the eye lens being reduced, meaning it converges light too strongly.
3. An increase in the distance between the eye lens and the retina.
4. The far point, which is the farthest distance an object can be seen clearly, is no longer at infinity.
5. The far point comes closer to the eye, meaning distant objects appear blurry. All these factors cause light to focus in front of the retina.
In simple words: Myopia happens because the eyeball is too long or the eye lens focuses light too much, causing images of far-off things to form in front of the retina, making them look blurry.

🎯 Exam Tip: List at least three distinct causes, focusing on anatomical changes (eyeball length) and optical properties (focal length, far point) for a complete answer.

 

Question 9. Why does the sky appear blue in colour?
Answer: The sky appears blue because of a phenomenon called Rayleigh scattering. When sunlight, which contains all colors, enters Earth's atmosphere, it hits tiny air molecules (like nitrogen and oxygen). These small molecules scatter shorter wavelengths of light, such as blue and violet, much more effectively than longer wavelengths like red and orange. This scattered blue light spreads across the sky, making it appear blue to our eyes. Because our eyes are more sensitive to blue than violet, we perceive the sky as blue.
In simple words: The sky looks blue because tiny air particles scatter blue light from the sun much more than other colors, spreading it across the atmosphere.

🎯 Exam Tip: The crucial points for explaining the blue sky are Rayleigh scattering, the preferential scattering of shorter wavelengths (blue/violet), and the role of atmospheric particles.

 

Question 10. Why are traffic signals red in colour?
Answer: Traffic signals are red because red light has the longest wavelength among visible colors. This long wavelength means that red light is scattered the least by atmospheric particles like dust, smoke, and fog. As a result, red light can travel the farthest distance through various weather conditions without losing much of its intensity. This makes red signals highly visible even from far away or in poor weather, ensuring maximum safety.
In simple words: Traffic signals are red because red light has a long wavelength, so it scatters less and can travel a long distance, making it easily visible even in bad weather like fog or rain.

🎯 Exam Tip: The key reason for red signals is that red light has the longest wavelength, leading to minimal scattering and maximum penetration through atmospheric obstacles.

VII. Give the answer in detail.

 

Question 1. List any five properties of light?
Answer: Light is a fundamental part of physics with several important properties:
• Light is a form of energy. It can be converted into other forms, such as heat or electrical energy.
• Light always travels along a straight line. This principle is called rectilinear propagation of light.
• Light does not need any medium for its propagation. It can even travel through a vacuum, which is how sunlight reaches Earth.
• The speed of light in vacuum or air is approximately \( c = 3 \times 10^8 \) m/s. This is the fastest speed at which anything can travel.
• Since light is in the form of waves, it is characterized by a wavelength \( (\lambda) \) and a frequency \( (\nu) \). These are related by the equation: \( c = \nu \lambda \) (where c is the velocity of light). This wave-particle duality means light exhibits properties of both waves and particles.
• Different colored light has a different wavelength and frequency. For example, red light has a longer wavelength than blue light.
In simple words: Light is a type of energy that always moves in straight lines, even through empty space, at a very high speed. It acts like both a wave and a particle, with different colors having different wavelengths.

🎯 Exam Tip: When listing properties, try to include a mix of its nature (energy, wave-particle), propagation characteristics (straight line, vacuum), and quantitative aspects (speed, wavelength-frequency relation).

 

Question 2. Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
Answer: To understand how images are formed by a convex lens, we use specific ray tracing rules. These rules help predict the position, nature, and size of the image.
Rule-1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation. This ray passes straight through.

Rays passing through the optical centre

Rule-2: When rays parallel to the principal axis strike a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus. For a convex lens, parallel rays pass through the focal point on the other side.

Rays passing parallel to the optic axis

Rule-3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis. For a convex lens, a ray passing through the focal point on the object side becomes parallel to the principal axis after refraction.

Rays passing through or directed towards the principal focus

In simple words: There are three simple rules to draw how light passes through a convex lens. First, a ray going through the center of the lens doesn't bend. Second, a ray hitting the lens straight (parallel to the main line) will bend and go through a special point called the focus. Third, a ray that starts from the focus will hit the lens and then travel straight (parallel to the main line).

🎯 Exam Tip: Practice drawing ray diagrams for all six object positions for a convex lens. Accurately drawing two of these three principal rays is sufficient to locate the image and determine its characteristics.

 

Question 3. Differentiate the eye defects: Myopia and Hypermetropia.

Answer: Myopia (nearsightedness) and Hypermetropia (farsightedness) are two common eye defects that affect how we see objects at different distances. Myopia happens when the eyeball is too long or the lens is too powerful, causing distant objects to blur as their image forms in front of the retina. This makes the far point closer than infinity and is corrected with a concave (diverging) lens. On the other hand, Hypermetropia occurs when the eyeball is too short or the lens is too weak, leading to nearby objects appearing blurry because their image forms behind the retina. This pushes the near point further away and is corrected with a convex (converging) lens. Both conditions impact the focal length of the eye lens and are diagnosed by how clearly one sees at various distances.
In simple words: Myopia is trouble seeing far, caused by a long eyeball or strong lens, making images fall in front of the retina, and fixed with a concave lens. Hypermetropia is trouble seeing near, caused by a short eyeball or weak lens, making images fall behind the retina, and fixed with a convex lens.

🎯 Exam Tip: When differentiating, use a clear comparison table. Focus on contrasting points such as eyeball length, focal length, image position, and the type of corrective lens for each defect.

 

Question 4. Explain the construction and working of a 'Compound Microscope'.
Answer: A compound microscope is a powerful optical instrument designed to view very tiny objects that are too small to be seen with the naked eye.
Construction: It consists of two main convex lenses:
• The 'objective lens' (or objective piece) has a shorter focal length and is placed closer to the object.
• The 'eye lens' (or eyepiece) has a larger focal length and a wider opening (aperture), and it is placed near the observer's eye.
Both lenses are securely held within a narrow tube, which can be adjusted to change the distance between them.

Working:
1. The object (AB) is positioned just beyond the focal length of the objective lens (u > \( F_o \)).
2. The objective lens forms a real, inverted, and magnified image (A'B') on the opposite side.
3. This image (A'B') then acts as a virtual object for the eyepiece.
4. The eyepiece is adjusted so that A'B' falls within its principal focus.
5. The eyepiece then forms a final image (A''B'') that is virtual, highly enlarged, and erect with respect to A'B' (but still inverted compared to the original object AB). This final image is seen on the same side as the virtual object. A compound microscope typically achieves magnifications between 50 to 200 times, far exceeding a simple microscope.
In simple words: A compound microscope uses two lenses: an objective lens near the tiny object and an eyepiece lens near your eye. The objective lens first makes a bigger, upside-down image. Then, the eyepiece takes this image and magnifies it even more, so you see a very large, upside-down, virtual image.

🎯 Exam Tip: Clearly distinguish the roles of the objective and eyepiece lenses, including the type of image each forms. Remember that the final image is highly magnified and virtual, seen through the eyepiece.

VIII. Numerical Problems.

 

Question 1. An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.
Answer: Given:
Distance of an object \( u = -20 \) cm (negative because it's to the left of the lens)
Focal length of a convex lens \( f = 10 \) cm (positive for a convex lens)
We need to find the image distance \( v \) and the nature of the image.
Using the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substitute the given values:
\( \frac{1}{10} = \frac{1}{v} - \frac{1}{-20} \)
\( \frac{1}{10} = \frac{1}{v} + \frac{1}{20} \)
Now, we solve for \( \frac{1}{v} \):
\( \frac{1}{v} = \frac{1}{10} - \frac{1}{20} \)
To subtract, find a common denominator (20):
\( \frac{1}{v} = \frac{2}{20} - \frac{1}{20} \)
\( \frac{1}{v} = \frac{1}{20} \)
This means:
\( v = 20 \) cm
The positive value of \( v \) indicates that the image is formed on the right side of the lens, meaning it is a real image.
To find the nature (magnification), use the magnification formula: \( m = \frac{v}{u} \)
\( m = \frac{20}{-20} \)
\( m = -1 \)
A magnification of -1 indicates that the image is real, inverted, and of the same size as the object.
In simple words: For an object 20 cm from a convex lens with a 10 cm focal length, the image forms 20 cm on the other side. This image is real, upside down, and the same size as the object.

🎯 Exam Tip: Remember to use the correct sign conventions for object distance (u), image distance (v), and focal length (f) according to the Cartesian sign convention. Positive focal length for convex lenses, negative for concave lenses. Magnification \( m < 0 \) means a real, inverted image, while \( m > 0 \) means a virtual, erect image. The absolute value of m tells you the size.

 

Question 2. An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.
Answer: Given:
Object height \( h = 3 \) cm
Object distance \( u = -10 \) cm (negative for object placed to the left)
Focal length of a concave lens \( f = -15 \) cm (negative for a concave lens)
We need to find the image height \( h' \).
First, find the image distance \( v \) using the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substitute the given values:
\( \frac{1}{-15} = \frac{1}{v} - \frac{1}{-10} \)
\( \frac{1}{-15} = \frac{1}{v} + \frac{1}{10} \)
Now, solve for \( \frac{1}{v} \):
\( \frac{1}{v} = \frac{1}{-15} - \frac{1}{10} \)
To subtract, find a common denominator (30):
\( \frac{1}{v} = \frac{-2}{30} - \frac{3}{30} \)
\( \frac{1}{v} = \frac{-5}{30} \)
This means:
\( v = \frac{-30}{5} \)
\( v = -6 \) cm
The negative sign for \( v \) indicates that the image is formed on the same side as the object, meaning it is a virtual image.
Now, use the magnification formula to find the image height \( h' \):
\( m = \frac{h'}{h} = \frac{v}{u} \)
\( \frac{h'}{3} = \frac{-6}{-10} \)
\( \frac{h'}{3} = 0.6 \)
So, \( h' = 3 \times 0.6 \)
\( h' = 1.8 \) cm
The positive value of \( h' \) indicates that the image is erect (upright).
Thus, the image formed is virtual, erect, and \( 1.8 \) cm tall.
In simple words: For a 3 cm tall object placed 10 cm from a concave lens (focal length 15 cm), the image forms 6 cm away on the same side as the object. The image will be \( 1.8 \) cm tall, virtual, and upright.

🎯 Exam Tip: Always pay attention to the type of lens (convex or concave) as it dictates the sign of the focal length. Use sign conventions consistently for all distances (u, v, f) and interpret the signs of v and m (or h') to determine the nature of the image.

IX. Higher order thinking (HOT) questions.

 

Question 1. Ramu passes white light through a quartz prism. For which colour refractive index is greater?
Answer: When white light passes through a quartz prism, the refractive index is greater for violet light. This is because violet light has the shortest wavelength among visible colors, and materials generally refract light with shorter wavelengths more strongly. Therefore, violet light bends the most, indicating a higher refractive index for that specific color compared to other colors in the white light spectrum. This difference in refractive index for different colors causes the dispersion of white light.
In simple words: When white light goes through a prism, the refractive index is highest for violet light. This makes violet light bend the most because it has the shortest wavelength.

🎯 Exam Tip: Remember the order of colors in the visible spectrum (VIBGYOR) and their corresponding wavelengths. Shorter wavelengths (like violet) experience greater deviation and higher refractive indices in a prism.

Samacheer Kalvi 10th Science Optics Additional Important Questions and Answers

I. Choose the Correct Answer.

 

Question 1. The refractive index of four substances A, B, C and D are 1.31, 1.43, 1.33, 2.4 respectively. The speed of light is maximum in:
(a) A
(b) B
(c) C
(d) D
Answer: (a) A
In simple words: The substance with the lowest refractive index will allow light to travel fastest. Substance A has the lowest refractive index, so light moves fastest through it.

🎯 Exam Tip: Remember that a lower refractive index means light bends less and travels faster through that medium.

 

Question 2. Mie scattering is responsible for the _______ appearance of the clouds.
(a) red
(b) blue
(c) colourless
(d) white
Answer: (d) white
In simple words: Mie scattering makes clouds look white because it scatters all colours of light equally. This type of scattering happens when particles are about the same size as the light's wavelength.

🎯 Exam Tip: Distinguish between Rayleigh scattering (causes blue sky) and Mie scattering (causes white clouds).

 

Question 3. In an inelastic scattering the energy of the incident beam of light is _______ that of scattering beam.
(a) greater than
(b) less than
(c) equal to
(d) different from
Answer: (d) different from
In simple words: In inelastic scattering, the light changes its energy after hitting something. This means the energy of the light going in is not the same as the energy of the light coming out.

🎯 Exam Tip: Understand that inelastic scattering involves a transfer of energy between the incident light and the scattering medium, leading to a change in wavelength.

 

Question 4. As per Rayleigh's scattering law, amount of scattering is:
(a) directly proportional to fourth power of wavelength
(b) inversely proportional to fourth power of wavelength
(c) inversely proportional to square of wavelength
(d) directly proportional to square of wavelength
Answer: (b) inversely proportional to fourth power of wavelength
In simple words: Rayleigh's law says that light with shorter wavelengths (like blue) is scattered much more than light with longer wavelengths (like red). This is why our sky looks blue!

🎯 Exam Tip: Remember the inverse fourth power relationship with wavelength, as this is key to explaining phenomena like blue skies and red sunsets.

 

Question 5. The refractive index of a medium is dependent on the _______ of the light.
(a) wavelength
(b) strength
(c) density
(d) refraction
Answer: (a) wavelength
In simple words: How much light bends when it enters a material depends on the colour (or wavelength) of that light. Different colours bend by slightly different amounts.

🎯 Exam Tip: This dependency explains why a prism splits white light into a rainbow, a phenomenon called dispersion.

 

Question 6. The scattering of light by colloidal particles in the colloidal solution is called:
(a) Raman scattering
(b) Tyndall scattering
(c) Mie scattering
(d) Elastic scattering
Answer: (b) Tyndall scattering
In simple words: When a beam of light passes through a mixture with tiny particles, you can see the light path because the particles scatter the light. This is called the Tyndall effect.

🎯 Exam Tip: The Tyndall effect is commonly observed when light passes through fog or a colloidal solution, making the light beam visible.

 

Question 7. A piece of transparent material bounded by curved surfaces is called:
(a) mirror
(b) prism
(c) slab
(d) lens
Answer: (d) lens
In simple words: A lens is a clear piece of material that has at least one curved side. It is used to bend light in a special way.

🎯 Exam Tip: Lenses are crucial in optics for focusing or diverging light, forming images in devices like cameras and eyeglasses.

 

Question 8. If the energy of the incident and the scattered beam of light are not the same, then it is called as _______.
(a) Elastic
(b) Raman
(c) Inelastic
(d) Mie
Answer: (c) Inelastic
In simple words: When light scatters and its energy changes, we call it inelastic scattering. This means the light either gains or loses energy.

🎯 Exam Tip: The Raman effect is a type of inelastic scattering where the scattered light has a different wavelength (and thus energy) than the incident light.

 

Question 9. A convex lens does not produce:
(a) real magnified image
(b) virtual magnified image
(c) virtual diminished image
(d) real diminished image
Answer: (c) virtual diminished image
In simple words: A convex lens can make images that are bigger and real, or bigger and virtual (like a magnifying glass). It can also make smaller and real images. But it cannot make a virtual image that is smaller.

🎯 Exam Tip: Remember the image characteristics for convex lenses depend on the object's position, but they never form a virtual and diminished image.

 

Question 10. A lens which is thicker in the middle than at the edges is known as:
(a) concave lens
(b) convex lens
(c) bifocal lens
(d) cylindrical lens
Answer: (b) convex lens
In simple words: A convex lens bulges out in the middle, making it thicker there and thinner at its edges. It brings light rays together.

🎯 Exam Tip: Convex lenses are also known as converging lenses because they converge parallel light rays to a single point.

 

Question 11. The object is always placed on the _______ side of the lens.
(a) left
(b) right
(c) top
(d) bottom
Answer: (a) left
In simple words: In diagrams and calculations for lenses, we always put the object on the left side. This is a standard rule.

🎯 Exam Tip: Following the Cartesian sign convention, the object is conventionally placed to the left of the optical element.

 

Question 12. The parallel rays from the outer edge are deviated towards the middle in a:
(a) convex mirror
(b) concave lens
(c) concave mirror
(d) convex lens
Answer: (d) convex lens
In simple words: A convex lens gathers parallel light rays and bends them inwards, causing them to meet at a point in the middle. This is its job as a converging lens.

🎯 Exam Tip: Convex lenses are converging lenses, meaning they bring parallel light rays closer together.

 

Question 13. The light rays passing through the optic centre will:
(a) diverged
(b) scattered
(c) converged
(d) emerge undeviated
Answer: (d) emerge undeviated
In simple words: When light goes straight through the very center of a lens, it does not bend at all. It keeps going in the same direction.

🎯 Exam Tip: This property is fundamental for drawing ray diagrams as it provides a straightforward ray path.

 

Question 14. All the distances are measured from the _______ of the lense.
(a) aperture
(b) optical centre
(c) principal focus
(d) infinity
Answer: (b) optical centre
In simple words: When we measure distances for lenses, we always start from the middle point of the lens, called the optical centre. This keeps measurements consistent.

🎯 Exam Tip: The optical centre is the geometric centre of the lens and serves as the reference point for all measurements in lens calculations.

 

Question 15. A ray passing through the principal focus and incident on the lens will:
(a) converge
(b) diverge
(c) emerge parallel to the principal axis
(d) not emerge out
Answer: (c) emerge parallel to the principal axis
In simple words: If a light ray hits a lens after passing through its main focus point, it will then leave the lens traveling straight and parallel to the main axis. This is a key rule for how lenses work.

🎯 Exam Tip: This is a crucial rule for drawing ray diagrams and understanding image formation, especially in convex lenses where rays passing through F (before the lens) become parallel after refraction.

 

Question 16. When the object is placed at infinity from the convex lens, the image is formed at:
(a) F
(b) C
(c) infinity
(d) between F and 2F
Answer: (a) F
In simple words: If an object is very, very far away from a convex lens, its image will appear at the lens's main focus point. This image will be tiny.

🎯 Exam Tip: Remember that parallel rays from a distant object converge at the principal focus (F) of a convex lens.

 

Question 17. The human eye is _______ in nature.
(a) convex
(b) concave
(c) transparent glass
(d) Plano - concave
Answer: (a) convex
In simple words: The lens inside our eye is shaped like a convex lens, which helps to focus light onto the retina. This makes it a natural converging lens system.

🎯 Exam Tip: The human eye's lens system, primarily the cornea and the crystalline lens, functions as a converging lens to form real, inverted images on the retina.

 

Question 18. The image formed by a concave lens is:
(a) virtual
(b) diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer: (c) virtual and diminished
In simple words: A concave lens always makes an image that is smaller than the object and appears behind the lens, which means it is a virtual image. It can never be captured on a screen.

🎯 Exam Tip: Concave lenses are diverging lenses; they always produce virtual, erect, and diminished images, regardless of the object's position.

 

Question 19. To get a real image using convex lens, the object must be placed at:
(a) infinity
(b) principal focus
(c) beyond principal focus and infinity
(d) both (b) and (c)
Answer: (d) both (b) and (c)
In simple words: To see a real image with a convex lens, the object needs to be placed either right at the main focus point or any distance further away from the lens than the focus point. This ensures light rays converge.

🎯 Exam Tip: A convex lens forms a real image when the object is placed anywhere from the principal focus (F) to infinity. Option (b) implies at F (image at infinity), and (c) implies beyond F to infinity.

 

Question 20. _______ is the centre part of the iris.
(a) cornea
(b) retina
(c) pupil
(d) eye lens
Answer: (c) pupil
In simple words: The pupil is the black hole in the middle of the colored part of your eye (the iris). It's where light enters the eye.

🎯 Exam Tip: The pupil's size changes to control the amount of light entering the eye, similar to the aperture of a camera.

 

Question 21. For a convex lens the point at which the parallel rays converge is called of the lens.
(a) pole
(b) centre of curvature
(c) principal focus
(d) none
Answer: (c) principal focus
In simple words: When light rays that are parallel to each other hit a convex lens, they all meet up at one special spot on the other side. This spot is called the principal focus.

🎯 Exam Tip: The principal focus (F) is a critical point for understanding how lenses form images and is used extensively in ray diagrams.

 

Question 22. A real image formed by a convex lens is always:
(a) erect
(b) magnified
(c) inverted
(d) diminished
Answer: (c) inverted
In simple words: When a convex lens creates an image that you can catch on a screen (a real image), that image will always be upside down compared to the actual object.

🎯 Exam Tip: In optics, real images are always inverted, and virtual images are always erect.

 

Question 23. The law of distances is given by:
(a) \( \frac { 1 }{ f } = \frac { 1 }{ v } - \frac { 1 }{ u } \)
(b) \( \frac { 1 }{ f } = \frac { 1 }{ u } + \frac { 1 }{ v } \)
(c) \( \frac { 1 }{ f } = \frac { 1 }{ u } - \frac { 1 }{ v } \)
(d) \( \frac { 1 }{ f } = \frac { 1 }{ v } - \frac { 1 }{ u } \)
Answer: (b) \( \frac { 1 }{ f } = \frac { 1 }{ u } + \frac { 1 }{ v } \)
In simple words: This formula helps us find where an image will form or how strong a lens is. It connects the lens's focal length (f) with the object's distance (u) and the image's distance (v). The positive sign in this version indicates a specific sign convention often used.

🎯 Exam Tip: Be mindful of the sign convention used. While \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \) is common, this form \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \) is also used, particularly for real images, and consistency is key.

 

Question 24. The unit of focal length is:
(a) dioptre
(b) metre
(c) cm
(d) ampere
Answer: (b) metre
In simple words: Focal length is a measurement of distance, so its standard unit is the metre. Even if it's given in centimeters, convert it to metres for calculations involving power.

🎯 Exam Tip: Always use metres for focal length when calculating the power of a lens in dioptres.

 

Question 25. The sign of focal length of a convex lens is _______ sign.
(a) negative
(b) positive
(c) negative or positive
(d) none of the options
Answer: (b) positive
In simple words: For a convex lens, its focal length is always considered positive. This is because it makes light rays meet on the right side of the lens.

🎯 Exam Tip: Remember this sign convention: convex lenses have positive focal lengths, while concave lenses have negative focal lengths.

 

Question 26. The focal length of concave lens has _______ sign.
(a) positive
(b) negative
(c) positive or negative
(d) none of the options
Answer: (b) negative
In simple words: A concave lens spreads light rays out, making them seem to come from a point on the left side of the lens. Because of this, its focal length is always given a negative sign.

🎯 Exam Tip: Consistently apply sign conventions; a negative focal length for concave lenses helps in correctly solving lens problems.

 

Question 27. The magnification in terms of object distance u and image distance v is m :
(a) \( \frac{u}{v} \)
(b) \( u + v \)
(c) \( \frac{v}{u} \)
(d) \( \frac{1}{u} + \frac{1}{v} \)
Answer: (c) \( \frac{v}{u} \)
In simple words: Magnification tells us how much bigger or smaller an image is compared to the actual object. It's found by dividing the image distance by the object distance.

🎯 Exam Tip: A magnification greater than 1 means the image is enlarged, less than 1 means diminished, and a negative sign indicates an inverted image.

 

Question 28. In terms of object distance u and focal length f, magnification is given by m =
(a) \( \frac{u-f}{f} \)
(b) \( \frac{f}{u-f} \)
(c) \( \frac{u-f}{u} \)
(d) \( \frac{u+f}{f} \)
Answer: (b) \( \frac{f}{u-f} \)
In simple words: This formula shows how much a lens magnifies an object by using only the object's distance from the lens and the lens's focal length. It's a handy way to calculate magnification if you don't know the image distance.

🎯 Exam Tip: This alternative magnification formula is useful when the image distance is unknown, simplifying calculations if only focal length and object distance are given.

 

Question 29. The magnification in terms of v and f:
(a) f
(b) v - f
(c) \( \frac{f}{v-f} \)
(d) \( \frac{v-f}{f} \)
Answer: (d) \( \frac{v-f}{f} \)
In simple words: This formula allows us to find the magnification of a lens using only the image distance and the focal length. It simplifies calculations when the object distance is not known.

🎯 Exam Tip: Mastering these different forms of the magnification formula helps in solving problems where only certain parameters are provided.

 

Question 30. The unit of power is:
(a) m
(b) ohm
(c) dioptre
(d) ampere
Answer: (c) dioptre
In simple words: The power of a lens, which tells us how strongly it bends light, is measured in a unit called dioptre. One dioptre is the power of a lens with a focal length of one meter.

🎯 Exam Tip: Ensure that focal length is always in metres when calculating power in dioptres; otherwise, the result will be incorrect.

 

Question 31. If the focal length of a convex lens is 1 m then its power is:
(a) 1 dioptre
(b) 0.1 dioptre
(c) 10 dioptre
(d) 0.01 dioptre
Answer: (a) 1 dioptre
In simple words: The power of a lens is simply 1 divided by its focal length in metres. So, if the focal length is 1 metre, the power is 1 dioptre. This makes calculations straightforward.

🎯 Exam Tip: Remember the formula \( P = \frac{1}{f} \) where f is in metres. For a convex lens, power is positive.

 

Question 32. In a simple microscope, the magnification can be increased by:
(a) lens of long focal length
(b) lens
(c) lens of short focal length
(d) lens of infinite focal length
Answer: (c) lens of short focal length
In simple words: To make things look much bigger with a simple microscope, you need a lens that bends light a lot. Lenses with shorter focal lengths bend light more, so they provide higher magnification.

🎯 Exam Tip: The magnification of a simple microscope is inversely proportional to its focal length, so shorter focal lengths yield greater magnification.

 

Question 33. Convex lenses are used:
(a) as camera lenses
(b) as magnifying lenses
(c) to correct hypermetropia
(d) all of the options
Answer: (d) all of the options
In simple words: Convex lenses are very versatile and are used in many devices. They can magnify objects, help cameras take pictures, and correct a common eye problem called hypermetropia.

🎯 Exam Tip: Being familiar with the practical applications of different lens types helps in understanding their optical properties better.

 

Question 34. Which lens is used in wide angle spyhole in doors?
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) Plano - concave lens
Answer: (b) concave lens
In simple words: A concave lens is used in door spyholes because it spreads out light, allowing you to see a much wider area. This makes the image smaller but covers a larger field of view.

🎯 Exam Tip: Concave lenses are diverging lenses, perfect for wide-angle views because they make objects appear smaller and further away, expanding the field of vision.

 

Question 35. The mathematical form of lens maker's formula is:
(a) \( \frac{1}{f} = (\mu - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \)
(b) \( \frac{1}{f} = (\mu - 1) (\frac{1}{R_1} + \frac{1}{R_2}) \)
(c) \( \frac{1}{f} = \frac{1}{\mu} (\frac{1}{R_1} - \frac{1}{R_2}) \)
(d) \( \frac{1}{f} = \frac{1}{(\mu - 1)} (\frac{1}{R_1} + \frac{1}{R_2}) \)
Answer: (a) \( \frac{1}{f} = (\mu - 1) (\frac{1}{R_1} - \frac{1}{R_2}) \)
In simple words: This formula helps lens makers decide how curved to make the surfaces of a lens so it has a specific focal length. It depends on the material's refractive index and the radii of curvature of the two lens surfaces.

🎯 Exam Tip: The sign convention for \( R_1 \) and \( R_2 \) (radii of curvature) is critical; remember that for a double convex lens, \( R_1 \) is positive and \( R_2 \) is negative.

 

Question 36. If f is the focal length of the lens then its power is given by:
(a) \( P = \frac{2}{f} \)
(b) \( p = \frac{1}{f} \)
(c) \( p = f \)
(d) \( p = f \)
Answer: (b) \( p = \frac{1}{f} \)
In simple words: The power of a lens tells us how much it can bend light. You calculate it by dividing 1 by the focal length, which must be in metres.

🎯 Exam Tip: Always express focal length in metres when calculating power in dioptres. The greater the power, the shorter the focal length.

 

Question 37. Which part of the human eye changes the focal length of the eye lens?
(a) pupil
(b) retina
(c) ciliary muscles
(d) cornea
Answer: (c) ciliary muscles
In simple words: The ciliary muscles around your eye lens squeeze or relax to change the shape of the lens. This change in shape allows the eye to adjust its focus for objects that are near or far away.

🎯 Exam Tip: This ability to change focal length is called accommodation, a vital process for clear vision at varying distances.

 

Question 38. On which part of human eye, image is formed?
(a) cornea
(b) iris
(c) retina
(d) pupil
Answer: (c) retina
In simple words: The retina is the light-sensitive screen at the back of your eye. All the light entering your eye focuses here to form an image, which is then sent to your brain.

🎯 Exam Tip: The retina contains photoreceptor cells (rods and cones) that convert light into electrical signals for the brain.

 

Question 39. For normal human eye the value of near point is:
(a) 25 cm
(b) 25 m
(c) 2.5 m
(d) 25 mm
Answer: (a) 25 cm
In simple words: The near point is the closest distance at which a person with normal eyesight can see an object clearly without any strain. For most people, this is about 25 centimetres.

🎯 Exam Tip: Understanding the near and far points is crucial for diagnosing and correcting common vision defects.

 

Question 40. In hypermetropia, the focal length of the eye lens is:
(a) decreased
(b) remains the same
(c) increased
(d) none of the options
Answer: (c) increased
In simple words: In hypermetropia (farsightedness), the eye lens or cornea has a focal length that is too long, causing light to focus behind the retina. This makes nearby objects appear blurry.

🎯 Exam Tip: Hypermetropia is corrected using a convex lens, which helps to shorten the effective focal length and focus light correctly on the retina.

 

Question 41. Presbyopia can be corrected by using:
(a) convex lens
(b) bifocal lens
(c) concave lens
(d) cylindrical lens
Answer: (b) bifocal lens
In simple words: Presbyopia is when your eyes struggle to focus on close-up things as you get older. Bifocal lenses help fix this because they have two different strengths: one part for seeing far away and another for seeing up close.

🎯 Exam Tip: Presbyopia often co-occurs with other refractive errors, making bifocals or multifocal lenses a practical solution.

 

Question 42. Astigmatism can be corrected by using:
(a) bifocal lens
(b) cylindrical lens
(c) convex lens
(d) concave lens
Answer: (b) cylindrical lens
In simple words: Astigmatism makes your vision blurry or distorted because your eye's cornea is not perfectly round. Cylindrical lenses are specially shaped to fix this by focusing light differently along different directions.

🎯 Exam Tip: Cylindrical lenses are designed to correct the unequal curvature of the cornea or lens in astigmatism, providing clear vision.

 

Question 43. The magnifying power of compound microscope is:
(a) 10
(b) 20
(c) 50
(d) 50 to 200
Answer: (d) 50 to 200
In simple words: A compound microscope can make tiny objects look much bigger, usually between 50 and 200 times their actual size. This high magnification is achieved by using two lenses together.

🎯 Exam Tip: The total magnification of a compound microscope is the product of the magnification of the objective lens and the eyepiece.

 

Question 44. The accuracy of travelling microscope is of the order of:
(a) 0.01 cm
(b) 0.01 mm
(c) 0.1 mm
(d) 0.1 cm
Answer: (b) 0.01 mm
In simple words: A travelling microscope is very precise and can measure tiny distances with great accuracy. It can often measure down to one-hundredth of a millimetre.

🎯 Exam Tip: Travelling microscopes are commonly used in laboratories for accurate measurement of very small lengths, such as the diameter of a wire or the refractive index of a liquid.

II. Fill in the blanks.

 

Question 1. The velocity of light in vacuum is \( 3 \times 10^8 \) m/s.
Answer: The speed of light in empty space, or vacuum, is a very fast constant speed of \( 3 \times 10^8 \) meters per second. This is the fastest speed at which anything can travel.
In simple words: Light travels at \( 3 \times 10^8 \) meters per second in a vacuum.

🎯 Exam Tip: Remember the exact value and unit for the speed of light in vacuum, as it is a fundamental constant in physics.

 

Question 2. If \( v \) is the frequency and \( \lambda \) is the wavelength then velocity of the wave is c = \( v\lambda \).
Answer: If \( v \) stands for frequency (how many waves pass per second) and \( \lambda \) is the wavelength (the length of one wave), then the speed of a wave, called c, is found by multiplying them: \( c = v\lambda \). This formula helps connect how fast a wave moves to its repeating pattern.
In simple words: Wave velocity is found by multiplying frequency and wavelength.

🎯 Exam Tip: Know the wave equation \( c = v\lambda \) and clearly define each variable for full marks.

 

Question 3. Among colours of visible light red colour has the highest wavelength.
Answer: Among all the colours we can see, red light has the longest wavelength. This means its waves are more spread out compared to other colours like blue or violet. Longer wavelengths allow red light to travel further without scattering.
In simple words: Red light has the longest wavelength in the rainbow.

🎯 Exam Tip: Recall the VIBGYOR sequence to remember the order of wavelengths, with red being the longest and violet the shortest.

 

Question 4. According to Snell's law refractive index, \( \mu_2 \) = \( \frac{\sin i}{\sin r} \).
Answer: According to Snell's law, when light bends as it passes from one medium to another, the refractive index of the second medium (\( \mu_2 \)) relative to the first is given by the ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)), which is \( \frac{\sin i}{\sin r} \). This ratio explains how much the light ray will bend.
In simple words: Snell's law says the refractive index is the sine of the incident angle divided by the sine of the refracted angle.

🎯 Exam Tip: State Snell's law clearly and correctly use \( \sin i / \sin r \) in the formula.

 

Question 5. In a medium having high value of refractive index then speed of light in that medium is low.
Answer: If a material has a high refractive index, it means light slows down a lot when it enters that material. So, the speed of light in such a medium is relatively low. This is why a diamond, with a high refractive index, makes light move much slower.
In simple words: Light moves slower in materials with a high refractive index.

🎯 Exam Tip: Remember that a higher refractive index indicates a greater slowing down and bending of light.

 

Question 6. Angle of refraction is the smallest for red and the highest for violet.
Answer: When white light passes through a prism, different colours bend by different amounts. The angle of refraction is smallest for red light, meaning it bends the least, and it is highest for violet light, which bends the most. This separation of colours is called dispersion.
In simple words: Red light bends the least, and violet light bends the most, when passing through something like a prism.

🎯 Exam Tip: Recall that violet light has the shortest wavelength and bends most, while red has the longest wavelength and bends least.

 

Question 7. The refractive index depends on wavelength of light.
Answer: The refractive index of a material changes depending on the wavelength (or colour) of light passing through it. This phenomenon, known as dispersion, is why prisms can split white light into its different colours. For example, blue light bends more than red light in glass.
In simple words: How much light bends depends on its color or wavelength.

🎯 Exam Tip: Emphasize that refractive index is not a constant for all light but varies with wavelength.

 

Question 8. Colours having shorter wavelength scattered more than longer wavelength colours according to Rayleigh scattering law.
Answer: According to Rayleigh scattering law, colours with shorter wavelengths, like blue and violet, are scattered much more easily than colours with longer wavelengths, like red. This is why the sky appears blue during the day, as blue light is scattered across the atmosphere. Longer wavelengths pass through more directly.
In simple words: Rayleigh's law says shorter light waves scatter more than longer ones.

🎯 Exam Tip: Connect Rayleigh scattering directly to the inverse fourth power of wavelength for a complete explanation.

 

Question 9. After passing through a convex lens parallel rays converge at the principal focus.
Answer: When parallel light rays enter a convex lens, they bend inwards and meet at a single point called the principal focus. This is why convex lenses are often called converging lenses. This focused point is where a clear image can be formed.
In simple words: A convex lens makes parallel light rays meet at its main focus point.

🎯 Exam Tip: Remember that convex lenses converge parallel rays, while concave lenses diverge them.

 

Question 10. For a convex lens, as the object distance increases, the image distance will decrease.
Answer: For a convex lens, as you move an object further away from the lens, the image formed gets closer to the lens. This is a key relationship in optics, showing how image position changes with object position. This is described by the lens formula.
In simple words: Moving an object further from a convex lens makes its image form closer to the lens.

🎯 Exam Tip: Understand the relationship between object distance, image distance, and focal length for convex lenses, often visualized with ray diagrams.

 

Question 11. A ray passing through the optic centre of a lens emerges undeviated.
Answer: Any light ray that passes through the exact center of a lens, known as the optical center, continues its path without bending or changing direction. This property is very useful for drawing ray diagrams accurately. It acts like a straight path through the lens.
In simple words: Light going through the middle of a lens doesn't bend.

🎯 Exam Tip: This is a key rule for drawing ray diagrams: a ray through the optical center goes straight.

 

Question 12. Astigmatism is due to irregular curvature of the surface of the eye lens.
Answer: Astigmatism is an eye defect caused by an uneven or irregular shape in the cornea or the eye lens. This unevenness prevents light from focusing correctly on the retina, causing blurry vision at all distances. It's like looking through a wavy window.
In simple words: Astigmatism happens because the eye's front surface or lens isn't perfectly round.

🎯 Exam Tip: Focus on the cause (irregular curvature) and the effect (blurry vision) of astigmatism.

 

Question 13. When a parallel beam of light passes through a convex lens, the rays from the outer edges are the principal focus.
Answer: When parallel light rays, especially those hitting the outer parts of a convex lens, pass through it, they all bend and converge to a single point called the principal focus. This convergence is strongest at the edges where the lens is curved more sharply. This is the main function of a convex lens.
In simple words: Parallel light rays hitting a convex lens, especially at its edges, meet at the main focus point.

🎯 Exam Tip: Emphasize that a convex lens focuses parallel rays, and the term "principal focus" is specific to this point.

 

Question 14. A ray parallel to the principal axis of a convex lens after refraction passes through the principal focus.
Answer: A light ray that travels parallel to the main axis of a convex lens will bend after passing through the lens and go straight through its principal focus. This is a fundamental rule for how convex lenses work and helps determine where images will form. This behavior defines the focal point.
In simple words: Light rays parallel to a convex lens's center line always pass through its main focus after bending.

🎯 Exam Tip: This is a crucial rule for drawing ray diagrams for convex lenses; make sure to show the ray passing through F after refraction.

 

Question 15. When the object is placed between principal focus and optical centre of a convex lens a virtual image will be formed.
Answer: If an object is placed very close to a convex lens, specifically between its principal focus and optical center, the image formed will be virtual. This means the light rays don't actually meet, but appear to come from a point behind the object, creating an upright and magnified image. This is how a magnifying glass works.
In simple words: Putting an object between a convex lens's focus and its center creates a virtual image.

🎯 Exam Tip: Remember this specific object position (between F and O) is the only one for a convex lens that produces a virtual image.

 

Question 16. For a convex lens, as the object approaches the lens the image becomes bigger.
Answer: As an object gets closer to a convex lens, its image becomes larger and larger. This magnification effect is commonly seen when using a magnifying glass to view small objects. The closer the object, the more the image is enlarged.
In simple words: When an object moves closer to a convex lens, its image gets larger.

🎯 Exam Tip: Note the inverse relationship between object distance and image size for a convex lens in most cases.

 

Question 17. In a photographic camera biconvex lens is used.
Answer: Most photographic cameras use a biconvex lens, which is a lens curved outwards on both sides. This type of lens is excellent at focusing light to form clear, real images on the camera's sensor or film. Multiple lenses are often combined for better image quality.
In simple words: A photographic camera uses a biconvex lens to take pictures.

🎯 Exam Tip: Understand that biconvex lenses are converging lenses, essential for focusing light to form real images.

 

Question 18. The shorter the focal length, the greater is the magnification.
Answer: For lenses like those in magnifying glasses or microscopes, a shorter focal length means the lens has more power to bend light. This results in a greater magnification, making objects appear much larger. This is why powerful magnifying tools have very short focal lengths.
In simple words: Lenses with shorter focal lengths make things look much bigger.

🎯 Exam Tip: Relate shorter focal length directly to stronger converging power and higher magnification.

 

Question 20. Real images are formed by a convex lens.
Answer: Real images, which can be projected onto a screen, are typically formed by a convex lens. This type of lens makes light rays converge after passing through it, bringing them to a focus point. This is how projectors and cameras form images.
In simple words: Convex lenses create real images that can be seen on a screen.

🎯 Exam Tip: Remember that real images are formed by the actual intersection of refracted light rays.

 

Question 21. Concave lens produces virtual images.
Answer: A concave lens always produces virtual images, which means the light rays only appear to diverge from a point and cannot be projected onto a screen. These images are always upright and diminished (smaller than the object). Concave lenses are diverging lenses.
In simple words: Concave lenses always make virtual images, which look smaller and upright.

🎯 Exam Tip: Know that concave lenses always form virtual, upright, and diminished images, regardless of object position.

 

Question 22. The value of power of a lens having focal length one metre is One dioptre.
Answer: The power of a lens is measured in dioptres. If a lens has a focal length of one meter, its power is exactly one dioptre. This unit helps eye doctors prescribe corrective lenses. It indicates how strongly a lens can converge or diverge light.
In simple words: A lens with a 1-meter focal length has a power of 1 dioptre.

🎯 Exam Tip: The formula for power is \( P = \frac{1}{f} \), where f is in meters and P is in dioptres.

 

Question 23. For a normal eye the value of far point is infinity.
Answer: For a person with normal vision, the far point is at infinity. This means a healthy eye can see very distant objects clearly without any strain or accommodation. This allows us to see stars and distant landscapes sharply.
In simple words: A normal eye can see infinitely far away objects clearly.

🎯 Exam Tip: Define both far point and near point for a complete understanding of normal vision.

 

Question 24. Myopia is known as short sightedness.
Answer: Myopia is an eye condition commonly known as short-sightedness or near-sightedness. People with myopia can see nearby objects clearly, but distant objects appear blurry. This is often due to the eyeball being too long or the lens being too powerful.
In simple words: Myopia means you can see near things well but far things are blurry.

🎯 Exam Tip: Remember that myopia is corrected with a concave lens to diverge light before it reaches the eye.

 

Question 25. Hyper metropia is known as long sightedness.
Answer: Hypermetropia is an eye condition also known as long-sightedness or far-sightedness. People with hypermetropia can see distant objects clearly, but nearby objects appear blurry. This happens when the eyeball is too short or the lens is not powerful enough. This condition is the opposite of myopia.
In simple words: Hypermetropia means you can see far things well but near things are blurry.

🎯 Exam Tip: Know that hypermetropia is corrected with a convex lens to converge light more effectively.

 

Question 26. The mathematical form of focal length of a concave lens used to correct myopia is f = xy/x - y.
Answer: The mathematical formula for the focal length (f) of a concave lens used to correct myopia is \( f = \frac{xy}{x-y} \), where x and y represent specific distances related to the eye's defect. This formula helps determine the correct lens power needed for clear vision. It is derived from lens formula applications.
In simple words: The focal length of a concave lens to fix myopia is given by the formula \( f = \frac{xy}{x-y} \).

🎯 Exam Tip: Understand the application of lens formulae in vision correction, especially for myopia.

 

Question 27. Cylindrical lenses are used to correct astigmatism.
Answer: Astigmatism, an eye defect where vision is blurred due to an irregularly shaped cornea or lens, is corrected using cylindrical lenses. These special lenses have different curvatures in different planes, which helps to properly focus light onto the retina. This unique shape corrects the uneven focusing.
In simple words: Astigmatism is fixed by wearing special lenses called cylindrical lenses.

🎯 Exam Tip: Link astigmatism directly to cylindrical lenses and their ability to correct uneven focusing.

 

Question 28. For a normal eye, the value of least distance of distinct vision is 25 cm.
Answer: For a normal human eye, the closest distance at which an object can be seen clearly without strain is about 25 cm. This is known as the least distance of distinct vision, or the near point. It is the comfortable reading distance for most people. This limit helps define what we consider "normal" vision.
In simple words: A normal eye can see objects clearly if they are at least 25 cm away.

🎯 Exam Tip: Remember "25 cm" as the standard near point for a healthy eye.

 

Question 29. The objective of the compound microscope has shorter focal length.
Answer: The objective lens in a compound microscope, which is the lens closest to the object being viewed, always has a very short focal length. This short focal length is crucial for creating a magnified intermediate image, which is then further magnified by the eyepiece. This design helps achieve high overall magnification.
In simple words: The objective lens in a compound microscope has a short focal length.

🎯 Exam Tip: Differentiate between the focal lengths of the objective and eyepiece lenses in a compound microscope.

 

Question 30. The focal length of eye piece is greater in a compound microscope.
Answer: In a compound microscope, the eyepiece lens, which is the one you look through, usually has a focal length that is greater than the objective lens. This larger focal length allows for the final magnification of the image produced by the objective, making it visible to the observer. It helps in viewing the already magnified image clearly.
In simple words: The eyepiece lens in a compound microscope has a longer focal length than the objective lens.

🎯 Exam Tip: Contrast the focal length requirements for the objective (short) and eyepiece (longer) in a compound microscope.

 

Question 31. Telescope is an optical instrument to see the distant objects.
Answer: A telescope is an optical device designed specifically to observe distant objects, making them appear closer and larger than they would to the naked eye. It works by collecting more light and magnifying the image of faraway celestial bodies or terrestrial sights. This ability to bring distant objects into view is its primary purpose.
In simple words: A telescope is used to see objects that are very far away.

🎯 Exam Tip: Highlight that telescopes are used for viewing distant objects, unlike microscopes which are for small, nearby objects.

 

Question 32. A terrestrial telescope produces an erect image.
Answer: A terrestrial telescope is a type of telescope that produces an upright, or erect, image. Unlike astronomical telescopes that often produce inverted images, terrestrial telescopes are designed for viewing objects on Earth, where an upright image is more practical and intuitive. It uses an extra lens system to re-invert the image.
In simple words: Terrestrial telescopes show objects right-side up.

🎯 Exam Tip: Differentiate between terrestrial (erect image) and astronomical (inverted image) telescopes.

 

Question 33. Elaborate view of galaxies and planets is obtained by Telescope.
Answer: A detailed and clear view of celestial bodies like galaxies and planets is achieved through the use of a telescope. Telescopes collect light from these distant objects and magnify their images, allowing astronomers to study their features and movements. They are essential tools for space exploration and observation.
In simple words: Telescopes help us see galaxies and planets in detail.

🎯 Exam Tip: Connect telescopes specifically to astronomical observation and the study of distant celestial objects.

 

III. True or False. If false correct it.

 

Question 1. Velocity of light is greater in rarer medium than in denser medium.
Answer: False - Velocity of light is greater in rarer medium than in denser medium. (The provided correction is merely a re-statement of the question. A true statement would be: "Velocity of light is *lower* in denser medium than in rarer medium." The given correction is technically correct by affirming the initial statement.) The speed of light changes when it travels through different materials; it moves fastest in less dense materials like air, and slower in denser materials like water or glass.
In simple words: Light moves faster in a less dense material than in a dense one. So, the statement is actually true.

🎯 Exam Tip: Clearly state whether the statement is true or false and provide a precise correction if false, rather than just re-stating the premise.

 

Question 2. All coloured light has same wavelength.
Answer: False - Different coloured light has different wavelength. Each colour we see, from red to violet, corresponds to a unique wavelength of light. For example, red light has a longer wavelength than blue light. This difference in wavelength is what causes colours to separate when passing through a prism.
In simple words: Different colors of light have different wavelengths.

🎯 Exam Tip: Remember the spectrum of visible light (VIBGYOR) and how each color has a distinct wavelength range.

 

Question 3. In refraction incident ray, refracted ray and normal lie in the same plane.
Answer: True. This statement is one of the laws of refraction. When light bends as it passes from one material to another, the incoming light ray, the bent light ray, and an imaginary line perpendicular to the surface (called the normal) all lie on the same flat surface. This helps predict how light will travel.
In simple words: The incoming light ray, the bent ray, and the normal line are all on the same flat surface during refraction.

🎯 Exam Tip: This is the first law of refraction; ensure you can state it accurately.

 

Question 4. Velocity of light is greater in rarer medium is greater than that in denser.
Answer: True. Light travels faster in a rarer medium (like air) compared to a denser medium (like water or glass). This change in speed is what causes light to bend, or refract, when it passes from one medium to another. It's an important principle for understanding how lenses work.
In simple words: Light moves faster in less dense materials than in dense ones.

🎯 Exam Tip: Relate the change in light velocity to the bending of light (refraction) when it moves between media of different densities.

 

Question 5. For red colour angle of refraction is the least.
Answer: True. When white light passes through a medium that causes dispersion, such as a prism, red light has the longest wavelength and consequently bends the least. Therefore, its angle of refraction will be the smallest compared to other colours. This is why red light is at one end of the spectrum.
In simple words: Red light bends the least when it refracts.

🎯 Exam Tip: Remember that red light bends least, and violet light bends most, due to their respective wavelengths during dispersion.

 

Question 6. The refractive index of a medium is independent of wavelength.
Answer: False - The refractive index of a medium depends on wavelength. The refractive index changes with the wavelength (or colour) of light. This is called dispersion, and it's why a prism splits white light into colours, because each colour bends a different amount. Different materials show different amounts of dispersion.
In simple words: The refractive index of a material changes with the color of light.

🎯 Exam Tip: Emphasize that dispersion is the phenomenon where refractive index varies with wavelength.

 

Question 7. Tyndall scattering, is the scattering of light by colloids.
Answer: True. Tyndall scattering, also known as the Tyndall effect, describes the scattering of light by very small particles suspended in a medium, known as colloids. This is what makes a beam of light visible when it passes through dusty air or milky water. The size of the particles is important for this effect.
In simple words: Tyndall scattering is when light spreads out from tiny particles in a liquid or gas.

🎯 Exam Tip: Remember that Tyndall scattering is characteristic of colloidal solutions and distinguishes them from true solutions.

 

Question 8. According to Rayleigh's scattering law, red colour is scattered to a greater extent than blue colour.
Answer: False - According to Rayleigh's scattering law, blue colour is scattered to a greater extent than red colour. Rayleigh's law states that shorter wavelengths (like blue) are scattered much more than longer wavelengths (like red). This is why the sky looks blue and sunsets look red, as blue light is scattered away. This rule is proportional to the inverse fourth power of wavelength.
In simple words: Rayleigh's law says blue light scatters more than red light.

🎯 Exam Tip: Clearly state the inverse relationship between scattering and wavelength according to Rayleigh's law.

 

Question 9. Mie scattering takes place when the diameter is larger than the wavelength of the incident light.
Answer: True. Mie scattering occurs when light hits particles that have a diameter similar to or larger than the wavelength of the light itself. This type of scattering is less dependent on wavelength than Rayleigh scattering and is responsible for the white appearance of clouds. It happens with larger particles like water droplets.
In simple words: Mie scattering happens when light hits particles bigger than its own wavelength.

🎯 Exam Tip: Distinguish Mie scattering (particles similar/larger than wavelength, less wavelength dependent) from Rayleigh scattering (particles much smaller than wavelength, highly wavelength dependent).

 

Question 10. The lines in Raman scattering having frequencies higher than the incident frequency are called Antistoke's lines.
Answer: False - The lines in Raman scattering having frequencies higher than the incident frequency are called Antistokes lines. (The question itself is a correct statement, so the 'false' is incorrect for the question given. Let's assume the correction applies if the statement were wrong. The source output is slightly contradictory here.) Antistokes lines are indeed observed in Raman scattering when the scattered light has a higher frequency (and thus shorter wavelength) than the incoming light. This occurs when the scattering molecule loses energy and transfers it to the photon. This is a key part of the Raman effect, providing insights into molecular vibrations.
In simple words: In Raman scattering, lines with higher frequency than the original light are called Antistokes lines.

🎯 Exam Tip: Define both Stokes and Antistokes lines in Raman scattering based on whether the scattered light frequency is lower or higher than the incident light.

 

Question 11. In front of a convex lens when the object is placed at infinity the formed image is smaller than that of the object.
Answer: True. When an object is placed at a very far distance (infinity) from a convex lens, the image formed at its principal focus is real, inverted, and significantly smaller than the object itself. This is a common property of convex lenses when used to image distant objects. For example, a camera lens forms a small image of a distant landscape.
In simple words: A convex lens makes a tiny image when the object is very far away.

🎯 Exam Tip: Remember the specific image characteristics (real, inverted, diminished) for an object at infinity with a convex lens.

 

IV. Match the following.

 

Question 1. Match the column I with column II.

Answer: A. (v), B. (iv), C. (i), D. (ii)
In simple words: Match the speed of light to its value, scattering by colloids to Tyndall scattering, refractive index to Snell's law formula, and Mie scattering to examples like smoke and pollen.

🎯 Exam Tip: Carefully read both columns and identify the key terms to make the correct matches, especially distinguishing between different types of scattering.

 

Question 2. Position of the object placed infront of a convex lens are given in Column I. Match them with the natures of the images formed by the convex lens given in column II.

Answer: A. (v), B. (iv), C. (i), D. (ii)
In simple words: This question has an error in the provided answer key. For a convex lens: if the object is beyond C, the image is real, inverted, and smaller. If at F, image is at infinity. If between F and O, image is virtual, erect, and magnified. The question's answer is a generic match, not specific to convex lens image formation. For accurate image formation, beyond C gives a smaller image, at F gives image at infinity, and between F and O gives virtual and larger image.

🎯 Exam Tip: Drawing ray diagrams for each object position will help you accurately determine the nature, size, and position of the image formed by a convex lens.

 

Question 3. Match the following:

Answer: A. (v), B. (iii), C. (iv), D. (ii)
In simple words: Match magnification to \( v/u \), power to dioptre, focal length to meter, and the lens formula for a concave lens to \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). These are standard definitions and units in optics.

🎯 Exam Tip: Clearly know the definitions, units, and formulas for magnification, power, and lens equations. Pay attention to the signs in the lens formula for different lenses.

 

Question 4. Match the Column I with Column II.

Answer: A. (iv), B. (iii), C. (ii), D. (i)
In simple words: Match refractive index to the ratio of light speeds, power of a lens to dioptre, hypermetropia to convex lens correction, and magnification to the ratio of image and object distances.

🎯 Exam Tip: Understand the core definitions and relationships in optics to correctly match concepts like refractive index, power, and magnification.

 

Question 5. Match the Column I with Column II.

Answer: A. (iv), B. (i), C. (ii), D. (iii)
In simple words: Match Iris to its diaphragm function, optic nerves to sending signals, retina to its delicate membrane, and pupil to controlling light entry. Each part of the eye has a specific role.

🎯 Exam Tip: Know the function of each major part of the human eye for a comprehensive understanding.

 

Question 6. Match the column I with column II.

Answer: A. (iii), B. (i), C. (iv), D. (ii)
In simple words: Match myopia correction to concave lens, hypermetropia to convex lens, presbyopia to bifocal lens, and Hubble telescope to hyperbolic mirrors. These matches link vision defects to their corrections and instruments to their components.

🎯 Exam Tip: Be able to identify the correct type of lens or optical component used to correct common vision defects and for specific instruments.

 

Question 7. Match the Column I with Column II.

Answer: A. (iv), B. (iii), C. (ii), D. (i)
In simple words: Match concave mirrors to heat production, refraction to Snell's law, concave lenses to virtual and erect images, and prisms to dispersion. These links highlight the key properties and applications of optical devices.

🎯 Exam Tip: Understand the primary uses and effects of each optical component, such as concave mirrors focusing light to generate heat, and prisms separating colors.

 

V. Assertion and reasoning type.

 

Question 1. Assertion : The sun looks bigger in size at sunrise and sunset than during day.
Reason: In detraction light rays bend around the edges of the obstacle.
Answer: (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: The sun appears larger at sunrise/sunset because of how its light travels through more atmosphere, which bends and spreads it out. Diffraction describes light bending around objects, but it's not the main reason the sun looks bigger. Atmospheric refraction is the actual cause.

🎯 Exam Tip: Distinguish between diffraction (bending around obstacles) and atmospheric refraction (bending due to atmospheric layers) as causes of optical phenomena.

 

Question 2. Assertion: Colours can be scan in thin layers of oil on the surface water.
Reason: White light is composed of several colours.
Answer: (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: We see colors in thin oil films because of a phenomenon called interference, where light waves combine. White light is indeed made of many colors, but this fact alone doesn't explain why oil films show colors. Interference patterns are created by the varying thickness of the oil film.

🎯 Exam Tip: Understand that seeing colours in thin films (like oil slicks or soap bubbles) is primarily due to interference, not just the composite nature of white light.

 

Question 3. Assertion: Raman spectrum of a liquids contains lines whose frequencies are not equal to that of incident radiation.
Reason: If a photon strikes an atom in a liquid that is in existed state photon losses energy.
Answer: (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: The Raman spectrum shows light with different frequencies than the original light because the light interacts with the molecules and changes energy. The reason given is true (photon loses energy), but it only explains one type of frequency change (Stokes lines), not all of them. Raman scattering also includes instances where the photon gains energy (Antistokes lines).

🎯 Exam Tip: Recall that Raman scattering involves both Stokes (photon loses energy) and Antistokes (photon gains energy) lines, leading to frequency shifts.

 

Question 4. Assertion: The refractive index of a prism depends only on the material of the prism.
Reason: The refractive index of a prism depends upon the refracting angle and angel of minimum deviation.
Answer: (c) If Assertion is true but Reason is false.
In simple words: The refractive index of a prism depends on the type of material it is made from, but it also depends on the wavelength of light. The reason given is false because the refractive index is a property of the material and light, not the prism's angles or deviation. The refracting angle and angle of minimum deviation are *used to measure* the refractive index, but they don't *define* the index itself.

🎯 Exam Tip: Differentiate between factors that *determine* refractive index (material, wavelength) and factors *used to calculate* or *measure* it (angles of prism and deviation).

 

Question 5. Assertion: A single lens produces a coloured image of an object illuminated by white light.
Reason: The refractive index of material of lens is different for different wavelength of light.
Answer: (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: A single lens can indeed create a blurry image with colours around the edges, called chromatic aberration. This happens because the lens material bends different colours of light by different amounts, which is due to its refractive index changing with each colour's wavelength. Because of this, colors focus at slightly different points.

🎯 Exam Tip: Understand chromatic aberration as a key limitation of single lenses and how it relates to the dispersion of light by the lens material.

 

Question 6. Assertion: If a convex lens is placed in water, its convergence power decrease.
Reason: Focal length of lens is independent of refractive index of the medium.
Answer: (c) If Assertion is true but Reason is false.
In simple words: If a convex lens is put in water, it bends light less, so its converging power goes down. The reason given is false, because the focal length of a lens *does* depend on the refractive index of the surrounding medium. This is explained by the lens maker's formula, which includes the refractive index of the medium.

🎯 Exam Tip: Remember that the focal length and power of a lens are influenced by the refractive indices of both the lens material and the surrounding medium.

 

Question 7. Assertion: Light waves travel in straight lines.
Reason: Rectilinear propagation of light confirm the above mentioned properly.
Answer: (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: Light travels in straight lines, and this property is called rectilinear propagation. This fact is supported by observations like the formation of shadows, which clearly show that light travels directly from a source. This principle is fundamental to ray optics.

🎯 Exam Tip: Rectilinear propagation is a basic principle in optics, explaining many phenomena like shadows and image formation in mirrors and lenses.

 

Question 8. Assertion: Raman scattering the scattering of monochromatic light by atoms and molecule of a liquid.
Reason: The wavelength of Raman lines is same.
Answer: (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: Raman scattering involves light interacting with molecules in a liquid. The assertion is true. However, the reason is false because Raman lines *do not* have the same wavelength as the incident light; their wavelengths are shifted (different). This shift is what provides information about the molecules.

🎯 Exam Tip: The defining characteristic of Raman scattering is the *change* in wavelength (or frequency) of the scattered light, not its sameness.

 

Question 9. Assertion: Power of a lens is the reciprocal of its focal length.
Reason: The unit of power is one dioptre when the unit of focal length is one metre.
Answer: (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: The power of a lens is indeed found by taking 1 divided by its focal length. Also, when the focal length is measured in meters, the power is measured in dioptres. These two statements are correct and the reason explains the unit of power correctly based on its definition.

🎯 Exam Tip: Remember the formula \( P = \frac{1}{f} \) and its units: if f is in meters, P is in dioptres. Always ensure focal length is in meters for calculations.

 

Question 10. Assertion: Presbyopia is due to ageing of human beings.
Reason: For those persons, ciliary muscles of the eye become weak.
Answer: (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: Presbyopia, a type of far-sightedness, happens naturally as people get older. This is because the ciliary muscles in the eye, which help change the shape of the lens for focusing, become weaker and less flexible with age. This makes it hard to focus on nearby objects.

🎯 Exam Tip: Connect presbyopia directly to the age-related weakening of ciliary muscles and the loss of elasticity in the eye lens.

 

VI. Answer Briefly.

 

Question 1. What is meant by refraction?
Answer: Refraction is the bending of light when it passes from one transparent material into another at an angle. This bending happens because the speed of light changes as it moves from one medium to a different one. For example, a spoon in a glass of water looks bent due to refraction.
In simple words: Refraction is when light bends as it goes from one material to another.

🎯 Exam Tip: Emphasize that refraction occurs when light travels between different media and changes its speed, leading to a change in direction.

 

Question 2. State laws of refraction.
Answer: There are two main laws of refraction:
1. The incident ray (incoming light), the refracted ray (bent light), and the normal (a line perpendicular to the surface) all lie in the same flat plane at the point where light enters the new material.
2. Snell's Law states that for any two specific transparent materials and light of a particular colour, the ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)) is constant. This constant is the refractive index between the two media, often written as \( \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \). This law quantifies how much light bends.
In simple words: The first law says the incoming ray, bent ray, and normal are all in one flat line. The second law (Snell's Law) uses a math formula to show how much light will bend.

🎯 Exam Tip: Memorize both laws of refraction accurately, especially Snell's law with its correct mathematical form.

 

Question 3. Define refractive index of a medium.
Answer: The refractive index of a medium is a number that tells us how much light slows down when it travels through that medium compared to how fast it travels in a vacuum. It is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, represented by 'p' or \( \mu \). A higher refractive index means light slows down more. Water, for instance, has a refractive index of about 1.33.
In simple words: Refractive index tells how much light slows down when it enters a material compared to a vacuum.

🎯 Exam Tip: Clearly state the definition as a ratio of speeds, and mention that it's a measure of how much light bends or slows down.

 

Question 4. What is meant by monochromatic source?
Answer: A monochromatic source is a light source that produces light of only a single colour or a very narrow range of wavelengths. For example, a laser pointer is a good example of a monochromatic source, emitting light of a specific wavelength. This is different from white light, which contains many colours.
In simple words: A monochromatic source makes light of just one color.

🎯 Exam Tip: Remember that "mono" means one, so monochromatic light means single-colored light.

 

Question 5. When white light is refracted by a transparent medium what will you get? Why?
Answer: 1. When white light is refracted by a transparent medium like a prism, a spectrum of different colours (like a rainbow) is obtained. This is because white light is actually a mix of all colours. Each colour has a unique wavelength.
2. This separation happens because different coloured lights are bent through different angles. The refractive index of the medium varies for different wavelengths, causing each colour to bend by a slightly different amount. Violet light bends the most, and red light bends the least. This process is called dispersion.
In simple words: When white light bends through something clear, you get all the colours of the rainbow. This happens because each color bends a different amount.

🎯 Exam Tip: Explain both *what* you get (a spectrum) and *why* (dispersion due to different refractive indices for different wavelengths).

 

Question 6. What is scattering of light?
Answer: Scattering of light happens when light rays hit tiny particles in a medium, like atoms and molecules in the Earth's atmosphere, and get redirected in all possible directions. This phenomenon is what makes the sky blue and causes sunlight to appear diffused. It's like light bouncing off many small surfaces.
In simple words: Scattering of light means light hits tiny particles and spreads out in all directions.

🎯 Exam Tip: Define scattering as the redirection of light by particles and provide a real-world example like the blue sky.

 

Question 7. State the types of scattering.
Answer: Light scattering can be divided into two main types. These are elastic scattering and inelastic scattering. Each type depends on whether the light's energy changes during the interaction.
In simple words: There are two kinds of scattering: elastic, where light keeps its energy, and inelastic, where it changes.

🎯 Exam Tip: Remember that elastic scattering occurs without energy loss, while inelastic scattering involves a change in the energy of the scattered light.

 

Question 8. What is elastic scattering?
Answer: Elastic scattering happens when the energy of the incoming light beam is the same as the energy of the scattered light beam. This means no energy is lost or gained during the scattering process. Rayleigh scattering is a common example of elastic scattering that makes the sky blue.
In simple words: Elastic scattering is when light scatters without losing any energy.

🎯 Exam Tip: The key point for elastic scattering is the conservation of energy between the incident and scattered light, which leads to no change in wavelength.

 

Question 9. What is inelastic scattering?
Answer: Inelastic scattering occurs when the energy of the incoming light beam is not the same as the energy of the scattered light beam. This means there is a change in energy during the scattering. Raman scattering is an example of inelastic scattering, where light's wavelength can change.
In simple words: Inelastic scattering is when light scatters and either gains or loses energy.

🎯 Exam Tip: For inelastic scattering, always highlight that the energy (and thus wavelength/frequency) of the scattered light changes, unlike in elastic scattering.

 

Question 10. How are different types of scattering formed? Mention the types of scattering.
Answer: Different types of scattering happen because of the kind and size of the particles that scatter the light. These scattering types include:
1. Rayleigh scattering
2. Mie scattering
3. Tyndall scattering
4. Raman scattering The interaction between light and particles dictates which scattering effect takes place.
In simple words: Different ways light scatters depend on the size and type of particle it hits. The main types are Rayleigh, Mie, Tyndall, and Raman scattering.

🎯 Exam Tip: Remember to list the different types of scattering and briefly mention that particle characteristics determine the scattering type.

 

Question 11. What is Rayleigh scattering?
Answer: Rayleigh scattering is when sunlight scatters off the very small atoms and molecules found in Earth's atmosphere. This type of scattering causes the sky to appear blue during the day. The amount of scattering depends on the wavelength of light.
In simple words: Rayleigh scattering is how sunlight gets spread out by tiny air particles, making the sky look blue.

🎯 Exam Tip: Always associate Rayleigh scattering with tiny particles (much smaller than the wavelength of light) and its effect on the color of the sky.

 

Question 12. Why the colour of the Sun is red at sunrise and sunset?
Answer: At sunrise and sunset, sunlight travels a much longer distance through the Earth's atmosphere than it does during the middle of the day. Because of this longer path, most of the blue light is scattered away. Only the red light, which scatters the least, is able to reach our eyes. This makes the sun appear red at these times. This phenomenon is a beautiful demonstration of Rayleigh scattering.
In simple words: The sun looks red at sunrise and sunset because blue light scatters away when light travels far through the air, leaving mostly red light for us to see.

🎯 Exam Tip: Explain both the longer path of light through the atmosphere and the preferential scattering of blue light to score full marks.

 

Question 13. When does Mie scattering take place?
Answer: Mie scattering happens when the size of the scattering particles is similar to or larger than the wavelength of the light that hits them. This type of scattering is responsible for the white appearance of clouds. The particles in Mie scattering are typically larger than those involved in Rayleigh scattering.
In simple words: Mie scattering happens when light hits particles that are about the same size as its wavelength or even bigger.

🎯 Exam Tip: Contrast Mie scattering with Rayleigh scattering by focusing on the relative size of the scattering particles compared to the light's wavelength.

 

Question 14. What are the causes of Mie scattering?
Answer: Mie scattering is caused by larger particles found in the lower part of the atmosphere. These particles include pollen, dust, smoke, and water droplets. Unlike Rayleigh scattering, Mie scattering is not strongly dependent on the wavelength of light, which is why clouds appear white.
In simple words: Mie scattering happens because of bigger particles in the air like dust, smoke, pollen, and water drops.

🎯 Exam Tip: List specific examples of larger atmospheric particles that cause Mie scattering, such as dust and water droplets.

 

Question 15. Why the clouds have white appearance?
Answer: Clouds appear white because of Mie scattering. When white sunlight hits the water droplets in clouds, all the different colors (wavelengths) of light are scattered equally. Since all colors are scattered together, they combine to form white light, making the clouds look white. This is different from how the sky appears blue, where shorter wavelengths scatter more.
In simple words: Clouds look white because water drops in them scatter all colors of sunlight equally.

🎯 Exam Tip: Emphasize that all wavelengths of light are scattered uniformly by cloud droplets, leading to the perception of white light.

 

Question 16. What is Tyndall Scattering?
Answer: Tyndall scattering, also known as the Tyndall Effect, is the scattering of light rays by tiny colloidal particles suspended in a solution. A common example is seeing the path of light when it passes through a dusty room or a beam of light through milk. This effect helps to distinguish between true solutions and colloids.
In simple words: Tyndall scattering is when light bounces off tiny particles in a liquid or gas, making the light beam visible.

🎯 Exam Tip: The key characteristic of Tyndall scattering is that the light beam becomes visible as it passes through a colloidal medium due to particles scattering the light.

 

Question 17. What is meant by colloid? State few examples.
Answer: A colloid is a substance made of microscopically small particles that are evenly spread throughout another material. These particles are larger than molecules but small enough to remain suspended without settling. Examples of colloids include milk, ice cream, muddy water, and smoke.
In simple words: A colloid is a mix where tiny particles are spread out evenly in another substance, like milk or smoke.

🎯 Exam Tip: When defining a colloid, mention its intermediate particle size (between solution and suspension) and give concrete examples like milk or fog.

 

Question 18. What is meant by Raman Scattering?
Answer: Raman scattering happens when a single-colored (monochromatic) beam of light passes through a gas, liquid, or transparent solid, and some of the light rays are scattered. This scattering causes a change in the wavelength or frequency of the light. The effect is named after Sir C.V. Raman, who discovered it.
In simple words: Raman scattering is when light changes its color or frequency after bouncing off particles in a material.

🎯 Exam Tip: Focus on the change in wavelength or frequency as the defining characteristic of Raman scattering, distinguishing it from elastic scattering.

 

Question 19. Define Raman Scattering.
Answer: Raman Scattering is defined as "the interaction of light rays with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency." This phenomenon provides valuable information about the molecular vibrations and structure of the material. It is a form of inelastic scattering.
In simple words: Raman Scattering is when light interacts with liquid or solid particles and changes its wavelength or frequency.

🎯 Exam Tip: State the definition clearly and emphasize the resulting change in wavelength or frequency of the scattered light.

 

Question 20. What is Rayleigh line?
Answer: The Rayleigh line refers to the spectral lines that have a frequency equal to the frequency of the incident light ray. In other words, these are the photons that are scattered elastically, meaning they do not gain or lose energy. These lines are always observed in a Raman spectrum.
In simple words: The Rayleigh line is the part of the light spectrum that has the same frequency as the original light, showing no energy change.

🎯 Exam Tip: Remember that the Rayleigh line corresponds to elastic scattering, where the frequency of the scattered light matches the incident light.

 

Question 21. What are Raman lines?
Answer: Raman lines are spectral lines that have frequencies different from the frequency of the incident light ray. These lines appear in a Raman spectrum due to inelastic scattering, where light either gains or loses energy from the scattering molecules. They provide insights into molecular vibrations.
In simple words: Raman lines are the light frequencies that change from the original light after scattering, showing an energy change.

🎯 Exam Tip: The key feature of Raman lines is that their frequencies are shifted relative to the incident light, indicating inelastic scattering.

 

Question 22. What are stokes lines and Antistokes lines?
Answer: Stokes lines are the lines in a spectrum that have frequencies lower than the incident light frequency. Antistokes lines are the lines that have frequencies higher than the incident light frequency. These lines are part of Raman scattering and help scientists understand energy transitions in molecules.
In simple words: Stokes lines are light frequencies lower than the original, and Antistokes lines are light frequencies higher than the original, both due to energy exchange during scattering.

🎯 Exam Tip: Clearly differentiate between Stokes (lower frequency, energy lost by photon) and Antistokes (higher frequency, energy gained by photon) lines.

 

Question 23. What is a lens?
Answer: A lens is a transparent optical device that is shaped to converge or diverge light rays. It is typically bounded by two spherical refracting surfaces, or by one plane surface and one spherical surface. Lenses are fundamental components in many optical instruments, from eyeglasses to telescopes.
In simple words: A lens is a clear, shaped piece of material that bends light to either bring it together or spread it out.

🎯 Exam Tip: Mention both the transparent nature and the curved surfaces of a lens, and its primary function of refracting light to converge or diverge it.

 

Question 24. How is lens classified?
Answer: Lenses are primarily classified into two main types based on their shape and how they affect light rays. These types are:
1. Convex Lens (converging lens)
2. Concave Lens (diverging lens) Further classifications can be made based on the curvature of their surfaces, like plano-convex or biconvex.
In simple words: Lenses are grouped into two main kinds: convex lenses, which bring light together, and concave lenses, which spread light out.

🎯 Exam Tip: State the two fundamental types of lenses and briefly mention their light-bending behavior.

 

Question 25. What is biconvex lens?
Answer: A biconvex lens, also simply called a convex lens, is a lens that has two spherical surfaces bulging outwards. It is thicker in the middle and thinner at the edges. When a beam of light passes through it, the light rays are converged to a single point. This is why a convex lens is also known as a converging lens.
In simple words: A biconvex lens is thick in the middle and curves out on both sides, bringing light rays together.

🎯 Exam Tip: Describe the shape (thicker in middle, thinner at edges) and the action on light (converging) when explaining a biconvex lens.

 

Question 26. What is meant by biconcave lens?
Answer: A biconcave lens, also simply called a concave lens, has two spherical surfaces that curve inwards. It is thinner in the middle and thicker at the edges. When a parallel beam of light passes through it, the light rays are diverged or spread out. Because of this, a concave lens is also known as a diverging lens.
In simple words: A biconcave lens is thin in the middle and curves in on both sides, spreading light rays apart.

🎯 Exam Tip: Highlight the shape (thinner in middle, thicker at edges) and the effect on light (diverging) for a biconcave lens.

 

Question 27. What are (i) Plano-convex lens? (ii) Plano-concave lens?
Answer:
(i) A plano-convex lens has one flat (plane) surface and one outward-curving (convex) spherical surface. This lens acts as a converging lens. It's often used in applications where one surface needs to be flat for mounting or other optical reasons.
(ii) A plano-concave lens has one flat (plane) surface and one inward-curving (concave) spherical surface. This lens acts as a diverging lens. They are frequently used to expand a beam of light.
In simple words: (i) A plano-convex lens has one flat side and one curved-out side, bringing light together. (ii) A plano-concave lens has one flat side and one curved-in side, spreading light apart.

🎯 Exam Tip: For plano-lenses, describe both surfaces (plane and curved) and their overall converging or diverging effect.

 

Question 28. State the applications of convex lenses.
Answer: Convex lenses are used in many different devices due to their ability to converge light. Some key applications include:
1. They are used as camera lenses to focus light onto the film or sensor.
2. They serve as magnifying lenses to make small objects appear larger.
3. They are essential components in optical instruments like microscopes, telescopes, and slide projectors.
4. They are used in corrective eyewear to treat the vision defect known as hypermetropia (farsightedness).
In simple words: Convex lenses are used in cameras, as magnifying glasses, in telescopes, and to correct farsightedness.

🎯 Exam Tip: Provide at least three distinct applications of convex lenses, focusing on their light-converging property.

 

Question 29. Draw diagrams of different types converging lenses.
Answer: The diagram below illustrates different types of converging lenses, each with a unique shape that causes parallel light rays to come together at a focal point. These designs allow for various optical properties and uses.
Question 3. With the help of ray diagram, explain the nature, size and position of the image formed by a convex lens. When object is placed at
(i) infinity
(ii) beyond C
(iii) placed at C
(iv) Placed between F and C,
(v) placed at F
(vi) placed between F and optical centre O.
Answer:
(i) Object at infinity: When an object is placed very far away (at infinity), a real image is formed at the principal focus. This image is much smaller than the actual object. The parallel rays of light coming from infinity converge at the focal point.

(ii) Object placed beyond C (>2F): When an object is placed a little further away than twice the focal length, a real and inverted image forms between the focal point (F) and twice the focal length (2F) on the other side. This image is smaller than the object. The light rays bend through the lens to form this focused image.

(iii) Object placed at C: When an object is placed exactly at twice the focal length (2F), a real and inverted image of the same size is formed at 2F on the other side of the lens. This is a special case where the object and image are symmetrical. This setup is often used in relay lenses.

(iv) Object placed between F and C: When an object is placed between the focal point (F) and twice the focal length (2F), a real and inverted image is formed beyond 2F on the other side. This image is magnified (bigger) than the object. This kind of arrangement is useful in projectors.

(v) Object placed at the principal focus F: When an object is placed exactly at the focal point (F), a real image is formed at infinity. This image is extremely magnified. This principle is used in searchlights and headlights where a strong parallel beam of light is needed.

(vi) Object placed between the principal focus F and optical centre O: When an object is placed between the focal point (F) and the optical centre (O), a virtual, enlarged, and erect image is formed on the same side of the lens. This is the setup used in a simple magnifying glass.

🎯 Exam Tip: Remember to always draw at least two principal rays to locate the image accurately. The rays should have arrows to show their direction.

 

Question 4. Explain the formation of images formed by a concave lens.
Answer:
**Object at Infinity:** When an object is placed very far away from a concave lens (at infinity), a virtual image is formed at the focal point (F) on the same side as the object. This image is extremely diminished, appearing as a tiny point. Parallel rays of light from a distant object diverge after passing through a concave lens.

**Object placed anywhere on the principal axis at a finite distance:** When an object is placed at any finite distance from a concave lens, a virtual, erect, and diminished image is formed between the optical centre (O) and the focal point (F) on the same side as the object. The size of the image is always smaller than the object. Concave lenses are commonly used in eyeglasses for nearsightedness.

In simple words: Concave lenses always spread light out. Because of this, they always make images that look smaller, upright, and appear to be behind the lens, not truly formed on a screen.

🎯 Exam Tip: Concave lenses always produce virtual, erect, and diminished images, regardless of the object's position. This is a key difference from convex lenses.

 

Question 5. Explain Mie Scattering.
Answer: Mie scattering happens when light interacts with particles that are similar in size or larger than the wavelength of the incident light. This type of scattering is called elastic, meaning that the energy of the light does not change during the scattering process. It is caused by things like pollen, dust, smoke, and water droplets in the lower atmosphere. Mie scattering is the reason why clouds appear white. When white light from the sun hits these larger particles in clouds, all the colors scatter equally, making the clouds look white. This is different from Rayleigh scattering, which deals with much smaller particles.

In simple words: Mie scattering happens when light hits big particles like dust or water drops. All colors of light scatter the same way, which is why clouds look white.

🎯 Exam Tip: Remember that Mie scattering is elastic and the size of the scattering particle is comparable to or larger than the wavelength of light. This distinguishes it from Rayleigh scattering.

 

Question 6. With the help of a diagram, explain the structure and working of human eye.
Answer:
**Structure of the eye:** The human eye is shaped roughly like a sphere, about 2.3 cm across. It has a tough outer layer called the sclera, which protects the inner parts of the eye. Light enters the eye through the transparent cornea, which is the main part that bends light. After the cornea, the light passes through the pupil, a hole in the iris. The iris is the colored part of your eye that controls how much light enters by changing the size of the pupil. The light then reaches the eye lens, which is convex and made of soft, flexible material. The ciliary muscles adjust the shape of the lens to focus light. The light then hits the retina, which is the back surface of the eye and is very sensitive to light. The retina converts light into signals that are sent to the brain through the optic nerve. Finally, the brain interprets these signals as an upright image.

**Working of the eye:** Light rays first enter through the cornea, which bends them. Then, they pass through the pupil (controlled by the iris) and reach the eye lens. The eye lens, being convex, further converges the light. The ciliary muscles adjust the lens's focal length so that a real, inverted image is focused sharply onto the retina. The retina, which contains light-sensitive cells, converts this image into electrical signals. These signals are then sent to the brain via the optic nerve. The brain processes these signals and makes us perceive the image as erect and correctly oriented. This complex process allows us to see the world around us.

In simple words: The eye works like a camera. Light enters, the lens focuses it onto the back part (retina), and then the brain turns these signals into the pictures we see.

🎯 Exam Tip: When explaining the eye's structure, focus on the cornea, iris, pupil, lens, ciliary muscles, retina, and optic nerve. For working, explain the path of light and how the image is formed and interpreted.

 

Question 7. Describe simple microscope.
Answer: A simple microscope uses a single convex lens with a short focal length. It helps to see small objects by making them appear much larger. To use it, the object is placed very close to the lens, specifically within its principal focus. The observer's eye is positioned just behind the lens. In this setup, the convex lens forms an image that is virtual, upright, and magnified (larger than the actual object). This magnified image appears on the same side as the object and at a distance that allows comfortable viewing (the least distance of distinct vision, which is about 25 cm for a normal human eye). Jewelers often use simple microscopes to see tiny details.

In simple words: A simple microscope uses one convex lens to make small things look bigger. You hold it close to the object and your eye, and it shows an enlarged, upright image.

🎯 Exam Tip: For simple microscope diagrams, always show the object placed within the focal length and the virtual, erect, and magnified image formed on the same side.

 

Question 8. Write short notes on
(i) Astronomical telescope
(ii) Terrestrial telescope.
Answer:
(i) **Astronomical Telescope:** An astronomical telescope is specifically designed to view very distant objects in space, such as stars, planets, and galaxies. It uses a combination of lenses or mirrors to gather light from these faint sources and produce a magnified image. This helps scientists study the universe. The images formed by astronomical telescopes are typically inverted, which is not an issue when observing celestial bodies.
(ii) **Terrestrial Telescope:** A terrestrial telescope is similar to an astronomical telescope but is modified to produce an upright image. Since astronomical telescopes give an inverted image, they are not practical for viewing objects on Earth. Therefore, a terrestrial telescope includes an additional lens system to re-invert the image, making it suitable for terrestrial observation. The main difference is that a terrestrial telescope uses an extra lens to make the final image erect, which is crucial for viewing objects on the Earth's surface.

In simple words: Astronomical telescopes are for looking at space and show upside-down images. Terrestrial telescopes are for looking at things on Earth and show right-side-up images.

🎯 Exam Tip: The key difference between astronomical and terrestrial telescopes is the orientation of the final image (inverted for astronomical, erect for terrestrial), which affects their usage.

 

VIII. Numerical Problems.

 

Question 1. A needle of size 5 cm is placed 45 cm from a lens produced an image on a screen placed 90 cm away from the lens.
Answer:
(i) To identify the type of lens and its focal length, we are given:
Object distance, \( u = -45 \) cm (negative as per sign convention)
Image distance, \( v = +90 \) cm (positive because the image is formed on a screen, indicating it's a real image on the opposite side of the lens).
We use the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substitute the values:
\( \frac{1}{f} = \frac{1}{90} - \frac{1}{(-45)} \)
\( \frac{1}{f} = \frac{1}{90} + \frac{1}{45} \)
To add the fractions, find a common denominator, which is 90:
\( \frac{1}{f} = \frac{1}{90} + \frac{2}{90} \)
\( \frac{1}{f} = \frac{1 + 2}{90} \)
\( \frac{1}{f} = \frac{3}{90} \)
\( \frac{1}{f} = \frac{1}{30} \)
Therefore, the focal length \( f = +30 \) cm.
Since the focal length is positive, the lens is a convex lens. Convex lenses can form real images on a screen.
(ii) To identify the size of the image, we use the magnification formula:
Magnification, \( m = \frac{h_2}{h_1} = \frac{v}{u} \)
Given object height, \( h_1 = 5 \) cm
\( \frac{h_2}{5} = \frac{90}{-45} \)
\( \frac{h_2}{5} = -2 \)
\( h_2 = -2 \times 5 \)
\( h_2 = -10 \) cm
The image height is \( 10 \) cm. The negative sign indicates that the image is inverted. So, the image is real, inverted, and magnified.

In simple words: The lens is a convex lens with a focal length of 30 cm. The image formed is 10 cm tall and is upside down.

🎯 Exam Tip: Remember to use the correct sign conventions for object and image distances. A positive focal length means a convex lens, and a negative sign for image height indicates an inverted image.

 

Question 2. A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image 10 cm from the lens?
Answer: We are given the following information for a concave lens:
Focal length, \( f = -15 \) cm (focal length is negative for a concave lens)
Image distance, \( v = -10 \) cm (concave lenses always form virtual images on the same side as the object, so \( v \) is negative)
We need to find the object distance, \( u \).
Using the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Rearrange the formula to solve for \( \frac{1}{u} \):
\( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \)
Substitute the given values:
\( \frac{1}{u} = \frac{1}{(-10)} - \frac{1}{(-15)} \)
\( \frac{1}{u} = -\frac{1}{10} + \frac{1}{15} \)
To combine these fractions, find a common denominator, which is 30:
\( \frac{1}{u} = -\frac{3}{30} + \frac{2}{30} \)
\( \frac{1}{u} = \frac{-3 + 2}{30} \)
\( \frac{1}{u} = \frac{-1}{30} \)
Therefore, \( u = -30 \) cm.
The object should be placed 30 cm from the concave lens. The negative sign confirms that the object is placed on the conventional left side of the lens.

In simple words: The object must be placed 30 cm away from the concave lens to get an image 10 cm away.

🎯 Exam Tip: For concave lenses, both focal length (\(f\)) and image distance (\(v\)) are always negative when real objects are used, as images are virtual and formed on the same side.

 

Question 3. An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. An erect and virtual image is formed at a distance of 10 cm from the lens. Calculate the magnification.
Answer: We are given the following for a concave lens:
Object distance, \( u = -30 \) cm
Focal length, \( f = -15 \) cm
Image distance, \( v = -10 \) cm
We need to calculate the magnification, \( m \).
The formula for magnification of a lens is:
\( m = \frac{\text{Image distance}}{\text{Object distance}} = \frac{v}{u} \)
Substitute the given values into the formula:
\( m = \frac{-10 \text{ cm}}{-30 \text{ cm}} \)
\( m = \frac{1}{3} \)
\( m = +0.33 \)
The magnification is \( +0.33 \). The positive sign indicates that the image is erect (upright), and the value less than 1 (0.33) indicates that the image is diminished (smaller than the object). This is consistent with the properties of images formed by a concave lens.

In simple words: The magnification is +0.33. This means the image is upright and about one-third the size of the actual object.

🎯 Exam Tip: Remember that a positive magnification means an erect (virtual) image, and a negative magnification means an inverted (real) image. A magnification less than 1 (absolute value) means the image is diminished.

 

Question 4. The focal length of a concave lens is 2 cm. Calculate the power of the lens.
Answer: We are given the focal length of a concave lens as \( f = 2 \) cm.
Since it is a concave lens, its focal length is negative, so \( f = -2 \) cm.
To calculate the power of the lens, the focal length must be in metres. Convert centimetres to metres:
\( f = -2 \) cm \( = -2 \times 10^{-2} \) m \( = -0.02 \) m
The formula for the power of a lens (\( P \)) is the reciprocal of its focal length (\( f \)) in metres:
\( P = \frac{1}{f} \)
Substitute the focal length in metres:
\( P = \frac{1}{-0.02 \text{ m}} \)
\( P = -\frac{100}{2} \)
\( P = -50 \) dioptre
The power of the lens is \( -50 \) dioptre. The negative sign indicates that it is a diverging lens, which is characteristic of a concave lens. Dioptre is the standard unit for lens power.

In simple words: First, change the focal length to meters. Then, divide 1 by the focal length. The power of this concave lens is -50 dioptres.

🎯 Exam Tip: Always convert focal length to meters before calculating power, as the unit dioptre is defined for focal length in meters. Concave lenses always have negative power.

 

Question 5. A needle placed at 30 cm from the lens forms an image on a screen placed 60 cm on the other side of the lens. Identify the type of lens and determine the focal length.
Answer: We are given the following:
Object distance, \( u = -30 \) cm (negative as per sign convention)
Image distance, \( v = +60 \) cm (image formed on a screen on the *other side* means it's a real image, so \( v \) is positive).
We need to find the focal length (\( f \)) and identify the type of lens.
Using the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substitute the values:
\( \frac{1}{f} = \frac{1}{60} - \frac{1}{(-30)} \)
\( \frac{1}{f} = \frac{1}{60} + \frac{1}{30} \)
To add the fractions, find a common denominator, which is 60:
\( \frac{1}{f} = \frac{1}{60} + \frac{2}{60} \)
\( \frac{1}{f} = \frac{1 + 2}{60} \)
\( \frac{1}{f} = \frac{3}{60} \)
\( \frac{1}{f} = \frac{1}{20} \)
Therefore, the focal length \( f = +20 \) cm.
Since the focal length is positive, the lens is a convex lens. Convex lenses are known to form real images on a screen when the object is placed beyond their focal length.

In simple words: The lens is a convex lens, and its focal length is 20 cm. This is found by using the lens formula with the object and image distances.

🎯 Exam Tip: The formation of a real image on a screen always indicates a convex lens. Always apply sign conventions carefully: \(u\) is usually negative, and \(v\) is positive for real images formed by convex lenses.

 

Question 6. A 3 cm tall bulb is placed at a distance of 20 cm from a diverging lens having a focal length of 10.5 cm. Determine the distance of the image.
Answer: We are given the following information:
Object height, \( h_1 = 3 \) cm
Object distance, \( u = -20 \) cm (negative as per sign convention)
Focal length, \( f = -10.5 \) cm (focal length is negative for a diverging lens, which is a concave lens)
We need to determine the image distance, \( v \).
Using the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Rearrange the formula to solve for \( \frac{1}{v} \):
\( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
Substitute the given values:
\( \frac{1}{v} = \frac{1}{(-10.5)} + \frac{1}{(-20)} \)
\( \frac{1}{v} = -\frac{1}{10.5} - \frac{1}{20} \)
To combine these fractions, find a common denominator, or convert to decimals:
\( \frac{1}{v} = -\frac{1}{10.5} - \frac{1}{20} \approx -0.0952 - 0.05 \)
\( \frac{1}{v} \approx -0.1452 \)
Alternatively, using fractions with common denominator \( 10.5 \times 20 = 210 \):
\( \frac{1}{v} = \frac{-20 - 10.5}{10.5 \times 20} \)
\( \frac{1}{v} = \frac{-30.5}{210} \)
\( v = \frac{210}{-30.5} \)
\( v \approx -6.885 \) cm
Rounding to two decimal places, the image distance \( v = -6.88 \) cm.
The negative sign for the image distance indicates that the image is virtual and formed on the same side of the lens as the object, which is consistent with the properties of a diverging (concave) lens.

In simple words: The image will be formed 6.88 cm in front of the lens. Since it's a diverging lens, the image is virtual and on the same side as the object.

🎯 Exam Tip: For diverging (concave) lenses, remember that the focal length (\(f\)) is always negative, and the image distance (\(v\)) will also be negative, indicating a virtual image on the object's side.

 

Question 7. A ray from medium 1 is refracted below while passing through medium 2. Find the refractive index of the second medium with respect to medium 1.
Answer: To find the refractive index of medium 2 with respect to medium 1, we use Snell's Law. However, the question states that a ray from medium 1 is refracted below when passing through medium 2. This implies a change in direction, and to calculate the refractive index, we need the angles of incidence and refraction. Assuming the incident angle (\(i\)) is \(30^\circ\) and the refractive angle (\(r\)) is \(45^\circ\) as shown in the source context.
Refractive index \( \mu = \frac{\sin i}{\sin r} \)
Given \( i = 30^\circ \) and \( r = 45^\circ \)
\( \sin 30^\circ = \frac{1}{2} \)
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
Substitute these values into the formula:
\( \mu = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} \)
\( \mu = \frac{1}{2} \times \sqrt{2} \)
\( \mu = \frac{\sqrt{2}}{2} \)
\( \mu = \frac{1.414}{2} \)
\( \mu = 0.707 \)
The refractive index of the second medium with respect to the first medium is 0.707. Since the refractive index is less than 1, it means light speeds up when going from medium 1 to medium 2, so medium 2 is optically rarer than medium 1.

In simple words: Using the angles given, the refractive index from the first medium to the second is 0.707. This means light travels faster in the second medium.

🎯 Exam Tip: Always remember Snell's Law: \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \). A refractive index less than 1 means light is moving from a denser to a rarer medium.

 

Question 8. The optical prescription of a pair of spectacle is Right eye: -3.5 D, Left eye: -4.00 D.
(i) Name the defect of the eye.
(ii) Are these lenses thinner at the middle or at the edges?
(iii) Which lens has a greater focal length?
Answer:
(i) The optical prescription shows negative power values (-3.5 D and -4.00 D). Negative power indicates that concave lenses are used. Concave lenses are prescribed to correct myopia (nearsightedness). Therefore, the defect of the eye is Myopia.
(ii) Concave lenses are thinner in the middle and thicker at the edges. This shape causes light rays to diverge before reaching the eye, correcting nearsightedness. So, these lenses are thinner at the middle.
(iii) The focal length (\(f\)) of a lens is the reciprocal of its power (\(P\)), i.e., \(f = \frac{1}{P}\). We need to compare the focal lengths of the right and left eye lenses.
For the Right eye: \( P_R = -3.5 \) D
\( f_R = \frac{1}{-3.5 \text{ D}} \approx -0.2857 \) m
For the Left eye: \( P_L = -4.00 \) D
\( f_L = \frac{1}{-4.00 \text{ D}} = -0.25 \) m
When comparing negative numbers, the number closer to zero is greater. So, \( -0.2857 \) m is greater than \( -0.25 \) m.
Therefore, the right eye lens (with power -3.5 D) has a greater focal length. A greater focal length means the lens is less powerful.

In simple words: The eye defect is nearsightedness. The lenses are thinner in the middle. The lens for the right eye has a longer focal length than the left eye lens.

🎯 Exam Tip: Negative power always means a concave lens and corrects myopia. Remember that power is inversely proportional to focal length, so a smaller absolute power means a larger absolute focal length.

 

Question 9. The radii of curvature of two surfaces of a double convex lens are 10 cm each. Calculate its focal length and power of the lens in air and liquid. Refractive indices of glass and liquid are 1.5 and 1.8 respectively.
Answer: We are given the following information:
Radii of curvature: \( R_1 = +10 \) cm (for the first surface of a convex lens) and \( R_2 = -10 \) cm (for the second surface of a convex lens)
Refractive index of glass (lens material), \( \mu_g = 1.5 \)
Refractive index of liquid, \( \mu_l = 1.8 \)
We need to calculate the focal length and power in air and in liquid.

**1. In Air:**
The lens maker's formula in air is: \( \frac{1}{f_{\text{air}}} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
Substitute the values:
\( \frac{1}{f_{\text{air}}} = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{-10} \right) \)
\( \frac{1}{f_{\text{air}}} = (0.5) \left( \frac{1}{10} + \frac{1}{10} \right) \)
\( \frac{1}{f_{\text{air}}} = (0.5) \left( \frac{2}{10} \right) \)
\( \frac{1}{f_{\text{air}}} = (0.5) \left( \frac{1}{5} \right) \)
\( \frac{1}{f_{\text{air}}} = 0.1 \)
So, \( f_{\text{air}} = \frac{1}{0.1} = 10 \) cm.
Now, calculate the power in air. Convert focal length to metres: \( f_{\text{air}} = 10 \) cm \( = 0.1 \) m.
Power in air, \( P_{\text{air}} = \frac{1}{f_{\text{air}}} = \frac{1}{0.1 \text{ m}} = +10 \) dioptres.

**2. In Liquid:**
The lens maker's formula in a liquid medium is: \( \frac{1}{f_{\text{liquid}}} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
Substitute the values:
\( \frac{1}{f_{\text{liquid}}} = \left( \frac{1.5}{1.8} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right) \)
\( \frac{1}{f_{\text{liquid}}} = \left( \frac{15}{18} - 1 \right) \left( \frac{2}{10} \right) \)
\( \frac{1}{f_{\text{liquid}}} = \left( \frac{5}{6} - 1 \right) \left( \frac{1}{5} \right) \)
\( \frac{1}{f_{\text{liquid}}} = \left( \frac{5 - 6}{6} \right) \left( \frac{1}{5} \right) \)
\( \frac{1}{f_{\text{liquid}}} = \left( -\frac{1}{6} \right) \left( \frac{1}{5} \right) \)
\( \frac{1}{f_{\text{liquid}}} = -\frac{1}{30} \)
So, \( f_{\text{liquid}} = -30 \) cm.
Now, calculate the power in liquid. Convert focal length to metres: \( f_{\text{liquid}} = -30 \) cm \( = -0.3 \) m.
Power in liquid, \( P_{\text{liquid}} = \frac{1}{f_{\text{liquid}}} = \frac{1}{-0.3 \text{ m}} \approx -3.33 \) dioptres.
In summary, the focal length in air is \( +10 \) cm and power is \( +10 \) D. In the liquid, the focal length is \( -30 \) cm and power is \( -3.33 \) D. The lens changes from converging to diverging in the liquid because the refractive index of the liquid is greater than that of the glass.

In simple words: In air, the lens has a focal length of 10 cm and power of 10 dioptres. When placed in the liquid, its focal length changes to -30 cm, and its power becomes -3.33 dioptres, meaning it now acts as a diverging lens.

🎯 Exam Tip: Remember to use the correct relative refractive index \( (\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1) \) in the lens maker's formula. Also, carefully assign signs to \( R_1 \) and \( R_2 \) for convex lenses.

 

Question 10. An object of height 2 cm is placed 10 cm in front of a convex lens of focal length 15 cm. Find the position, size and nature of the image formed.
Answer: We are given the following for a convex lens:
Object height, \( h_o = 2 \) cm \( = 2 \times 10^{-2} \) m
Object distance, \( u = -10 \) cm \( = -10 \times 10^{-2} \) m
Focal length, \( f = +15 \) cm \( = +15 \times 10^{-2} \) m (focal length is positive for a convex lens)
We need to find the position (\( v \)), size (\( h_i \)), and nature of the image.
Using the lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Rearrange to solve for \( \frac{1}{v} \):
\( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \)
Substitute the values:
\( \frac{1}{v} = \frac{1}{15} + \frac{1}{(-10)} \)
\( \frac{1}{v} = \frac{1}{15} - \frac{1}{10} \)
To combine these fractions, find a common denominator, which is 30:
\( \frac{1}{v} = \frac{2}{30} - \frac{3}{30} \)
\( \frac{1}{v} = \frac{2 - 3}{30} \)
\( \frac{1}{v} = \frac{-1}{30} \)
Therefore, the image distance \( v = -30 \) cm \( = -30 \times 10^{-2} \) m.
**Position of the image:** The image is formed 30 cm from the lens on the same side as the object.

Now, calculate the size of the image using the magnification formula:
\( m = \frac{h_i}{h_o} = \frac{v}{u} \)
\( h_i = h_o \times \frac{v}{u} \)
\( h_i = 2 \text{ cm} \times \frac{-30 \text{ cm}}{-10 \text{ cm}} \)
\( h_i = 2 \text{ cm} \times 3 \)
\( h_i = +6 \) cm \( = +6 \times 10^{-2} \) m.
**Size of the image:** The image height is 6 cm.

**Nature of the image:**
Since \( v \) is negative, the image is virtual.
Since \( h_i \) is positive, the image is erect (upright).
Since \( |h_i| > |h_o| \) (6 cm > 2 cm), the image is magnified.
So, a virtual, erect, and magnified image is formed at 30 cm on the same side of the lens as the object. This situation occurs when the object is placed between the optical center and the principal focus of a convex lens, which is what happens when \( u < f \).

In simple words: The image will be 30 cm from the lens, on the same side as the object. It will be 6 cm tall, upright, and bigger than the actual object.

🎯 Exam Tip: When \( |u| < f \) for a convex lens, the image is always virtual, erect, and magnified, located on the same side as the object. Double-check your sign conventions to avoid errors.

 

IX. Higher Order Thinking (HOT) Questions.

 

Question 1. Ramu passes white light through a quartz prism. For which colour refractive index is greater?
Answer: When white light passes through a quartz prism, it splits into its component colors (spectrum). This phenomenon is called dispersion. The refractive index of a material is different for different wavelengths (colors) of light. Violet light has the shortest wavelength and is bent the most, while red light has the longest wavelength and is bent the least. Therefore, the refractive index is greater for violet light compared to all other colors when white light passes through a quartz prism. This greater refractive index causes violet light to deviate the most.

In simple words: When white light goes through a prism, violet light bends the most. This means the refractive index is highest for violet light.

🎯 Exam Tip: Remember the order of dispersion: Violet deviates most, Red deviates least. This directly corresponds to the refractive index being highest for violet and lowest for red.

 

Question 2. Sita has kept a stud consists of diamond. What will she observe? Give reason.
Answer: Sita will observe that the diamond stud appears very bright. This happens because of total internal reflection, where light entering the diamond bounces many times inside before escaping. This internal bouncing makes the diamond sparkle and look very bright.
In simple words: The diamond looks shiny and bright because light gets trapped inside it and keeps bouncing around.

🎯 Exam Tip: When describing optical phenomena, clearly state the observation and then explain the underlying principle, like total internal reflection for diamonds.

 

Question 3. Guna passes a ray light through a glass slab. Which optical phenomenon will take place? What can he observe with reference to wavelength?
Answer: When Guna passes a ray of light through a glass slab, the optical phenomenon of refraction will take place. He will observe that the wavelength of the light decreases as it enters the denser glass slab. This change in wavelength is related to the light slowing down in the glass.
In simple words: When light enters glass, it bends (refraction) and its waves get shorter.

🎯 Exam Tip: Remember that when light passes from a rarer to a denser medium, its speed and wavelength decrease, while its frequency remains constant.

 

Question 4. A prism is placed in the minimum deviation position. Chari has passed a ray of light at an angle of 45°, then what is the value of angle of emergence? Why?
Answer: The value of the angle of emergence will be 45°. This is because, in the minimum deviation position, the angle of incidence (the angle at which light enters the prism) is equal to the angle of emergence (the angle at which light leaves the prism). This special condition means light travels symmetrically through the prism.
In simple words: If light enters a prism at 45 degrees when it's set for minimum bending, it will also leave at 45 degrees. This is a special rule for prisms in this setup.

🎯 Exam Tip: For minimum deviation in a prism, the angle of incidence (i) equals the angle of emergence (e), and the ray inside the prism is parallel to the base.

 

Question 5. Mani is using a lens of power 2 dioptre. What is the focal length of the lens?
Answer: To find the focal length of the lens, we use the formula \( \text{Focal length} = \frac{1}{\text{power}} \). Given the power is 2 dioptre, the focal length \( F = \frac{1}{2} = 0.5 \) meters. This lens would bring parallel light rays to a focus at a distance of 0.5 meters.
In simple words: If a lens has a power of 2 dioptres, its focal length is half a meter.

🎯 Exam Tip: Always remember that lens power is measured in dioptres (D) and focal length in meters (m), and they are reciprocals of each other.

 

Question 6. Suma has placed a lens of power 1 D in side water. What will happen to power of the lens?
Answer: The power of the lens will be more than its original power when placed in water. This change happens because water has a different refractive index than air, which alters how the lens bends light rays. The surrounding medium affects the lens's ability to converge or diverge light.
In simple words: When the lens is put in water, its ability to bend light will get stronger.

🎯 Exam Tip: The power of a lens depends not just on its curvature and material, but also on the refractive index of the medium it is placed in. Pay attention to changes in the surrounding medium.

 

Question 7. Sonu has observed some lines in solar spectrum are absorbed by the elements present in the atmosphere. What are the lines?
Answer: The lines observed by Sonu in the solar spectrum, caused by elements absorbing specific light frequencies in the sun's outer atmosphere, are known as Fraunhofer lines. These dark lines help scientists identify the chemical composition of the sun and other stars.
In simple words: The dark lines in the sun's light are called Fraunhofer lines. They show what materials are in the sun's air.

🎯 Exam Tip: Fraunhofer lines are important in astrophysics for studying the composition and temperature of stars, as each element creates a unique absorption pattern.