GSEB Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

 

Question 1. Define the principal focus of a concave mirror.
Answer: It is a specific point on the main axis where light rays, initially traveling parallel to that axis, gather or meet after reflecting from the concave mirror.
In simple words: For a concave mirror, the principal focus is the spot on the central line where all parallel light rays come together after bouncing off the mirror.

Exam Tip: Remember to mention both "principal axis" and "parallel rays" for a complete definition of principal focus for a concave mirror.

 

Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer: The radius of curvature \( R = 20 \) cm. The relationship between the radius of curvature \( R \) and the focal length \( f \) is \( R = 2f \).
Therefore, the focal length \( f = \frac{R}{2} \).
So, \( f = \frac{20}{2} = 10 \) cm.
The focal length of the mirror is 10 cm.
In simple words: The focal length is half of the radius of curvature. If the mirror's curve radius is 20 cm, its focal length is 10 cm.

Exam Tip: Always state the formula \(f = R/2\) and include units in your final answer to score full marks.

 

Question 3. Which mirror can give an erect and enlarged image of an object?
Answer: A concave mirror produces an erect and enlarged image when the object is placed between its pole and principal focus.
In simple words: A concave mirror can make an object look bigger and upright if you place the object very close to the mirror, between its center and the focal point.

Exam Tip: Remember that a concave mirror is the only spherical mirror that can form both real/inverted and virtual/erect images, depending on the object's position.

 

Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer: A convex mirror can cover a wider range and always gives an erect and diminished image. Hence, a convex mirror is used as a rear-view mirror to offer a wider field of view.
In simple words: We choose convex mirrors for rear-view because they show a large area behind the vehicle and the images are always upright, though smaller, making driving safer.

Exam Tip: Key points for convex mirrors as rear-view mirrors are "wider field of view," "erect image," and "diminished image."

 

Question 5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer: Radius of curvature \( R = 32 \) cm.
The relationship between radius of curvature \( R \) and focal length \( f \) is \( R = 2f \).
So, \( f = \frac{R}{2} \).
Therefore, \( f = \frac{32}{2} = 16 \) cm.
The focal length of the convex mirror is 16 cm.
In simple words: The focal length is always half of the radius of curvature. So, for a convex mirror with a 32 cm radius, its focal length will be 16 cm.

Exam Tip: Always use the formula \(f = R/2\) for both concave and convex mirrors, but remember that for convex mirrors, the focal length is taken as positive.

 

Question 6. An object is placed at 10 cm in front of a concave mirror. Where is the image located, given that the magnification is -3?
Answer: Object distance \( u = -10 \) cm (for a concave mirror, object is in front).
Magnification \( m = -3 \).
The magnification formula for a mirror is \( m = \frac{-v}{u} \).
Substituting the values: \( -3 = \frac{-v}{-10} \).
\( -3 = \frac{v}{10} \).
So, \( v = -3 \times 10 = -30 \) cm.
The image is formed 30 cm in front of the concave mirror. The negative sign for \( v \) indicates that the image is formed on the same side as the object, which is a real and inverted image.
In simple words: If an object is 10 cm from a concave mirror and the image is three times bigger and upside down (magnification -3), the image forms 30 cm in front of the mirror, on the same side as the object.

Exam Tip: Pay close attention to the sign conventions for \(u\), \(v\), and \(m\) when solving numerical problems for mirrors. A negative \(m\) indicates a real and inverted image.

 

Question 7. A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer: The light bends towards the normal upon entry into the water because water is optically denser than air. Light slows down as it moves from a rarer medium (air) to a denser medium (water), causing it to bend towards the normal line.
In simple words: When light goes from air into water at an angle, it bends closer to the normal line. This happens because water is thicker for light than air, causing the light to slow down and change its path.

Exam Tip: Remember the rule: light bends towards the normal when going from a rarer to a denser medium, and away from the normal when going from a denser to a rarer medium.

 

Question 8. Light enters from air to glass having a refractive index 1.50. What is the speed of light in glass if the speed of light in a vacuum is \(3 \times 10^8\) m/s?
Answer: The speed of light in a vacuum \( c = 3 \times 10^8 \) m/s.
Refractive index of glass \( n_g = 1.50 \).
The formula for refractive index is \( n = \frac{\text{speed of light in vacuum}}{\text{speed of light in medium}} \).
So, \( n_g = \frac{c}{v_g} \), where \( v_g \) is the speed of light in glass.
Rearranging the formula to find \( v_g \): \( v_g = \frac{c}{n_g} \).
Substituting the given values: \( v_g = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \) m/s.
The speed of light in glass is \( 2 \times 10^8 \) m/s.
In simple words: To find how fast light travels in glass, divide the speed of light in empty space by the glass's refractive index. If light moves at \(3 \times 10^8\) m/s in vacuum and glass has a refractive index of 1.5, then light will move at \(2 \times 10^8\) m/s in glass.

Exam Tip: The speed of light in any medium is always less than or equal to the speed of light in a vacuum. Always use the formula \(v = c/n\).

 

Question 9. Find out from the following table the medium having highest optical density. Also find the medium with lowest optical density.

Answer: Diamond has the highest optical density with a refractive index of 2.42. Air has the lowest optical density with a refractive index of 1.0003.
In simple words: From the table, diamond is the "thickest" for light with the biggest refractive index (2.42), and air is the "thinnest" with the smallest refractive index (1.0003).

Exam Tip: A higher refractive index means higher optical density and slower speed of light in that medium, while a lower refractive index means lower optical density and faster speed of light.

 

Question 10. You are given kerosene, turpentine, and water. In which of these does the light travel fastest? Use the information given in the table above.
Answer: Refractive index of kerosene = 1.44.
Refractive index of turpentine = 1.47.
Refractive index of water = 1.33.
Light travels faster in the medium with a lower refractive index. Comparing the values, water has the lowest refractive index (1.33). Hence, light will travel fastest in water.
In simple words: Light travels quickest in the material with the lowest refractive index. Out of kerosene (1.44), turpentine (1.47), and water (1.33), light will move fastest through water.

Exam Tip: Remember that a lower refractive index corresponds to a faster speed of light in that medium. This is a crucial concept for understanding refraction.

 

Question 11. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer: The refractive index being 2.42 for diamond means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is 2.42. This indicates that light travels 2.42 times slower in diamond compared to its speed in air or a vacuum. It also suggests that diamond is an optically very dense medium.
In simple words: If a diamond's refractive index is 2.42, it means light slows down by 2.42 times when it passes from air into the diamond. This makes diamond an optically dense material.

Exam Tip: When explaining refractive index, always mention it as a ratio of speeds of light in two media, and clarify the effect on the speed of light within the material.

 

Question 12. Define 1 dioptre of power of a lens.
Answer: One dioptre (1D) is the power of a lens whose focal length is 1 meter. It is the reciprocal of the focal length when the focal length is expressed in meters. So, \( 1D = 1m^{-1} \).
In simple words: A lens has a power of 1 dioptre if its focal length is exactly 1 meter.

Exam Tip: Always remember that focal length must be in meters for power to be in dioptres. State the unit correctly.

 

Question 13. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer: For a convex lens, if a real and inverted image is formed at the same size as the object, the object must be placed at \( 2F \) (twice the focal length) on one side, and the image will also form at \( 2F \) on the other side.
Given, image distance \( v = +50 \) cm (real image is formed on the opposite side of a convex lens, so positive).
Since the image size is equal to the object size, the object distance \( u \) must be \( -50 \) cm (at \( 2F \)).
Using the lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
Substituting the values: \( \frac{1}{50} - \frac{1}{(-50)} = \frac{1}{f} \).
\( \frac{1}{50} + \frac{1}{50} = \frac{1}{f} \).
\( \frac{2}{50} = \frac{1}{f} \).
\( \frac{1}{25} = \frac{1}{f} \).
So, \( f = 25 \) cm.
To find the power \( P \), convert focal length to meters: \( f = 25 \) cm \( = 0.25 \) m.
Power \( P = \frac{1}{f} = \frac{1}{0.25} = +4.0 \) D.
The needle is placed at 50 cm in front of the convex lens, and the power of the lens is \( +4.0 \) D.
In simple words: When a convex lens creates a real, upside-down image that's the same size as the object and 50 cm away, the object was also 50 cm in front of the lens. This means the lens has a focal length of 25 cm, and its power is +4.0 dioptres.

Exam Tip: For convex lenses, an image of the same size, real, and inverted is formed only when the object is at \(2F\). Always convert focal length to meters for power calculations.

 

Question 14. Find the power of a concave lens of focal length 2 m.
Answer: Focal length of a concave lens \( f = -2 \) m (focal length is negative for concave lenses).
The power of a lens \( P = \frac{1}{f} \).
Substituting the value: \( P = \frac{1}{-2} = -0.5 \) D.
The power of the concave lens is \( -0.5 \) D.
In simple words: To find the power of a lens, take the inverse of its focal length in meters. For a concave lens with a 2-meter focal length, its power is -0.5 dioptres.

Exam Tip: Always remember that concave lenses have negative focal lengths and thus negative power, indicating their diverging nature.

 

In-Text Activities Solved

 

Activity 10.1
Answer:

• Take a large shining spoon. Try to view your face on its curved surface.
• The image obtained is larger.
• Move the spoon slowly away from your face.
• The image becomes inverted.
• Reverse the spoon and repeat the activity.
• The image is diminished and virtual.
• The image formed is smaller and erect.

In simple words: When you look into a shiny spoon, the inside (concave) makes your face look bigger. If you move it back, the image turns upside down. The outside (convex) always makes your face look smaller and upright.

Exam Tip: This activity highlights the different image formations by concave (enlarged, inverted) and convex (diminished, erect) mirrors, depending on the object's position.

 

Activity 10.2
Answer:

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror onto a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find a bright, sharp spot of light on the paper sheet.
• Hold the mirror and the paper in the same position for a few minutes.

Observation: The paper initially turns blackish, then burns to produce smoke. Eventually, it catches fire.

Conclusion: The light from the sun is converged at a specific point on the paper. This point is the focus of the concave mirror.

In simple words: If you point a concave mirror at the sun, it gathers sunlight into a bright, hot spot on a piece of paper. This spot is the mirror's focus, showing how it concentrates light and energy.

Exam Tip: This experiment demonstrates how concave mirrors converge parallel rays of light (from the sun) to their principal focus, which can generate enough heat to burn paper.

 

Activity 10.3
Answer:

• Take a concave mirror. Find out its approximate focal length. This might be around 10 cm.
• Mark a line on the table with chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with chalk two more lines parallel to the previous line such that the distance between two successive lines is equal to the focal length of the mirror.
• These lines will correspond to the positions of the points \( F \) and \( C \) respectively.
• Keep a bright object, like a burning candle, at a position far beyond \( C \). Place a paper screen and move it in front of the mirror until you obtain a sharp, bright image of the candle flame on it.
• Observe the image carefully. Note down its natural position and relative size with respect to the object size.
• Repeat the activity by placing the candle:

(a) just beyond \( C \)
(b) at \( C \)
(c) between \( F \) and \( C \)
(d) at \( F \)
(e) between \( P \) and \( F \)
• In one of the cases, the image is not obtained on the screen; specifically, when the object is between \( P \) and \( F \).

Observation:

Observation Table

In simple words: Use a concave mirror and a candle. By moving the candle to different spots (like very far, at 'C', at 'F', or between 'P' and 'F'), observe where the image forms on a screen and how big or small it is. Sometimes, you won't get an image on the screen at all.

Exam Tip: This activity systematically explores image formation by a concave mirror. Understanding the object positions relative to F and C is key to predicting image characteristics.

 

image formation by a concave mirror:

In simple words: This table summarizes where images form, their size, and whether they're real/inverted or virtual/erect for a concave mirror, depending on where the object is placed relative to the mirror's focus and center of curvature.

Exam Tip: Memorizing this table is critical for concave mirror problems. Pay attention to the critical points: infinity, C, F, and between P and F.

 

Activity 10.4
Answer:(a)
infinity
(b)
(c)
(d)
(e)
(f)
In simple words: These diagrams illustrate how a concave mirror forms images for different object positions. They show how light rays bounce off the mirror to create either real, inverted images (when the object is further away) or virtual, erect, and enlarged images (when the object is very close).

Exam Tip: Understanding and accurately drawing these ray diagrams is essential. Practice the three principal rays: parallel to PA, through F, and through C to locate the image for each case.

 

Activity 10.5
Answer:

• Take a convex mirror. Hold it in one hand.
• Hold a pencil in the upright position in the other hand.
• Observe the image of the pencil in the mirror.

Observation: The image formed is smaller and erect.

infinity
In simple words: When you look at a pencil in a convex mirror, the image you see is always smaller than the actual pencil and stands upright.

Exam Tip: Convex mirrors always form virtual, erect, and diminished images, regardless of the object's position, and this gives them a wide field of view.

 

Activity 10.6
Answer:

• Observe the image of a distant object, such as a distant tree, in a plane mirror.

Observation: The full length of the distant object is not seen in a small plane mirror.

• Now use a concave mirror to see the full-length image of the object.

Observation: The full-length image is not obtained in a concave mirror.

• Now use a convex mirror to see the same.

Observation: In this case, a full-length image of an object is obtained in a small mirror.

In simple words: This activity shows how different mirrors display a distant object like a tree. Plane and concave mirrors cannot show the whole tree in a small mirror, but a convex mirror can, even though the image is smaller.

Exam Tip: This activity demonstrates the field of view for different mirrors: convex mirrors have the widest field of view, while plane and concave mirrors have narrower views, making them less suitable for viewing large objects entirely.

 

Activity 10.7
Answer:

• Place a coin at the bottom of a bucket filled with water.
• With your eye to the side above the water, try to pick up the coin in one go.

Observation: One cannot pick up the coin in one attempt because of the refraction of light; the coin does not appear to be at its original position.

In simple words: When you try to grab a coin at the bottom of a water-filled bucket, you'll miss it on your first attempt. This is because light bends when it goes from water to air, making the coin look like it's in a different spot than it actually is.

Exam Tip: This activity clearly illustrates the phenomenon of refraction, where light bends as it passes from one medium to another, causing objects to appear at a different depth or position.

 

Activity 10.8
Answer:

• Place a large shallow bowl on a table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Keep looking for the coin from your position.

Observation: The coin becomes visible again and appears slightly raised above its actual position upon pouring water into the bowl. This occurs due to the refraction of light.

In simple words: This activity shows that if you put a coin in a bowl and then add water, the coin, which was previously hidden, reappears. It also looks higher than it actually is, all because light bends when it moves from water to air.

Exam Tip: This experiment vividly demonstrates the principle of apparent depth due to refraction. Always explain that the coin appears raised because light rays bend away from the normal when passing from water (denser) to air (rarer) before reaching the eye.

 

Activity 10.9
Answer:

• Draw a thick straight line in ink, over a sheet of white paper placed on a table.
• Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
• Look at the portion of the line under the slab from the sides. The line appears to be bent at the edges.
• Now, place the glass slab such that it is normal to the line. The part of the line under the glass slab appears to be bent.
• Look at the line from the top of the glass slab. The line appears to be raised. It is due to the refraction of light.

In simple words: If you draw a straight line and put a glass slab over it, especially at an angle, the line underneath will look bent or shifted when viewed from the side or top. This visual trick happens because light bends as it enters and leaves the glass.

Exam Tip: This activity demonstrates lateral displacement and apparent shift caused by refraction through a glass slab. The key concept is that light changes direction when entering and exiting the glass, making the line appear discontinuous.

 

Activity 10.10
Answer:

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of the tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of the tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O'.
• Join O and O'. Also, produce EF up to I as shown by a dotted line.

In simple words: This activity uses pins and a glass slab to trace how a light ray bends as it enters and leaves the glass. By connecting the pin marks, you can see the incident, refracted, and emergent rays, along with the amount of lateral shift.

Exam Tip: This experiment demonstrates the laws of refraction and lateral displacement. Ensure all pin positions are accurately marked, and the normals are drawn perpendicular to the glass surface for correct angle measurement.

 

Activity 10.11
Answer:

• Hold a convex lens in your hand. Direct it towards the Sun. (Do not look at the Sun through the lens, or else it may cause permanent damage to your eye).
• Focus the light from the Sun on a sheet of paper. Obtain a sharp, bright image of the sun.

Observation: The paper begins to burn to produce smoke, and it may catch fire after a while.

Condition: The light rays from the sun get converged when they pass through the lens, and a bright spot is formed on the paper, which is the focus of the Lens.

In simple words: This experiment shows that a convex lens can gather sunlight into a very hot, bright spot (its focus) on a piece of paper, proving its ability to converge light rays.

Exam Tip: This activity highlights the converging nature of a convex lens, where parallel rays (from the sun) meet at the principal focus, producing a real and highly diminished image.

 

Activity 10.12
Answer:

• Take a convex lens. Find its approximate focal length, similar to activity 10.11.
• Draw five parallel straight lines, using chalk, on a long table, such that the distance between successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Position it on the central line so that the optic center of the lens lies just over the line.
• Mark the two lines on either side of the lens corresponding to \( F \) and \( 2F \) of the lens respectively. Label them with appropriate letters such as \( 2F_1 \), \( F_1 \), \( F_2 \), and \( 2F_2 \).
• Place a burning candle, far beyond \( 2F_1 \) to the left. Obtain a clear, sharp image on a screen on the opposite side of the lens.
• Note down the nature, position, and relative size of the image.
• Repeat this activity by placing an object just behind \( 2F_1 \), at \( 2F_1 \), between \( F_1 \) at \( 2F_1 \), between \( F_1 \) and \( O \). Note down and tabulate the observations.

In simple words: This activity involves using a convex lens, lines on a table, and a candle to observe how images change. By placing the candle at various distances from the lens, you can see where the image forms, how big or small it is, and whether it's real or virtual.

Exam Tip: This is a comprehensive experiment for convex lenses. Systematically varying the object position and recording image characteristics is vital for understanding lens behavior.

 

Observations for a convex lens:

In simple words: This table summarizes how a convex lens forms images. Depending on where the object is placed (from infinity to very close), the image can be real and inverted (and vary in size) or virtual, erect, and enlarged.

Exam Tip: This table is fundamental for understanding convex lens behavior. Pay close attention to the image characteristics at key positions like \(2F\), \(F\), and between \(F\) and \(O\).

 

Activity 10.13
Answer:

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image.
• Try to get the image on a screen, if possible, or else observe the image directly through the lens.
• Note down the nature, relative size, and approximate position of the image.
• Move the candle away from the lens.
• Note the change in the size of the image. Record your observations by placing the candle at a position too far away from the lens.

Observations for a concave lens:

In simple words: This activity involves observing images formed by a concave lens using a candle. Regardless of how far the candle is, the concave lens always produces a smaller, upright, and virtual image, always positioned between its focus and the optical center.

Exam Tip: Concave lenses always form virtual, erect, and diminished images, irrespective of the object's position. The image is always located between the optical center and the principal focus on the same side as the object.

 

Gujarat Board Class 10 Science Light Reflection and Refraction Textbook Questions and Answers

 

Question 1. Which one of the following materials cannot be used to make a lens?
(a) Water
(C) Plastic
(d) Clay
Answer: (d) Clay
In simple words: Clay cannot be used to make a lens because lenses need to be transparent to let light pass through, and clay is opaque.

Exam Tip: Lenses require transparent materials that allow light to pass through and refract predictably. Opaque materials like clay cannot serve this purpose.

 

Question 2. The image formed by a concave mirror is observed to be virtual, erect, and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the center of curvature
(b) At the center of curvature
(c) Beyond the center of curvature
(d) Between the pole of the mirror and its principal focus
Answer: (d) Between the pole of the mirror and its principal focus.
In simple words: For a concave mirror to create a virtual, upright, and magnified image, the object must be placed very close to the mirror, specifically between its pole and focal point.

Exam Tip: This is a crucial case for concave mirrors. Remember that it's the only scenario where a concave mirror forms a virtual and erect image.

 

Question 3. Where should an object be placed in front of a convex lens to get a real image of the same size as the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical center of the lens and its principal focus
Answer: (b) At twice the focal length.
In simple words: To get a real image that is the same size as the object with a convex lens, you need to place the object exactly at twice the focal length from the lens.

Exam Tip: For both convex lenses and concave mirrors, an object placed at \(2F\) (or C for mirrors) results in a real, inverted image of the same size at \(2F\) (or C).

 

Question 4. A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave
(b) both convex
(c) the mirror is concave and the lens in convex.
(d) the mirror is convex but the lens is concave.
Answer: (a) Both concave.
In simple words: If both a spherical mirror and a spherical lens have a focal length of -15 cm, it means they are both concave because negative focal lengths are characteristic of concave mirrors and concave lenses.

Exam Tip: Negative focal length implies a converging mirror (concave) or a diverging lens (concave). Positive focal length implies a diverging mirror (convex) or a converging lens (convex).

 

Question 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane
(b) concave.
(c) convex
(d) either plane or convex.
Answer: (d) either plane or convex.
In simple words: If your reflection is always upright, no matter how far you are from the mirror, then the mirror must be either a flat (plane) mirror or a curved-outward (convex) mirror.

Exam Tip: Plane mirrors always form erect images. Convex mirrors always form erect and diminished images. Concave mirrors form erect images only when the object is between P and F.

 

Question 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer: (c) A convex lens of focal length 5 cm.
In simple words: To read small print, you need a magnifying glass, which is a convex lens. A shorter focal length means greater magnification, so a 5 cm convex lens would be best.

Exam Tip: For magnification, a convex lens is used as a simple microscope. A shorter focal length means higher power and greater magnification, which is ideal for reading small letters.

 

Question 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer: The focal length of the concave mirror = 15 cm.
To obtain an erect image using a concave mirror, the object should be placed in front of the given concave mirror at a distance of less than 15 cm. This means the object is between the pole (P) and the principal focus (F) of the mirror (i.e., \( 0 < u < 15 \) cm).
The image formed in this case is virtual and erect.
The image formed is larger than the object.

Ray diagram:

In simple words: To get an upright and larger image from a concave mirror (15 cm focal length), the object needs to be placed within 15 cm of the mirror. This will create a virtual, magnified image behind the mirror. The ray diagram shows the object between P and F, with rays appearing to come from an enlarged virtual image.

Exam Tip: This is a critical scenario for concave mirrors, often tested. Remember the object must be between P and F to form a virtual, erect, and magnified image behind the mirror.

 

Question 8. Name the type of mirror used in the following situations.
(a) The headlight of a car.
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) For car headlights, a concave mirror is used to get a powerful beam of light after reflection.
(b) A convex mirror is used for side/rearview mirrors in vehicles. This mirror forms an erect and diminished image of vehicles and gives a wider view of the rear.
(c) In a solar furnace, a concave mirror is used as a reflector. It concentrates sunlight at a point where the temperature sharply increases to 180°C - 200°C.
In simple words: Headlights and solar furnaces use concave mirrors to focus light or create strong beams. Rearview mirrors use convex mirrors to provide a wide, clear view of the traffic behind.

Exam Tip: Remember specific applications for each mirror type. Concave mirrors are for focusing light (like headlights) or magnifying (shaving mirrors), while convex mirrors are for wide fields of view and diminished, erect images (rearview mirrors).

 

Question 9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally? Explain your observations.
Answer: Yes, even if one-half of a convex lens is covered with black paper, the lens will still produce a full image of an object. The image formed on the screen does not depend on the size of the lens; however, the brightness of the image decreases because fewer light rays pass through the lens.
In simple words: Even with half the lens covered, you will still see the whole picture. It just won't be as bright because less light gets through.

Exam Tip: Understand that each part of a lens can form a complete image, but the intensity (brightness) is reduced if a portion is blocked. This is a key concept in optics.

 

Question 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image.
Answer: We are given: Object height \( h_0 = 5 \) cm, Object distance \( u = -25 \) cm (for converging lens, object on left), Focal length \( f = 10 \) cm (for converging lens, positive).
Using the lens formula: \( \frac { 1 }{ v } - \frac { 1 }{ u } = \frac { 1 }{ f } \)
Substitute the known values:
\( \frac { 1 }{ v } - \frac { 1 }{ (-25) } = \frac { 1 }{ 10 } \)
\( \frac { 1 }{ v } + \frac { 1 }{ 25 } = \frac { 1 }{ 10 } \)
\( \frac { 1 }{ v } = \frac { 1 }{ 10 } - \frac { 1 }{ 25 } \)
To subtract these fractions, find a common denominator, which is 50:
\( \frac { 1 }{ v } = \frac { 5 }{ 50 } - \frac { 2 }{ 50 } \)
\( \frac { 1 }{ v } = \frac { 3 }{ 50 } \)
\( v = \frac { 50 }{ 3 } \)
\( v \approx 16.67 \) cm
The positive value of \( v \) means the image is formed on the other side of the lens.
Now, find the height of the image (\( h_i \)) using the magnification formula:
\( m = \frac { h_i }{ h_0 } = \frac { v }{ u } \)
\( \frac { h_i }{ 5 } = \frac { 16.67 }{ -25 } \)
\( h_i = 5 \times \frac { 16.67 }{ -25 } \)
\( h_i = \frac { 16.67 }{ -5 } \)
\( h_i \approx -3.33 \) cm
The negative value of \( h_i \) indicates that the image is inverted. The magnitude (3.33 cm) is less than the object's height (5 cm), so it is a reduced image.
Therefore, the image is formed at 16.67 cm from the lens on the other side. Its size is 3.33 cm, and it is real and inverted.
In simple words: When the object is placed 25 cm away from the lens, the image appears 16.67 cm on the other side. The image is upside down and smaller, about 3.33 cm tall.

Exam Tip: Always remember the sign conventions for object distance (u), image distance (v), and focal length (f) for both lenses and mirrors. For a converging (convex) lens, focal length \( f \) is positive, and for a diverging (concave) lens, \( f \) is negative.

 

Question 11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer: We are given: Concave lens. Focal length \( f = -15 \) cm (for concave lens, negative). Image distance \( v = -10 \) cm (for concave lens, image is always virtual and formed on the same side as the object, so \( v \) is negative).
Using the lens formula: \( \frac { 1 }{ v } - \frac { 1 }{ u } = \frac { 1 }{ f } \)
Substitute the known values:
\( \frac { 1 }{ -10 } - \frac { 1 }{ u } = \frac { 1 }{ -15 } \)
\( - \frac { 1 }{ 10 } - \frac { 1 }{ u } = - \frac { 1 }{ 15 } \)
Rearrange to solve for \( u \):
\( - \frac { 1 }{ u } = - \frac { 1 }{ 15 } + \frac { 1 }{ 10 } \)
Find a common denominator, which is 30:
\( - \frac { 1 }{ u } = - \frac { 2 }{ 30 } + \frac { 3 }{ 30 } \)
\( - \frac { 1 }{ u } = \frac { 1 }{ 30 } \)
\( u = -30 \) cm
The object is placed 30 cm in front of the concave lens.
In simple words: For a concave lens, if the image forms 10 cm away, the object must be 30 cm in front of the lens.

Exam Tip: Always remember that for a concave lens, the image is always virtual, erect, and diminished, and it is formed on the same side as the object.

 

Question 12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer: We are given: Convex mirror. Object distance \( u = -10 \) cm (object always on the left, so negative). Focal length \( f = +15 \) cm (for convex mirror, focal length is positive).
Using the mirror formula: \( \frac { 1 }{ v } + \frac { 1 }{ u } = \frac { 1 }{ f } \)
Substitute the known values:
\( \frac { 1 }{ v } + \frac { 1 }{ (-10) } = \frac { 1 }{ 15 } \)
\( \frac { 1 }{ v } - \frac { 1 }{ 10 } = \frac { 1 }{ 15 } \)
Rearrange to solve for \( v \):
\( \frac { 1 }{ v } = \frac { 1 }{ 15 } + \frac { 1 }{ 10 } \)
Find a common denominator, which is 30:
\( \frac { 1 }{ v } = \frac { 2 }{ 30 } + \frac { 3 }{ 30 } \)
\( \frac { 1 }{ v } = \frac { 5 }{ 30 } \)
\( \frac { 1 }{ v } = \frac { 1 }{ 6 } \)
\( v = +6 \) cm
The image is formed 6 cm behind the mirror (positive \( v \)), and it is virtual and erect.
In simple words: When an object is 10 cm from a convex mirror with a 15 cm focal length, the image forms 6 cm behind the mirror and appears virtual and upright.

Exam Tip: Remember that convex mirrors always form virtual, erect, and diminished images, regardless of the object's position. The image is always located behind the mirror.

 

Question 13. The magnification produced by a plane mirror is +1. What does this mean?
Answer: Magnification, \( m = +1 \). The positive sign (+) means the image is virtual and erect. The magnitude '1' indicates that the image size is equal to the object size.
In simple words: A magnification of +1 from a plane mirror means the image you see is upright, not real, and exactly the same size as the actual object.

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

 

Question 14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.
Answer: We are given: Object height \( h_1 = 5.0 \) cm. Object distance \( u = -20 \) cm. Radius of curvature \( R = 30 \) cm. For a convex mirror, the focal length \( f \) is positive and is half the radius of curvature:
\( f = \frac { R }{ 2 } = \frac { 30 }{ 2 } = 15 \) cm. So, \( f = +15 \) cm.
Using the mirror formula: \( \frac { 1 }{ v } + \frac { 1 }{ u } = \frac { 1 }{ f } \)
Substitute the known values:
\( \frac { 1 }{ v } + \frac { 1 }{ (-20) } = \frac { 1 }{ 15 } \)
\( \frac { 1 }{ v } - \frac { 1 }{ 20 } = \frac { 1 }{ 15 } \)
Rearrange to solve for \( v \):
\( \frac { 1 }{ v } = \frac { 1 }{ 15 } + \frac { 1 }{ 20 } \)
Find a common denominator, which is 60:
\( \frac { 1 }{ v } = \frac { 4 }{ 60 } + \frac { 3 }{ 60 } \)
\( \frac { 1 }{ v } = \frac { 7 }{ 60 } \)
\( v = \frac { 60 }{ 7 } \)
\( v \approx 8.57 \) cm
The positive value of \( v \) means the image is formed behind the mirror.
Now, find the height of the image (\( h_2 \)) using the magnification formula:
\( m = \frac { h_2 }{ h_1 } = - \frac { v }{ u } \)
\( \frac { h_2 }{ 5.0 } = - \frac { 8.57 }{ (-20) } \)
\( \frac { h_2 }{ 5.0 } = \frac { 8.57 }{ 20 } \)
\( h_2 = 5.0 \times \frac { 8.57 }{ 20 } \)
\( h_2 = \frac { 8.57 }{ 4 } \)
\( h_2 \approx 2.14 \) cm
Thus, a 2.14 cm high, virtual and erect image is formed at a distance of approximately 8.6 cm behind the mirror.
In simple words: For an object 5 cm tall placed 20 cm in front of a convex mirror, the image will be about 2.14 cm tall, upright, and appear roughly 8.6 cm behind the mirror.

Exam Tip: Convex mirrors always produce virtual, erect, and diminished images, which are formed behind the mirror. Ensure correct sign conventions for \( u, v, f \) and \( R \) for a convex mirror.

 

Question 15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focussed image can be obtained? Find the size and nature of the image.
Answer: We are given: Object height \( h_0 = 7.0 \) cm. Object distance \( u = -27 \) cm. Focal length \( f = -18 \) cm (for concave mirror, focal length is negative).
Using the mirror formula: \( \frac { 1 }{ v } + \frac { 1 }{ u } = \frac { 1 }{ f } \)
Substitute the known values:
\( \frac { 1 }{ v } + \frac { 1 }{ (-27) } = \frac { 1 }{ (-18) } \)
\( \frac { 1 }{ v } - \frac { 1 }{ 27 } = - \frac { 1 }{ 18 } \)
Rearrange to solve for \( v \):
\( \frac { 1 }{ v } = - \frac { 1 }{ 18 } + \frac { 1 }{ 27 } \)
Find a common denominator, which is 54:
\( \frac { 1 }{ v } = - \frac { 3 }{ 54 } + \frac { 2 }{ 54 } \)
\( \frac { 1 }{ v } = \frac { -1 }{ 54 } \)
\( v = -54 \) cm
A screen should be placed at 54 cm in front of the mirror (negative \( v \)) to obtain a sharply focused image. This indicates a real image.
Now, find the height of the image (\( h_i \)) using the magnification formula:
\( m = \frac { h_i }{ h_0 } = - \frac { v }{ u } \)
\( \frac { h_i }{ 7.0 } = - \frac { (-54) }{ (-27) } \)
\( \frac { h_i }{ 7.0 } = - (2) \)
\( h_i = 7.0 \times (-2) \)
\( h_i = -14 \) cm
The negative value of \( h_i \) means the image is inverted. The magnitude (14 cm) is greater than the object's height (7.0 cm), so it is an enlarged image.
Therefore, the image is real, inverted, and enlarged.
In simple words: Place a screen 54 cm in front of the concave mirror to get a clear image. This image will be real, upside down, and 14 cm tall, meaning it is larger than the object.

Exam Tip: For a concave mirror, if the object is placed between F and C (as in this case, 27 cm is between 18 cm and 36 cm (2F)), the image formed is real, inverted, and enlarged, and it's formed beyond C.

 

Question 16. Find the focal length of a lens of power - 2.0 D. What type of lens is this?
Answer: We are given: Power \( P = -2.0 \) D.
The formula relating power and focal length is \( P = \frac { 1 }{ f } \) (where \( f \) is in meters).
So, \( f = \frac { 1 }{ P } \)
\( f = \frac { 1 }{ -2.0 } \)
\( f = -0.5 \) m
The focal length is -0.5 m, or -50 cm. Since the focal length is negative, the lens is a concave lens.
In simple words: A lens with -2.0 D power has a focal length of -0.5 meters, which means it is a concave lens.

Exam Tip: Remember that negative power and focal length correspond to a concave (diverging) lens, while positive power and focal length correspond to a convex (converging) lens.

 

Question 17. A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer: We are given: Power \( P = +1.5 \) D.
The formula relating power and focal length is \( P = \frac { 1 }{ f } \) (where \( f \) is in meters).
So, \( f = \frac { 1 }{ P } \)
\( f = \frac { 1 }{ +1.5 } \)
\( f = \frac { 1 }{ \frac { 3 }{ 2 } } = \frac { 2 }{ 3 } \)
\( f \approx +0.67 \) m
The focal length of the lens is approximately +0.67 m. Since the power of the lens is positive, it is a converging lens (convex lens).
In simple words: A lens with +1.5 D power has a focal length of about 0.67 meters. Because the power is positive, it's a converging, or convex, lens.

Exam Tip: Always remember that a positive power indicates a converging (convex) lens, used to correct farsightedness, while a negative power indicates a diverging (concave) lens, used to correct nearsightedness.

Gujarat Board Class 10 Science Light Reflection and Refraction Additional Important Questions and Answers

Very Short Answer Type Questions

 

Question 1. What is light?
Answer: Light is a form of electromagnetic radiation that causes the sensation of sight. It does not require any material medium to travel.
In simple words: Light is energy that helps us see, and it can travel even through empty space without needing anything to carry it.

Exam Tip: When defining light, mention both its nature (electromagnetic radiation) and its primary function (sensation of sight), as well as its ability to travel in a vacuum.

 

Question 2. Name some phenomena associated with light during image formation by mirrors.
Answer: The main phenomenon associated with light during image formation by mirrors is reflection.
In simple words: When mirrors create images, it's mostly due to light bouncing off their surface, which we call reflection.

Exam Tip: Reflection is the fundamental principle for mirrors. For lenses, refraction is the key phenomenon.

 

Question 3. Define reflection of light.
Answer: The phenomenon of coming back of light in the same medium after striking a plane and polished surface is called reflection of light.
In simple words: Reflection of light means light hitting a smooth surface and bouncing back into the same area it came from.

Exam Tip: Emphasize "coming back in the same medium" and "striking a surface" as key parts of the definition of reflection.

 

Question 4. Define the following terms related to reflection: Incident ray, Reflected ray, Normal, Angle of Incidence, and Angle of Reflection.
Answer:
• Incident ray - The light that falls on the mirror/polished surface is called the incident ray.
• Reflected ray - The light ray that goes back into the same medium after striking the surface is called the reflected ray.
• Normal - The perpendicular line drawn to the reflecting surface at the point of incidence is called the normal.
• Angle of incidence - The angle formed between the incident ray and the normal is known as the angle of incidence.
• Angle of reflection - The angle formed between the reflected ray and the normal is known as the angle of reflection.
In simple words: The incident ray is the light hitting the mirror, and the reflected ray is the light bouncing off. The normal is an imaginary line straight out from the mirror's surface where the light hits. The angle of incidence is between the incoming light and the normal, and the angle of reflection is between the outgoing light and the normal.

Exam Tip: Be precise with definitions. Clearly distinguish between a ray and an angle, and ensure you mention the "point of incidence" and "reflecting surface" for context.

 

Question 5. State laws of reflection. (CBSE 2011)
Answer: The laws of reflection of light are:
1. The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.
2. The angle of incidence is equal to the angle of reflection.
In simple words: The first rule says the incoming light, the outgoing light, and the imaginary straight line from the surface all meet on one flat surface. The second rule says the angle the light comes in at is exactly the same as the angle it bounces out at.

Exam Tip: These two laws are fundamental to understanding reflection. Ensure you state them clearly and accurately, especially the "same plane" and "equal angles" aspects.

 

Question 6. What are the properties of an image formed by a plane mirror?
Answer: The properties of an image formed by a plane mirror are:
• The image is virtual and erect.
• The size of the image is equal to that of the object.
• The image is laterally inverted (left appears right, and vice-versa).
• The image formed by a plane mirror is always at the same distance behind the mirror as the object is in front of it.
In simple words: A plane mirror creates an image that seems to be behind it, stands upright, is the same size as you, and swaps left and right. It also looks like it's as far behind the mirror as you are in front.

Exam Tip: Remember these four key characteristics of plane mirror images: virtual/erect, same size, laterally inverted, and equidistant from the mirror.

 

Question 7. What are spherical mirrors?
Answer: Mirrors whose reflecting surface is part of a sphere are called spherical mirrors.
In simple words: Spherical mirrors are mirrors shaped like a piece cut from a larger ball or sphere.

Exam Tip: Spherical mirrors include both concave and convex mirrors, which differ in whether the inner or outer surface reflects light.

 

Question 8. Define pole, the center of curvature, the radius of curvature, principal axis, aperture, focus, and focal length of a spherical mirror.
Answer:
1. Pole: The center of the reflecting surface of a spherical mirror is called the pole. It lies on the surface of the mirror and is represented by "P".
2. Centre of Curvature: The center of the sphere of which the mirror forms a part is called the center of curvature. It is represented by "C".
3. Radius of Curvature: The radius of the sphere of which the mirror forms a part is called the radius of curvature. It is represented by "R".
4. Focal Length: The distance between the pole and the principal focus of a spherical mirror is called the focal length. It is represented by "f".
In simple words: The pole is the middle of the mirror's surface. The center of curvature is the center of the big imaginary ball the mirror comes from. The radius of curvature is the distance from the mirror to that center. The focal length is the distance from the mirror to its special focus point.

Exam Tip: Understanding these terms is crucial for drawing ray diagrams and applying mirror formulas correctly. Memorize their definitions and corresponding symbols.

 

Question 9. Give some uses of a concave mirror.
Answer: Uses of concave mirrors include:
(a) Used in torches, searchlights, and vehicle headlights to get powerful parallel beams of light.
(b) Used as a shaving mirror to see a large image of the face.
(c) Used by dentists to see large images of patients' teeth.
(d) Used in a solar furnace to concentrate sunlight at a point.
In simple words: Concave mirrors are used in car lights and searchlights for strong beams, as shaving mirrors to magnify faces, by dentists to see teeth up close, and in solar furnaces to gather sunlight.

Exam Tip: Connect the uses to the properties: concave mirrors can produce large, virtual images (shaving/dentist) or converge light into a strong beam (headlights/solar furnace).

 

Question 10. Give uses of a convex mirror.
Answer: Uses of convex mirrors include:
(a) Used as a rearview mirror in vehicles.
(b) Used to see a full-length image of a falling building (or wide view of a large area).
In simple words: Convex mirrors are used in cars as rearview mirrors to see a wide area behind you, and also to view large objects completely, like an entire building.

Exam Tip: The key property of convex mirrors for these uses is their ability to provide a wider field of view and always form erect, diminished images.

 

Question 11. Give the sign conventions for spherical mirrors.
Answer:

In simple words: These rules tell us whether to use a plus or minus sign for distances and heights when working with mirrors. For example, object distance is always negative, but image distance and focal length depend on the type of mirror and where the image forms.

Exam Tip: Master the sign conventions thoroughly. A single sign error can lead to an incorrect answer in numerical problems. Always apply the Cartesian sign convention.

 

Question 12. State mirror formula and write it mathematically.
Answer: The relationship between focal length of the mirror (\( f \)), distance of the object (\( u \)), and distance of the image (\( v \)) is known as the mirror formula. It is given mathematically as:
\( \frac { 1 }{ u } + \frac { 1 }{ v } = \frac { 1 }{ f } \)
Where:
\( v \) = Image distance
\( u \) = Object distance
\( f \) = Focal length
In simple words: The mirror formula is a simple equation that connects how far an object is from a mirror, how far its image appears, and the mirror's focal length.

Exam Tip: Clearly state the formula and define each variable. Always use consistent sign conventions when applying this formula.

 

Question 13. Give the relation between focal length and radius of curvature.
Answer: The focal length (\( f \)) of a spherical mirror is half its radius of curvature (\( R \)).
\( f = \frac { R }{ 2 } \)
In simple words: The focal length of a curved mirror is always half of its radius of curvature.

Exam Tip: This relationship holds true for both concave and convex spherical mirrors when the aperture is small. Remember to apply the correct sign convention for \( R \) and \( f \).

 

Question 14. Define the magnification of the mirror.
Answer: Magnification (\( m \)) for a mirror is defined as the ratio of the height of the image (\( h' \)) to the height of the object (\( h \)). It is also equal to the negative ratio of the image distance (\( v \)) to the object distance (\( u \)).
\( m = \frac { h' }{ h } = - \frac { v }{ u } \)
Magnification of a real image is negative, and of a virtual image is positive.
In simple words: Magnification tells us how much larger or smaller an image is compared to the object. It's found by dividing the image height by the object height, or by dividing the negative of the image distance by the object distance.

Exam Tip: Remember that magnification helps determine both the relative size and the nature (erect/inverted) of the image. Positive \( m \) means erect/virtual; negative \( m \) means inverted/real.

 

Question 15. Define the refraction of light.
Answer: The change in direction of light when it travels from one medium to another is called refraction of light.
In simple words: Refraction is when light bends as it moves from one material, like air, into a different material, like water.

Exam Tip: The key aspect of refraction is the "change in direction" and "travels from one medium to another." This bending is due to a change in the speed of light.

 

Question 16. State laws of refraction.
Answer: The laws of refraction of light are:
1. The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction for a light of a given color and for a given pair of media is constant. This is called Snell's law.
\( \frac { \sin i }{ \sin r } = \text{Constant} \)
In simple words: The first rule says the incoming light, the bent light, and the imaginary straight line from the surface all lie on the same flat surface. The second rule, Snell's Law, says that for specific colors of light going between two materials, the ratio of the sine of the incoming angle to the sine of the bent angle is always the same number.

Exam Tip: Snell's Law (the second law) is critical for numerical problems involving refraction. Remember the conditions: "given color" and "given pair of media."

 

Question 17. What do you observe when a light ray passes through a rectangular slab?
Answer: When a light ray passes through a rectangular glass slab, we observe the following:
(a) The angle of incidence is equal to the angle of emergence.
(b) The incident ray is parallel to the emergent ray.
(c) Lateral displacement is proportional to the thickness of the glass slab.
(d) Lateral displacement is proportional to the angle of incidence.
In simple words: When light goes through a glass block, the light coming out is parallel to the light that went in, and the angle it leaves at is the same as the angle it entered at. The amount it shifts sideways depends on how thick the glass is and how steeply the light hit it.

Exam Tip: Focus on lateral displacement and the parallelism between incident and emergent rays as key observations when light passes through a glass slab.

 

Question 18. Define lateral displacement.
Answer: Lateral displacement is the perpendicular distance between the incident ray and the emergent ray when light passes through a transparent slab.
In simple words: Lateral displacement is how much the light ray shifts sideways when it goes through something like a glass block, measured as the straight distance between the original path and the final path.

Exam Tip: Lateral displacement is the horizontal shift, not the bending itself. It only occurs when light passes through a medium with parallel surfaces.

 

Question 19. Define the refractive index.
Answer: The refractive index of a medium is defined as the ratio of the speed of light in medium 1 to the speed of light in medium 2. It is represented as \( n_{21} \) and is read as the refractive index of medium 2 with respect to medium 1.
\( n_{21} = \frac { \text{speed of light in medium 1} }{ \text{speed of light in medium 2} } \)
In simple words: Refractive index tells us how much light slows down and bends when it enters a new material. It's a ratio comparing the speed of light in one material to its speed in another.

Exam Tip: Remember that refractive index is a ratio of speeds. A higher refractive index means light travels slower and bends more in that medium.

 

Question 20. Define the absolute refractive index.
Answer: When medium 1 is a vacuum, then the refractive index of medium 2 with respect to the vacuum is considered as the absolute refractive index.
In simple words: The absolute refractive index measures how much light slows down and bends when it enters a material, compared to how it behaves in empty space (a vacuum).

Exam Tip: The absolute refractive index is a specific type of refractive index where the first medium is always a vacuum (or approximately air). It gives a standard measure for a material's optical density.

 

Question 21. What is the unit of the refractive index?
Answer: It has no unit.
In simple words: The refractive index is just a number; it doesn't have a unit of measurement.

Exam Tip: Since the refractive index is a ratio of two speeds (which have the same units), the units cancel out, making it a dimensionless quantity.

 

Question 22. Define optical density.
Answer: The ability of a medium to refract light is called optical density.
In simple words: Optical density describes how much a material can bend light.

Exam Tip: Optical density is not the same as mass density. A medium with higher optical density slows down light more and causes it to bend more when light enters it from a rarer medium.

 

Question 23. What is the relation between optical density, refractive index, and speed of light?
Answer: A medium with a higher refractive index in which the speed of light is less is known as an optically denser medium. Conversely, a medium with a lower refractive index in which the speed of light is more is known as an optically rarer medium.
In simple words: If a material has a high refractive index, light moves slower in it, making it optically denser. If it has a low refractive index, light moves faster, making it optically rarer.

Exam Tip: Understand the inverse relationship: higher refractive index means higher optical density and slower light speed. Optically denser materials cause light to bend more towards the normal.

 

Question 24. State lens formula and write it mathematically.
Answer: The relationship between object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) of a lens is known as the lens formula. It is given mathematically as:
\( \frac { 1 }{ v } - \frac { 1 }{ u } = \frac { 1 }{ f } \)
In simple words: The lens formula is an equation that connects the distance of an object from a lens, the distance of the image formed by that lens, and the lens's focal length.

Exam Tip: Distinguish the lens formula from the mirror formula. The lens formula has a minus sign between \( \frac{1}{v} \) and \( \frac{1}{u} \), while the mirror formula has a plus sign.

 

Question 25. Define the magnification of the lens.
Answer: Magnification (\( m \)) for a lens is defined as the ratio of the height of the image (\( h' \)) to the height of the object (\( h \)). It is also equal to the ratio of the image distance (\( v \)) to the object distance (\( u \)).
\( m = \frac { h' }{ h } = \frac { v }{ u } \)
For convex lenses, \( m \) can be more than, less than, or equal to one. For concave lenses, \( m \) is always less than one.
In simple words: Magnification tells us how much an image produced by a lens is bigger or smaller than the real object. It's found by comparing their heights or their distances from the lens.

Exam Tip: Note the difference in the magnification formula for lenses (\( m = \frac{v}{u} \)) versus mirrors (\( m = -\frac{v}{u} \)).

 

Question 26. Define a lens.
Answer: A transparent material bounded by two surfaces, of which one or both surfaces are spherical, forms a lens.
In simple words: A lens is a clear piece of material, usually curved on one or both sides, that can bend light.

Exam Tip: Key terms in the definition are "transparent material" and "bounded by two spherical surfaces" (or one spherical and one plane).

 

Question 27. What are the two types of lenses?
Answer: The two types of lenses are convex lenses and concave lenses.
In simple words: The two main kinds of lenses are convex, which are thicker in the middle, and concave, which are thinner in the middle.

Exam Tip: Be able to describe the basic shape and light-bending properties (converging/diverging) of each type of lens.

 

Question 28. Write the nature, position, and relative size of the image formed by a convex lens.
Answer: The nature, position, and relative size of the image formed by a convex lens for various positions of the object are shown in the table below.

In simple words: The type of image made by a convex lens changes depending on how close or far the object is. It can be small and upside down, or very big and upright, or even not form at all (at infinity).

Exam Tip: Practice drawing ray diagrams for different object positions to internalize these image properties. Pay attention to the critical positions: infinity, beyond 2F, at 2F, between F and 2F, at F, and between F and O.

 

Question 29. Write the nature, position and relative size of the image formed by a concave lens.
Answer: The nature, position, and relative size of the image formed by a concave lens for various positions of the object are shown in the table below.

In simple words: A concave lens always creates an image that is smaller than the object, appears upright, and seems to be on the same side as the object. It's always a virtual image.

Exam Tip: Concave lenses are simpler than convex lenses as they always form the same type of image (virtual, erect, diminished) regardless of the object's position (as long as it's not at infinity). The image is always between F\( _1 \) and O.

 

Question 6. Write the position, nature, and size of images formed by a concave mirror.
Answer: The table below shows the image formation by a concave mirror for various object positions.

In simple words: The type of image made by a concave mirror changes based on where the object is placed. It can be real, virtual, magnified, or diminished.

Exam Tip: Memorize the six key positions of objects for concave mirrors and the corresponding characteristics of the images formed.

 

Question 7. Give the laws of refraction of light.
Answer: The laws of refraction of light are as follows:
1. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for light of a given color and for a given pair of media. This law is also known as Snell's law of refraction.
In simple words: When light bends, it follows two main rules. First, the incoming light, the bent light, and a special line (normal) all sit on the same flat surface. Second, there's a fixed ratio between the angle of the incoming light and the angle of the bent light, which is always the same for a specific color of light and materials.

Exam Tip: Clearly state both laws and ensure you mention "same plane" for the first law and "constant ratio" or "Snell's Law" for the second.

 

Question 8. What is the refractive index?
Answer: If \( i \) is the angle of incidence and \( r \) is the angle of refraction, then the ratio \( \frac{ \sin i }{ \sin r } \) is a constant. This constant value is called the refractive index of the second medium concerning the first.
In simple words: Refractive index is a number that tells us how much light bends when it goes from one material into another. It's found by comparing the sine of the angle where light hits and the sine of the angle where it bends.

Exam Tip: Remember to define both \( i \) and \( r \) when stating the formula for refractive index.

 

Question 9. Define absolute refractive index of the medium.
Answer: The absolute refractive index of a medium is determined when medium 1 is a vacuum. In this case, the refractive index of medium 2 is considered concerning the vacuum. It measures how much light slows down when it enters a specific medium from a vacuum.
\( n = \frac{ \text{Speed of light in air} }{ \text{Speed of light in the medium} } \)
In simple words: Absolute refractive index tells us how much light bends when it moves from empty space (or air) into a material. It compares the speed of light in empty space to its speed in that material.

Exam Tip: Always specify that the first medium is a vacuum (or air for practical purposes) when defining absolute refractive index.

 

Question 10. Two medium with refractive index 1.31 and 1.50 are given. In which case
(1) bending of light is more?
(2) speed of light is more?
Answer:
1. The bending of light is more in the medium where the refractive index is 1.50. A higher refractive index means light bends more when it enters the medium.
2. The speed of light is more in the medium with a refractive index of 1.31. A lower refractive index indicates a faster speed of light in that medium.
In simple words: Light bends more in the material with the higher number (1.50), and it moves faster in the material with the smaller number (1.31).

Exam Tip: Remember that a higher refractive index means greater optical density, leading to more bending of light and a slower speed of light.

 

Question 11. The Refractive index of kerosene oil is 1.44 and that of water is 1.33. A ray of light enters from kerosene oil to water. Where would' light ray bend and why?
Answer: A ray of light entering from kerosene oil to water (i.e., from a refractive index of 1.44 to 1.33) moves from a denser to a rarer medium. Consequently, the light ray bends away from the normal.
In simple words: Because light travels from a thicker liquid (kerosene) to a thinner liquid (water), it speeds up and bends away from the imaginary normal line.

Exam Tip: Always relate the change in medium (denser to rarer or vice versa) to the bending direction (towards or away from the normal).

 

Question 12. Which is optically denser out of the two medium M1 = 1.71 (refractive index) and M2 = 1.36 (refractive index). How does speed of light change when it travels from optically rarer to denser medium.
Answer: Medium \( M_1 \) with a refractive index of 1.71 is optically denser than the other medium \( M_2 \). When light travels from an optically rarer medium to an optically denser medium, its speed of light decreases.
In simple words: Material M1 (refractive index 1.71) is denser than M2 (refractive index 1.36). When light goes from a less dense to a more dense material, it slows down.

Exam Tip: Understand that a higher refractive index indicates a denser medium, which causes light to slow down.

 

Question 13. Draw a ray diagram of the image formed when an object is placed in front of a convex lens
1. beyond 2F
2. between F and 2F.
Answer: You should draw two separate ray diagrams for a convex lens for each case:
1. **Object placed beyond \( 2F_1 \)**:
- Draw a convex lens with its principal axis, optical center (O), and focal points \( F_1, F_2 \) and \( 2F_1, 2F_2 \) marked.
- Place an object (arrow) beyond \( 2F_1 \) on the left side.
- Draw two principal rays from the top of the object:
- A ray parallel to the principal axis, which after refraction passes through \( F_2 \).
- A ray passing through the optical center (O), which goes undeviated.
- The intersection of these two refracted rays will form a real, inverted, and diminished image between \( F_2 \) and \( 2F_2 \) on the other side.
2. **Object placed between \( F_1 \) and \( 2F_1 \)**:
- Draw a convex lens with its principal axis, optical center (O), and focal points \( F_1, F_2 \) and \( 2F_1, 2F_2 \) marked.
- Place an object (arrow) between \( F_1 \) and \( 2F_1 \) on the left side.
- Draw two principal rays from the top of the object:
- A ray parallel to the principal axis, which after refraction passes through \( F_2 \).
- A ray passing through the optical center (O), which goes undeviated.
- The intersection of these two refracted rays will form a real, inverted, and enlarged image beyond \( 2F_2 \) on the other side.
In simple words: For a convex lens, draw two pictures. First, if the object is far away (past \( 2F \)), the image will be smaller and upside down between \( F \) and \( 2F \). Second, if the object is closer (between \( F \) and \( 2F \)), the image will be bigger and upside down, appearing past \( 2F \) on the other side.

Exam Tip: When drawing ray diagrams, always use a ruler for straight lines and ensure arrows indicate the direction of light. Mark the principal foci and optical center clearly.

 

Question 14. Comment on the size, the position of the image formed by a concave mirror of focal length 18 cm when an object is placed:
1. at 22 cm
2. 14 cm
3. 40 cm.
in front of the mirror without calculations.
Answer: Given focal length \( f = 18 \) cm for a concave mirror.
Therefore, the center of curvature \( C = 2f = 2 \times 18 = 36 \) cm.
1. **Object at 22 cm**: Since \( 18 \text{ cm} < 22 \text{ cm} < 36 \text{ cm} \), the object is placed between the principal focus (F) and the center of curvature (C). In this situation, the image is formed beyond C, it is real, inverted, and magnified.
2. **Object at 14 cm**: Since \( 14 \text{ cm} < 18 \text{ cm} \), the object is placed between the pole (P) and the principal focus (F). In this case, the image is formed behind the mirror, it is virtual, erect, and magnified.
3. **Object at 40 cm**: Since \( 40 \text{ cm} > 36 \text{ cm} \), the object is placed beyond the center of curvature (C). Here, the image is formed between F and C, it is real, inverted, and diminished.
In simple words: For a concave mirror with a focal length of 18 cm (so C is at 36 cm):
1. If the object is at 22 cm (between F and C), the image appears larger, upside down, and beyond C.
2. If the object is at 14 cm (inside F), the image appears larger, right-side up, and behind the mirror.
3. If the object is at 40 cm (beyond C), the image appears smaller, upside down, and between F and C.

Exam Tip: To quickly analyze image formation without calculations, compare the object distance with the focal length and radius of curvature. Remember the standard cases for concave mirrors.

 

Question 15. Complete the following ray diagrams:
Answer: You need to complete the ray diagrams based on the rules of reflection for mirrors. Here are the descriptions of the completed diagrams, focusing on the path of the rays:
(a) **Concave mirror, incident ray parallel to principal axis**: The ray, after striking the concave mirror, will pass through the principal focus (F).
(b) **Concave mirror, incident ray passing through the principal focus**: The ray, after striking the concave mirror, will become parallel to the principal axis.
(c) **Concave mirror, incident ray passing through the center of curvature**: The ray, after striking the concave mirror, will reflect back along the same path, passing through the center of curvature (C).
(d) **Convex mirror, incident ray parallel to principal axis**: The ray, after striking the convex mirror, will appear to diverge from the principal focus (F) behind the mirror. Extend the reflected ray backward to meet F.
(e) **Convex mirror, incident ray directed towards the principal focus**: The ray, after striking the convex mirror, will become parallel to the principal axis.
(f) **Convex mirror, incident ray directed towards the center of curvature**: The ray, after striking the convex mirror, will reflect back along the same path as if it were coming from the center of curvature (C) behind the mirror.
In simple words: For concave mirrors:
(a) Parallel rays hit and go through the focus.
(b) Rays through the focus hit and go parallel.
(c) Rays through the center hit and go straight back.
For convex mirrors:
(d) Parallel rays hit and bounce away as if from the focus behind the mirror.
(e) Rays aiming for the focus hit and go parallel.
(f) Rays aiming for the center hit and bounce straight back.

Exam Tip: Practice drawing these standard ray diagrams for both concave and convex mirrors until you can do them quickly and accurately. Pay attention to the use of solid and dashed lines.

 

Question 16. With the help of a ray diagram showing how a pencil appears when dipped in water.
Answer: When a pencil is dipped in water, it appears bent or broken at the water surface due to the refraction of light. A ray of light from the part of the pencil in water travels from a denser medium (water) to a rarer medium (air). As it moves from denser to rarer, the light ray bends away from the normal. When these refracted rays reach our eyes, our brain interprets them as coming from a straight line, making the pencil appear at a different, shallower position than its actual location, creating the illusion of bending. Your diagram should show the pencil partially submerged, with light rays from the submerged part bending away from the normal as they exit the water, causing the apparent position of the pencil to be shifted upwards.
Exam Tip: For ray diagrams, label all parts clearly, including the normal, incident ray, refracted ray, actual position, and apparent position. Ensure the bending direction is correct for water-to-air transition.

 

Question 17. Define the power of the lens. What is the S. I. unit of power of a lens? If power of lens is + 2D what is the nature and focal length of the lens?
Answer: The power of a lens is defined as the degree of convergence or divergence of light rays that a lens can produce. It is also the reciprocal of its focal length. The S.I. unit of power of a lens is 'dioptre', denoted by 'D'.
Given power \( P = +2D \).
Since the power is positive, the lens is a converging lens (convex lens).
The focal length \( f \) can be calculated as:
\( P = \frac{ 1 }{ f } \)
\( +2 = \frac{ 1 }{ f } \)
\( f = \frac{ 1 }{ 2 } \)
\( f = 0.5 \text{ m} \) or \( 50 \text{ cm} \).
Thus, the lens is a convex lens with a focal length of \( +0.50 \text{ m} \).
In simple words: The power of a lens tells us how strongly it bends light. Its unit is a dioptre. If a lens has a power of +2D, it means it's a convex lens (converging light), and its focal length is 0.5 meters.

Exam Tip: Remember that positive power implies a converging (convex) lens, and negative power implies a diverging (concave) lens. Ensure focal length is in meters for power calculations.

 

Question 18. If the speed of light in water is \( 2.25 \times 10^8 \text{ m/s} \) and the speed in a vacuum is \( 3 \times 10^8 \text{ m/s} \). Calculate the refractive index of water.
Answer: The refractive index of water (\( n_w \)) is given by the ratio of the speed of light in a vacuum (or air) to the speed of light in water.
\( n_w = \frac{ \text{Speed of light in 1 medium (air)} }{ \text{Speed of light in 2 medium (water)} } \)
\( n_w = \frac{ c }{ v } \)
\( n_w = \frac{ 3 \times 10^8 \text{ m/s} }{ 2.25 \times 10^8 \text{ m/s} } \)
\( n_w = 1.33 \)
Therefore, the refractive index of water is 1.33.
In simple words: To find how much water bends light, we divide how fast light moves in empty space by how fast it moves in water. When we do that with the given speeds, we get 1.33.

Exam Tip: Always write down the formula before substituting values. Ensure the units are consistent, and note that refractive index is a unitless quantity.

 

Question 19. The refractive index of water is 1.33 and kerosene is 1.44. Calculate the refractive index of kerosene with respect to water.
Answer: Given:
Refractive index of water \( n_w = 1.33 \)
Refractive index of kerosene \( n_k = 1.44 \)
The refractive index of kerosene with respect to water (\( n_{kw} \)) is calculated as the ratio of the refractive index of kerosene to the refractive index of water.
\( n_{kw} = \frac{ n_k }{ n_w } \)
\( n_{kw} = \frac{ 1.44 }{ 1.33 } \)
\( n_{kw} \approx 1.08 \)
Therefore, the refractive index of kerosene with respect to water is approximately 1.08.
In simple words: To find how much light bends going from water into kerosene, we divide the bending value for kerosene by the bending value for water. Doing this calculation gives us about 1.08.

Exam Tip: When calculating relative refractive index, remember that "refractive index of medium A with respect to medium B" is \( n_A / n_B \).

 

Question 20. Redraw the given diagram and show the path of the refracted ray.
Answer: To redraw the diagram and show the path of the refracted ray through a convex lens, follow these steps:
- **For a ray passing parallel to the principal axis**: This ray will refract and pass through the principal focus (\( F_2 \)) on the other side of the lens.
- **For a ray passing through the optical center**: This ray will pass undeviated through the lens.
- **For a ray passing through \( F_1 \)**: This ray will refract and become parallel to the principal axis after passing through the lens.
The diagram generally depicts a convex lens with its principal axis, two principal foci (\( F_1, F_2 \)), and points \( 2F_1, 2F_2 \). An incident ray is shown approaching the lens. You need to draw the refracted path correctly based on the lens rules. For example, if the incident ray is parallel to the principal axis, the refracted ray will pass through \( F_2 \).
In simple words: Draw the picture of the lens again. If a light ray comes in straight, it will bend and go through the focus on the other side. If it goes through the center, it won't bend. If it goes through the focus on the first side, it will come out straight.

Exam Tip: When redrawing, ensure that the lens, principal axis, and focal points are accurately positioned. Use a ruler to draw straight lines for the rays and an arrow to indicate the direction of light.

 

Question 21. Why does a ray of light bend when it travels from one medium into another?
Answer: A ray of light bends when it travels from one medium into another due to a change in its velocity. Light travels at different speeds in different media. When light enters a new medium at an angle, one part of the wavefront changes speed before the other part, causing the entire wavefront to pivot or bend. This change in direction minimizes the time it takes for light to travel the new path (Fermat's Principle).
In simple words: Light bends because its speed changes when it moves from one material to another. This speed change causes it to turn, similar to how a car changes direction when one wheel hits mud before the other.

Exam Tip: The key concept here is the change in the speed of light. Also, mentioning Fermat's Principle or the wavefront explanation can earn extra marks.

 

Question 22. Draw the given diagram in your answer book and complete it for the path of ray of light beyond the lens.
Answer: You need to draw a convex lens with an incident ray shown parallel to the principal axis. The path of the ray beyond the lens will be as follows:
- **Incident Ray**: Draw a ray of light approaching the convex lens, parallel to its principal axis.
- **Refracted Ray**: After striking the lens, this ray will refract and pass through the principal focus (\( F' \)) on the opposite side of the lens.
Ensure your diagram shows the lens, the principal axis, and the focal points clearly marked. The refracted ray should correctly converge at \( F' \).
In simple words: Draw the picture of the convex lens. A light ray coming straight into the lens will bend and pass through the focus point on the other side.

Exam Tip: Always use arrows on rays to show the direction of light. For a convex lens, parallel incident rays always converge at the focal point on the other side.

 

Question 23. What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror.
Answer: The minimum number of rays required for locating the image formed by a concave mirror for an object is two rays.
To show the formation of a virtual image by a concave mirror, the object must be placed between the pole (P) and the principal focus (F) of the mirror. Your ray diagram should illustrate this:
- **Concave Mirror and Axis**: Draw a concave mirror with its principal axis, pole (P), principal focus (F), and center of curvature (C).
- **Object Placement**: Place an object (e.g., an arrow AB) between P and F.
- **Ray 1**: Draw a ray from A parallel to the principal axis. After reflection, this ray will pass through F.
- **Ray 2**: Draw a ray from A passing through C. This ray will reflect back along the same path.
- **Image Formation**: The reflected rays appear to diverge. When extended backward (as dashed lines), they intersect behind the mirror, forming a virtual, erect, and magnified image (A'B').
In simple words: You need at least two light rays to find where an image forms. To see a fake (virtual) image from a curved mirror that bulges inward, put the object very close to the mirror (between the pole and the focus). The light rays will bounce off and seem to come from behind the mirror, making a bigger, upright image.

Exam Tip: Always use dashed lines for virtual rays and virtual images. Ensure the virtual image is erect and larger than the object for a concave mirror when the object is between P and F.

 

Question 24. Takedown this diagram on to your answer book and complete the path of the ray.
Answer: The diagram shows a concave lens with an incident ray directed towards its principal focus (\( F' \)) on the opposite side. The path of this ray beyond the lens will be as follows:
- **Incident Ray**: Draw a ray of light approaching the concave lens, directed towards the principal focus (\( F' \)) on the far side.
- **Refracted Ray**: After striking the concave lens, this ray will refract and emerge parallel to the principal axis.
Ensure your diagram clearly shows the concave lens, the principal axis, and the focal points. The refracted ray should correctly emerge parallel to the principal axis.
In simple words: Draw the picture of the concave lens. A light ray that tries to hit the far focus point will bend and come out straight, parallel to the main line.

Exam Tip: For concave lenses, rays directed towards the far focal point always become parallel to the principal axis after refraction. Remember that concave lenses are diverging lenses.

 

Question 25. What kind of mirrors are used in big shopping stores to watch the activities of customers?
Answer: Convex mirrors are used in big shopping stores to watch the activities of customers. This is because a convex mirror always forms a virtual, erect, and diminished image, regardless of the object's position. This property allows it to cover a wider field of view, making it ideal for surveillance purposes to monitor a large area.
In simple words: Large stores use convex mirrors to watch customers. These mirrors make objects look smaller and right-side up, but they let you see a much wider area, which is great for security.

Exam Tip: Highlight "wider field of view" and "virtual, erect, and diminished image" as the key reasons for using convex mirrors in surveillance.

 

Question 26. Draw a ray diagram to determine the position of the image formed of an object placed between the pole and the focus of a concave mirror.
Answer: To determine the position of the image formed when an object is placed between the pole (P) and the principal focus (F) of a concave mirror, draw the following ray diagram:
- **Concave Mirror and Axis**: Draw a concave mirror with its principal axis, pole (P), principal focus (F), and center of curvature (C).
- **Object Placement**: Place an object (e.g., an arrow AB) between P and F.
- **Ray 1**: Draw a ray from the top of the object (A) parallel to the principal axis. After reflection from the mirror, this ray will pass through the principal focus (F).
- **Ray 2**: Draw a ray from the top of the object (A) passing through the center of curvature (C). This ray will reflect back along the same path.
- **Image Formation**: The reflected rays appear to diverge. When these reflected rays are extended backward (using dashed lines) behind the mirror, they intersect at a point. This intersection point (A') gives the location of the top of the image. The image A'B' will be virtual, erect, and magnified, formed behind the mirror.
In simple words: Draw a curved mirror that bulges inward. Put an object very close to the mirror, between the center and the focus. Then, draw light rays bouncing off. You'll see that the reflected rays spread out, but if you trace them backward, they meet behind the mirror, making a bigger, upright, fake image.

Exam Tip: For a concave mirror, an object between P and F is the only case that produces a virtual image. This is a common ray diagram question, so practice it well.

 

Question 27. State the mirror formula, lens formula and power of lens.
Answer:
**Mirror Formula:** The mirror formula provides the relationship between the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) for spherical mirrors. It is expressed as:
\( \frac{ 1 }{ v } + \frac{ 1 }{ u } = \frac{ 1 }{ f } \)
Where:
\( v \) = Image distance
\( u \) = Object distance
\( f \) = Focal length

**Lens Formula:** The lens formula provides the relationship between the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) for spherical lenses. It is expressed as:
\( \frac{ 1 }{ v } - \frac{ 1 }{ u } = \frac{ 1 }{ f } \)

**Power of Lens:** The power of a lens (\( P \)) is defined as the reciprocal of its focal length (\( f \)) when the focal length is measured in meters. It indicates the degree of convergence or divergence of light rays produced by the lens.
\( P = \frac{ 1 }{ f } \) (where \( f \) is in meters)
In simple words: The mirror formula links how far an object is, how far its image is, and the mirror's focal length. The lens formula does the same for lenses, but with a minus sign. The power of a lens tells how much it bends light and is just one divided by its focal length in meters.

Exam Tip: Carefully distinguish between the mirror formula (addition) and the lens formula (subtraction). Always use meters for focal length when calculating lens power.

 

Question 28. Draw ray diagrams to show the image formed by a concave lens for the object placed at
1. infinity
2. Between F and 2F of the lens.
Answer: You need to draw two ray diagrams for a concave lens:
1. **Object at infinity**: For an object placed at infinity, the incident rays are considered parallel to the principal axis.
- **Concave Lens and Axis**: Draw a concave lens with its principal axis and focal points (\( F_1, F_2 \)).
- **Incident Rays**: Draw two parallel rays approaching the concave lens.
- **Refracted Rays**: After passing through the concave lens, these rays will diverge. When extended backward (dashed lines), they appear to originate from the principal focus (\( F_1 \)) on the same side as the object. The image formed is virtual, erect, and highly diminished (point-sized) at \( F_1 \).
2. **Object between F and 2F of the lens**: For an object placed between \( F_1 \) and \( 2F_1 \).
- **Concave Lens and Axis**: Draw a concave lens with its principal axis and focal points (\( F_1, F_2 \)) and \( 2F_1, 2F_2 \).
- **Object Placement**: Place an object (e.g., an arrow AB) between \( F_1 \) and \( 2F_1 \).
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After refraction, it diverges and appears to come from \( F_1 \).
- **Ray 2**: Draw a ray from the top of the object passing through the optical center (O). This ray goes undeviated.
- **Image Formation**: The intersection of the backward extension of the first refracted ray and the second ray forms a virtual, erect, and diminished image between O and \( F_1 \) on the same side as the object.
In simple words: Draw two diagrams for a concave lens:
1. When the object is super far away, the light rays spread out from the lens, and the image looks like a tiny dot at the first focus point.
2. When the object is between the focus and twice the focus, the light also spreads out, making a smaller, upright, fake image between the lens and the first focus point.

Exam Tip: Remember that concave lenses always form virtual, erect, and diminished images, regardless of the object's position. Practice drawing these diagrams with correct ray tracing.

 

Question 1. With the help of a ray, a diagram showing the type of images formed when the object is at the following positions in front of a concave mirror.
(a) at infinity
(b) beyond C
Answer: You need to draw two ray diagrams for a concave mirror:
(a) **Object at infinity**: For an object placed at infinity, incident rays are considered parallel to the principal axis.
- **Concave Mirror and Axis**: Draw a concave mirror with its principal axis, pole (P), principal focus (F), and center of curvature (C).
- **Incident Rays**: Draw two parallel rays approaching the concave mirror.
- **Reflected Rays**: After reflection, these rays converge at the principal focus (F). The image formed is real, inverted, and highly diminished (point-sized) at F.
(b) **Object beyond C**: For an object placed beyond the center of curvature (C).
- **Concave Mirror and Axis**: Draw a concave mirror with its principal axis, P, F, and C.
- **Object Placement**: Place an object (e.g., an arrow) beyond C.
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After reflection, it passes through F.
- **Ray 2**: Draw a ray from the top of the object passing through F. After reflection, it becomes parallel to the principal axis.
- **Image Formation**: The intersection of the reflected rays forms a real, inverted, and diminished image between F and C.
In simple words: You need to draw two pictures for a curved mirror that bulges inward:
(a) If the object is extremely far away, the light rays hit the mirror and all meet at one tiny spot, the focus, making a small, upside-down image.
(b) If the object is placed a bit far from the mirror (beyond C), the light rays hit, and the image formed is smaller and upside down, appearing between the focus and the center.

Exam Tip: Always use arrows to show ray direction. For objects at infinity, parallel rays are drawn. For objects at finite distances, use at least two principal rays.

 

Question 1. (d) atC
(e) between F and C
(f) between F and P
C = centre of curvature
P = pole of the mirror
F = focus
Answer: Continuing the ray diagrams for a concave mirror:
(d) **Object at C**: When an object is placed at the center of curvature (C).
- **Concave Mirror and Axis**: Draw a concave mirror with P, F, and C.
- **Object Placement**: Place an object (e.g., an arrow) at C.
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After reflection, it passes through F.
- **Ray 2**: Draw a ray from the top of the object passing through F. After reflection, it becomes parallel to the principal axis.
- **Image Formation**: The intersection of the reflected rays forms a real, inverted, and same-sized image at C.
(e) **Object between F and C**: When an object is placed between the principal focus (F) and the center of curvature (C).
- **Concave Mirror and Axis**: Draw a concave mirror with P, F, and C.
- **Object Placement**: Place an object (e.g., an arrow) between F and C.
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After reflection, it passes through F.
- **Ray 2**: Draw a ray from the top of the object passing through F. After reflection, it becomes parallel to the principal axis.
- **Image Formation**: The intersection of the reflected rays forms a real, inverted, and enlarged image beyond C.
(f) **Object between F and P**: When an object is placed between the principal focus (F) and the pole (P).
- **Concave Mirror and Axis**: Draw a concave mirror with P, F, and C.
- **Object Placement**: Place an object (e.g., an arrow) between F and P.
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After reflection, it passes through F.
- **Ray 2**: Draw a ray from the top of the object directed towards P, reflecting at an equal angle below the principal axis.
- **Image Formation**: The reflected rays appear to diverge. When extended backward (dashed lines), they intersect behind the mirror, forming a virtual, erect, and enlarged image behind the mirror.
In simple words: For a concave mirror, further diagrams are needed:
(d) If the object is at the center, the image is also at the center, the same size, and upside down.
(e) If the object is between the focus and the center, the image is bigger, upside down, and further away than the center.
(f) If the object is between the focus and the mirror's surface, the image is bigger, right-side up, and appears behind the mirror.

Exam Tip: Be precise with the angles and intersection points, especially for the object between F and P, where the image is virtual and magnified.

 

Question 2. With the help of a ray diagram show the position, size and the nature of the image formed by a convex lens for various positions of the object.
Answer: You need to draw ray diagrams for a convex lens for various object positions. Here are the descriptions of the image formation for key positions:
1. **Object at infinity**: Image forms at \( F_2 \), real, inverted, highly diminished.
2. **Object beyond \( 2F_1 \)**: Image forms between \( F_2 \) and \( 2F_2 \), real, inverted, diminished.
3. **Object at \( 2F_1 \)**: Image forms at \( 2F_2 \), real, inverted, same size.
4. **Object between \( F_1 \) and \( 2F_1 \)**: Image forms beyond \( 2F_2 \), real, inverted, enlarged.
5. **Object at \( F_1 \)**: Image forms at infinity, real, inverted, highly enlarged.
6. **Object between optical center (O) and \( F_1 \)**: Image forms on the same side as the object, virtual, erect, enlarged.
Each of these cases requires a separate ray diagram showing the lens, principal axis, object, two principal rays (e.g., parallel to axis passing through \( F_2 \), and through O undeviated), and the resulting image.
In simple words: Draw different pictures for a convex lens depending on where the object is:
1. Far away: Image is tiny, upside down at \( F_2 \).
2. Beyond \( 2F_1 \): Image is smaller, upside down between \( F_2 \) and \( 2F_2 \).
3. At \( 2F_1 \): Image is same size, upside down at \( 2F_2 \).
4. Between \( F_1 \) and \( 2F_1 \): Image is bigger, upside down beyond \( 2F_2 \).
5. At \( F_1 \): Image is super big, upside down, far away.
6. Close up (between O and \( F_1 \)): Image is bigger, right-side up, on the same side as the object.

Exam Tip: Practice drawing all six ray diagrams for a convex lens. This question often appears as a detailed explanation with diagrams for different scenarios.

 

Question 3. Name the type of mirror used in the following situations:
1. Rearview mirror in vehicles
2. Solar furnace
3. Torch
4. Solar cooker
5. To get the full-length image of a tall building.
Answer: The type of mirror used in each situation is:
1. **Rearview mirror in vehicles**: Convex mirror (because it gives a virtual, diminished image and a wider field of view).
2. **Solar furnace**: Concave mirror (to concentrate all parallel beams of light from the sun at its focus).
3. **Torch**: Concave mirror (to get a powerful, parallel beam of light when the bulb is placed at its focus).
4. **Solar cooker**: Concave mirror (to concentrate heat rays at a point, similar to a solar furnace).
5. **To get the full-length image of a tall building**: Convex mirror (as it produces a diminished, erect image, allowing a full view of large objects).
In simple words:
1. Car side mirrors use convex mirrors.
2. Solar furnaces use concave mirrors.
3. Torches use concave mirrors.
4. Solar cookers use concave mirrors.
5. To see a whole tall building, use a convex mirror.

Exam Tip: Understand the properties of concave (converging, can magnify, wider field for specific object positions) and convex (diverging, always virtual/erect/diminished, wide field of view) mirrors to correctly identify their uses.

 

Question 4. Draw and explain the ray diagram formed by a convex mirror when
(a) the object is at infinity
(b) the object is a finite distance from the mirror.
Answer: You need to draw two ray diagrams for a convex mirror:
(a) **Object at infinity**: For an object placed at infinity, incident rays are considered parallel to the principal axis.
- **Convex Mirror and Axis**: Draw a convex mirror with its principal axis, pole (P), principal focus (F), and center of curvature (C) behind the mirror.
- **Incident Rays**: Draw two parallel rays approaching the convex mirror.
- **Reflected Rays**: After striking the convex mirror, these rays diverge. When extended backward (dashed lines), they appear to originate from the principal focus (F) behind the mirror. The image formed is virtual, erect, and highly diminished (point-sized) at F.
(b) **Object is a finite distance from the mirror**: For an object placed anywhere between infinity and the pole (P) of the convex mirror.
- **Convex Mirror and Axis**: Draw a convex mirror with its principal axis, P, F, and C behind the mirror.
- **Object Placement**: Place an object (e.g., an arrow) in front of the mirror.
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After reflection, it diverges and appears to come from F.
- **Ray 2**: Draw a ray from the top of the object directed towards C. This ray reflects back along the same path.
- **Image Formation**: The reflected rays appear to diverge. When extended backward (dashed lines), they intersect behind the mirror, forming a virtual, erect, and diminished image between P and F.
In simple words: Draw two pictures for a curved mirror that bulges outward:
(a) If the object is super far away, the light rays hit the mirror and spread out, seeming to come from the focus point behind the mirror. The image is tiny and upright at the focus.
(b) If the object is any normal distance away, the light rays hit and spread out, seeming to come from a spot behind the mirror, between the mirror and its focus. The image is always smaller and upright.

Exam Tip: Remember that convex mirrors always form virtual, erect, and diminished images, regardless of the object's position. Practice drawing these two diagrams clearly.

 

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

Using the lens formula:
\( \frac{ 1 }{ v } - \frac{ 1 }{ u } = \frac{ 1 }{ f } \)
\( \frac{ 1 }{ 30 } - \frac{ 1 }{ u } = \frac{ 1 }{ 15 } \)
\( - \frac{ 1 }{ u } = \frac{ 1 }{ 15 } - \frac{ 1 }{ 30 } \)
\( - \frac{ 1 }{ u } = \frac{ 2 - 1 }{ 30 } \)
\( - \frac{ 1 }{ u } = \frac{ 1 }{ 30 } \)
\( u = -30 \text{ cm} \)
The object should be placed at 30 cm in front of the convex lens (which is at \( 2F_1 \), since \( 2f = 30 \text{ cm} \)).

To find the size of the image (\( h_i \)), use the magnification formula:
\( m = \frac{ h_i }{ h_o } = \frac{ v }{ u } \)
\( \frac{ h_i }{ 5 } = \frac{ 30 }{ -30 } \)
\( \frac{ h_i }{ 5 } = -1 \)
\( h_i = -5 \text{ cm} \)
The image size is 5 cm. The negative sign indicates that the image is inverted. Thus, the image is real, inverted, and the same size as the object.

**Ray Diagram**: You need to draw a ray diagram for an object placed at \( 2F_1 \) of a convex lens:
- **Convex Lens and Axis**: Draw a convex lens with its principal axis, optical center (O), and focal points \( F_1, F_2 \) and \( 2F_1, 2F_2 \). Mark \( F_1 \) and \( 2F_1 \) at 15 cm and 30 cm respectively on the left, and \( F_2 \) and \( 2F_2 \) similarly on the right.
- **Object Placement**: Place the object (arrow, 5 cm high) at \( 2F_1 \) (30 cm from the lens).
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After refraction, it passes through \( F_2 \).
- **Ray 2**: Draw a ray from the top of the object passing through the optical center (O). This ray goes undeviated.
- **Image Formation**: The intersection of these two refracted rays will form a real, inverted, and same-sized image at \( 2F_2 \) (30 cm from the lens on the other side). The image should also be 5 cm high and inverted.
In simple words: For a convex lens with a 15 cm focal length, to get a real, upside-down image 30 cm away on the other side, the object must be placed 30 cm in front of the lens. If the object is 5 cm tall, the image will also be 5 cm tall and upside down. Draw a picture showing the object at \( 2F \) and the image also at \( 2F \) on the other side, both the same size.

Exam Tip: Remember that for a convex lens, when an object is placed at \( 2F_1 \), the image is formed at \( 2F_2 \), is real, inverted, and of the same size. This is a crucial case to remember for both calculations and diagrams.

 

Question 6. Redraw the given diagram and show the path of the refracted ray.
Answer: The diagram shows a convex lens with an incident ray passing through its principal focus (\( F_1 \)) on the left side. The path of the ray beyond the lens will be as follows:
- **Incident Ray**: Draw a ray of light approaching the convex lens, passing through the principal focus (\( F_1 \)) on the same side as the object.
- **Refracted Ray**: After striking the convex lens, this ray will refract and emerge parallel to the principal axis.
Ensure your diagram clearly shows the convex lens, the principal axis, and the focal points. The refracted ray should correctly emerge parallel to the principal axis.
In simple words: Draw the picture of the convex lens again. A light ray that goes through the focus on the first side will bend and come out straight, parallel to the main line.

Exam Tip: For convex lenses, a ray passing through \( F_1 \) always emerges parallel to the principal axis after refraction. This is one of the three standard principal rays.

 

Question 7. A convex lens has a focal length of 12 cm. At what distance from the lens should an object of height 6 cm be placed so that on the other side of the lens its real and inverted image is formed 24 cm away from the lens? What would be the size of the image formed? Draw a ray diagram to show the image formed in this case.
Answer: Given:
Focal length of convex lens, \( f = +12 \text{ cm} \)
Image distance, \( v = +24 \text{ cm} \) (real and inverted image formed on the other side)
Object height, \( h_o = 6 \text{ cm} \)

Using the lens formula:
\( \frac{ 1 }{ v } - \frac{ 1 }{ u } = \frac{ 1 }{ f } \)
\( \frac{ 1 }{ 24 } - \frac{ 1 }{ u } = \frac{ 1 }{ 12 } \)
\( - \frac{ 1 }{ u } = \frac{ 1 }{ 12 } - \frac{ 1 }{ 24 } \)
\( - \frac{ 1 }{ u } = \frac{ 2 - 1 }{ 24 } \)
\( - \frac{ 1 }{ u } = \frac{ 1 }{ 24 } \)
\( u = -24 \text{ cm} \)
The object should be placed at 24 cm in front of the convex lens (which is at \( 2F_1 \), since \( 2f = 24 \text{ cm} \)).

To find the size of the image (\( h_i \)), use the magnification formula:
\( m = \frac{ h_i }{ h_o } = \frac{ v }{ u } \)
\( \frac{ h_i }{ 6 } = \frac{ 24 }{ -24 } \)
\( \frac{ h_i }{ 6 } = -1 \)
\( h_i = -6 \text{ cm} \)
The image size is 6 cm. The negative sign indicates that the image is inverted. Thus, the image is real, inverted, and the same size as the object.

**Ray Diagram**: You need to draw a ray diagram for an object placed at \( 2F_1 \) of a convex lens:
- **Convex Lens and Axis**: Draw a convex lens with its principal axis, optical center (O), and focal points \( F_1, F_2 \) and \( 2F_1, 2F_2 \). Mark \( F_1 \) and \( 2F_1 \) at 12 cm and 24 cm respectively on the left, and \( F_2 \) and \( 2F_2 \) similarly on the right.
- **Object Placement**: Place the object (arrow, 6 cm high) at \( 2F_1 \) (24 cm from the lens).
- **Ray 1**: Draw a ray from the top of the object parallel to the principal axis. After refraction, it passes through \( F_2 \).
- **Ray 2**: Draw a ray from the top of the object passing through the optical center (O). This ray goes undeviated.
- **Image Formation**: The intersection of these two refracted rays will form a real, inverted, and same-sized image at \( 2F_2 \) (24 cm from the lens on the other side). The image should also be 6 cm high and inverted.
In simple words: For a convex lens with a 12 cm focal length, to get a real, upside-down image 24 cm away on the other side, the object must be placed 24 cm in front of the lens. If the object is 6 cm tall, the image will also be 6 cm tall and upside down. Draw a picture showing the object at \( 2F \) and the image also at \( 2F \) on the other side, both the same size.

Exam Tip: This is a classic \( 2F \) case for a convex lens. Always confirm that if \( u = 2f \), then \( v \) should also be \( 2f \), and \( h_i = h_o \). Pay attention to the signs in the lens formula and magnification.

 

Question 1. Amit visited a fair and saw a mirror in which he got a very funny image. The above part of his body was big in size, the middle part was of normal size and the lower part of the body showed a very small size. What kind of mirror is this?
Answer: This mirror is a combination of different types of mirrors arranged vertically:
- **Upper part**: Concave mirror (to produce an enlarged image of the upper body).
- **Middle part**: Plane mirror (to produce a normal-sized image of the middle part).
- **Lower part**: Convex mirror (to produce a diminished image of the lower part).
This type of mirror is often called a funhouse mirror.
In simple words: The funny mirror Amit saw was made of three different mirror types put together: a concave mirror for the top (made it look big), a flat mirror for the middle (made it look normal), and a convex mirror for the bottom (made it look small).

Exam Tip: Relate the image characteristics (enlarged, normal, diminished) to the specific mirror types (concave, plane, convex).

 

Question 2. Nidhi wanted the image of her pencil to be double the size of its original size. Name the mirror used for getting such an image.
Answer: To obtain an image that is double the size of the original object, a concave mirror must be used. Specifically, the object should be placed between the principal focus (F) and the center of curvature (C) of the concave mirror to achieve a real, inverted, and enlarged image, or between the pole (P) and F to achieve a virtual, erect, and enlarged image.
In simple words: Nidhi needs a concave mirror to make her pencil look twice as big.

Exam Tip: Remember that only a concave mirror can produce an enlarged image (either real or virtual).

 

Question 3. Give the mirror image of ‘AMBULANCE”
Answer: The mirror image of 'AMBULANCE' will be 'ECNALUBMA'. This is due to lateral inversion, where the left-right orientation is reversed in a plane mirror.
In simple words: If you look at 'AMBULANCE' in a mirror, it will read 'ECNALUBMA' because the mirror flips it from left to right.

Exam Tip: Understand lateral inversion as the side-to-side reversal of an image formed by a plane mirror. Practice writing words backward for such questions.

 

Question 4. An incident ray makes an angle of 60 degrees with the mirror. What is the angle of reflection?
Answer: The angle of incidence is defined as the angle between the incident ray and the normal to the mirror surface. If the incident ray makes an angle of 60 degrees with the mirror, then the angle of incidence (\( i \)) will be \( 90^\circ - 60^\circ = 30^\circ \).
According to the law of reflection, the angle of incidence is always equal to the angle of reflection (\( r \)).
Therefore, \( r = i = 30^\circ \).
The angle of reflection is 30 degrees.
In simple words: The light ray hits the mirror at a 60-degree angle to the surface. But we measure the angle from an imaginary straight line called the normal, which is perpendicular to the mirror. So, the angle of incidence is \( 90 - 60 = 30 \) degrees. Because the light bounces off at the same angle it hit, the angle of reflection is also 30 degrees.

Exam Tip: Always remember to measure angles of incidence and reflection with respect to the normal, not the mirror surface itself. This is a common point of confusion.

 

Question 5. Define the following.
(a) What is a ray?
(b) What is the beam?
(c) What is the reflection of light?
(d) What is a reflector?
(e) What is the focal length?
(f) What is the principal focus?
(g) What is refraction?
(h) What is an optically rare medium?
(i) What is an optically denser medium?
(j) What is power?
(k) What is 1 dioptre?
Answer:
(a) **Ray**: A ray is the path along which light energy travels. It is generally represented by a straight line with an arrow indicating the direction of light.
(b) **Beam**: A beam is a group or collection of parallel light rays emitted from a source of light.
(c) **Reflection of light**: It is the phenomenon of light bouncing back into the same medium after striking a smooth and polished surface.
(d) **Reflector**: A reflector is any surface that reflects light. Mirrors are common examples of reflectors.
(e) **Focal length**: For a spherical mirror or lens, the focal length is the distance between the pole (or optical center) and the principal focus.
(f) **Principal focus**: The principal focus (F) is a point on the principal axis where rays of light parallel to the principal axis converge after reflection from a mirror or refraction through a lens (or appear to diverge from it).
(g) **Refraction**: Refraction is the bending of light rays as they pass from one medium to another, caused by a change in the speed of light.
(h) **Optically rare medium**: An optically rare medium is a medium in which the speed of light is faster compared to another medium. It has a lower refractive index.
(i) **Optically denser medium**: An optically denser medium is a medium in which the speed of light is slower compared to another medium. It has a higher refractive index.
(j) **Power**: The power of a lens is the degree of convergence or divergence of light rays achieved by the lens. It is the reciprocal of its focal length in meters.
(k) **1 dioptre**: One dioptre (1D) is the power of a lens whose focal length is 1 meter.
In simple words:
(a) A ray is the path light takes, like a straight line.
(b) A beam is a bunch of these light rays together.
(c) Reflection is when light bounces off a surface.
(d) A reflector is anything that bounces light.
(e) Focal length is the distance from the mirror/lens center to its focus point.
(f) Principal focus is the special point where parallel light rays meet (or seem to meet) after hitting a mirror or lens.
(g) Refraction is when light bends as it goes from one material to another.
(h) A rare medium is where light moves faster.
(i) A denser medium is where light moves slower.
(j) Power tells us how much a lens bends light.
(k) 1 dioptre is the power of a lens that has a focal length of 1 meter.

Exam Tip: For definitions, be concise and use key terms accurately. Distinguish clearly between reflection and refraction, and between focal length and principal focus.

 

Question 6. What are the two types of reflection?
Answer: The two types of reflection are:
(i) **Regular reflection**: This occurs when light reflects off a very smooth surface (like a plane mirror) and the reflected rays remain parallel, forming a clear image.
(ii) **Irregular (or diffuse) reflection**: This happens when light reflects off a rough or uneven surface, causing the reflected rays to scatter in various directions, which does not form a clear image.
In simple words: There are two kinds of reflection: regular, which happens on shiny, smooth surfaces and makes clear images; and irregular, which happens on rough surfaces and makes light scatter everywhere.

Exam Tip: Provide a brief description or an example for each type of reflection to ensure a complete answer.

 

Question 7. Write the laws of reflection.
Answer: The laws of reflection are:
(a) The angle of incidence is equal to the angle of reflection.
(b) The incident ray, the normal to the mirror at the point of incidence, and the reflected ray, all lie in the same plane.
In simple words: Light bounces off surfaces by following two rules: first, the angle at which it hits is the same as the angle at which it bounces off. Second, the incoming light, the bounced light, and a special imaginary line (the normal) are all on the same flat surface.

Exam Tip: Clearly state both laws without omitting any details. Make sure to mention "same plane" for the second law.

 

Question 8. Give characteristics of the image formed by a plane mirror
Answer: A plane mirror always creates a virtual and erect image. The size of the image is equal to the object's size. Also, the image appears at the same distance behind the mirror as the object is placed in front of it.
In simple words: A plane mirror makes an image that looks upright and appears behind the mirror. The image is the same size as the real object.

Exam Tip: Remember these three key properties for plane mirrors: virtual and erect, same size, and equidistant from the mirror.

 

Question 9. Give uses of a plane mirror.
Answer: Plane mirrors are commonly used as:

• Looking glasses for daily use.
• Optical tools in submarines, such as periscopes.
• Components in solar cookers to reflect light.
• Parts of kaleidoscopes to create beautiful patterns.

In simple words: Plane mirrors are used as regular mirrors, in periscopes for submarines, in solar cookers, and in kaleidoscopes.

Exam Tip: List a variety of applications, from everyday items to scientific instruments, to show comprehensive knowledge.

 

Question 10. Name two types of spherical mirrors.
Answer: The two main types of spherical mirrors are:
1. Concave mirror
2. Convex mirror
In simple words: Spherical mirrors are either concave (curved inward) or convex (curved outward).

Exam Tip: Understand the basic shapes of spherical mirrors as concave and convex, as their properties differ significantly.

 

Question 11. Give uses of a concave mirror.
Answer: Concave mirrors are used in various ways:

• As reflectors in vehicle headlights, searchlights, and torches to produce powerful parallel beams of light.
• As shaving mirrors because they give a larger image of the face.
• By dentists to view larger images of patients' teeth.
• In solar cookers and solar furnaces to concentrate sunlight at a single point, generating high temperatures.

In simple words: Concave mirrors are used in headlights, shaving mirrors, by dentists, and in solar cookers because they can focus light or make objects appear bigger.

Exam Tip: When listing uses, briefly explain *why* a concave mirror is suitable for each application (e.g., focusing light, magnifying images).

 

Question 12. Give uses of a convex mirror.
Answer: Convex mirrors are mainly used in:

• Rear-view mirrors in cars, which give a virtual image, reduce the size of objects, and offer a wider view of the rear traffic.
• Street lights, where they diverge light over a larger area to illuminate a broader region.

In simple words: Convex mirrors are used as car rearview mirrors to see more of the road behind, and in street lights to spread light widely.

Exam Tip: Focus on the two main characteristics of convex mirrors - wider field of view and diminished, erect image - when explaining their uses.

 

Question 13. What are the two types of refractive index?
Answer: The two primary types of refractive index are:
Relative refractive index - This is the ratio comparing the speed of light in one medium to its speed in another medium.
Absolute refractive index - This is the ratio comparing the speed of light in a vacuum to its speed in a specific medium.
In simple words: There's relative refractive index (compares light speed between two materials) and absolute refractive index (compares light speed in a material to light speed in empty space).

Exam Tip: Differentiate clearly between relative and absolute refractive indices by mentioning the reference medium for each (another medium vs. vacuum).

 

Question 14. Why do we prefer a convex mirror as a rearview mirror in vehicles?
Answer: We prefer convex mirrors for rear-view in vehicles because they produce an erect (upright) and diminished (smaller) image of the traffic behind the vehicle. This also gives a wider field of view, allowing the driver to see more of what is happening behind.
In simple words: Convex mirrors are chosen for car rear-view because they show an upright, smaller image and let you see a wider area behind the car.

Exam Tip: Highlight the "wider field of view" and "erect and diminished image" as the key reasons for using convex mirrors in vehicles.

 

Question 15. Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Answer:
(a) Concave mirror
(b) Convex mirror
(c) Concave mirror
In simple words: Headlights use concave mirrors, car side mirrors use convex mirrors, and solar furnaces use concave mirrors.

Exam Tip: Match each application with the correct mirror type based on the mirror's ability to converge, diverge, magnify, or diminish light.

 

Question 16. The magnification produced by a plane mirror is +1. What does this mean?
Answer: A magnification of \( +1 \) for a plane mirror means two things. The \( + \) sign indicates that the image formed is virtual and erect. The numerical value \( 1 \) shows that the image size is exactly equal to the object's size.
In simple words: \( +1 \) magnification from a plane mirror means the image is upright, not real, and the same size as the object.

Exam Tip: Clearly state what both the sign (positive/negative) and the magnitude of magnification indicate about the image.

 

Question 17. Find the focal length of a lens of power - 2.0 D. What type of lens is this?
Answer: Given, Power \( P = -2.0 \, D \).
The relationship between power and focal length is given by:
\( P = \frac {1}{f (in \, metre)} \)
So, \( -2.0 = \frac {1}{f} \)
This means, \( f = \frac {1}{-2.0} \, m \)
\( f = -0.5 \, m \)
Converting to centimeters: \( f = -50 \, cm \).
Since the focal length is negative, this type of lens is a concave lens.
In simple words: If a lens has a power of -2.0 D, its focal length is -50 cm. A negative focal length always means it is a concave lens.

Exam Tip: Remember that a negative focal length indicates a concave (diverging) lens, while a positive focal length indicates a convex (converging) lens. Always include units.

Practical Based Questions

 

Question 1. A student finds the focal length of the convex lens and records it as given below: 10 cm, 10.6 cm, 10.5 cm, and 10.5 cm. Which reading he should discard and why?
Answer: The focal length of a convex lens is defined as the distance between the lens and the image of a distant object. This value should be constant for a specific lens. Therefore, the reading \( 10 \, cm \) should be discarded because it is significantly different from the other three values (\( 10.6 \, cm, 10.5 \, cm, 10.5 \, cm \)) which are close to each other. Accurate readings should show good consistency.
In simple words: The student should ignore the 10 cm reading because it's too far off from the other, more consistent, readings of 10.5 cm and 10.6 cm. Focal length should be steady for one lens.

Exam Tip: In experiments, always look for consistency in readings. Values that deviate significantly from a cluster of consistent readings are usually outliers and should be re-evaluated or discarded.

 

Question 2. Suggest how can one avoid the errors during measuring the focal length of the given lens.
Answer: To avoid errors when measuring the focal length of a lens, several steps can be taken:

• Ensure the distant object used is truly distant enough so that light rays arriving at the lens are effectively parallel. If the object is too close, the focal length measurement may not be accurate.
• For accurate results, keep the distant object constant and take multiple readings (at least three). If these readings are very close to each other (e.g., within \( \pm 0.1 \, cm \)), they demonstrate precision, and their average can be used.
• Make sure the lens is perfectly perpendicular to the principal axis and the screen is adjusted carefully to get the sharpest possible image.

In simple words: To get better focal length results, use a very distant object, take several measurements, and make sure the lens and screen are set up correctly.

Exam Tip: Emphasize precision (multiple readings, consistent results) and proper experimental setup (distant object, alignment) as key to reducing errors.

 

Question 3. A student wants to find the refractive index of the glass slab given to him. What are the materials required to study the same in the lab? Give the formula to calculate the refractive index.
Answer: To find the refractive index of a glass slab in the lab, a student would need the following materials: the glass slab itself, four to six pins, a pencil, a protractor, a soft board, and a white sheet of paper with four board pins. The formula used to calculate the refractive index is based on Snell's Law:
\( n = \frac {sin \, i}{sin \, r} = Constant \)
Where \( n \) is the refractive index, \( i \) is the angle of incidence, and \( r \) is the angle of refraction.
In simple words: To measure a glass slab's refractive index, you need the slab, pins, a pencil, a protractor, and paper. You find it by dividing the sine of the angle of light entering by the sine of the angle of light bending inside.

Exam Tip: Be precise with the list of materials and correctly state Snell's Law, identifying the variables \( i \) and \( r \).

 

Question 4. Explain how practically you can identify the concave lens and convex lens in the lab. Justify your answer.
Answer: To practically identify a concave lens and a convex lens, one can try to obtain a real focus of the lens by focusing a distant object onto a screen. The lens that successfully forms a clear, focused image of a distant object on a screen is the convex lens. This is because a convex lens is a converging lens; it brings all parallel light rays from a distant source to meet at a single point, which is its principal focus. Conversely, the lens that does not produce a real, focused image on a screen is the concave lens. A concave lens is a diverging lens; it spreads out parallel light rays, making them appear to come from a virtual focus, and thus cannot form a real image on a screen.
In simple words: You can tell a convex lens from a concave one by trying to make a sharp image of something far away on a screen. The convex lens will make a clear image because it brings light together, but the concave lens won't because it spreads light out.

Exam Tip: The key differentiator is the ability to form a *real* image. Emphasize convergence for convex lenses and divergence for concave lenses when justifying the identification method.

 

Question 5. If you have to construct a microscope to view the minute objects and magnify them, suggest how will you do it practically and what is the material required for the same.
Answer: To prepare a simple microscope, you would generally require two convex lenses of different focal lengths. One lens, typically with a shorter focal length, serves as the objective lens, placed close to the object being viewed. The other lens, with a longer focal length, is the eyepiece, through which you observe. The setup involves placing the objective lens so that it forms a magnified real image of the small object. This real image then acts as the object for the eyepiece, which is positioned to form a further magnified virtual image. This arrangement helps in magnifying the image to a greater extent, allowing the viewing of minute objects.
In simple words: To build a simple microscope, you need two convex lenses, one with a short focal length for the object and one with a longer focal length for your eye. These lenses work together to make small things look much bigger.

Exam Tip: Specify that two convex lenses with different focal lengths are needed for a simple microscope and briefly explain the role of each lens (objective and eyepiece).

 

Question 6. The below diagram is used to study how light travels through a glass slab.
(a) On the figure draw the normal rayon side AB.
(b) Join dots P3 and P4, draw normal at side CD, and measure the angle formed here.
(d) At the side, AB draw angle of incidence with 30 degrees.
Answer:
(a) To draw the normal on side AB, place a protractor at the point where the incident ray strikes the glass slab's surface AB. Draw a line perpendicular to the surface AB at that point. This line represents the normal.
(b) First, connect the dots \( P_3 \) and \( P_4 \) to draw the emergent ray. Then, at the point where the emergent ray leaves the glass slab (on side CD), draw a line perpendicular to the surface CD. This is the normal at the point of emergence. Use the protractor to measure the angle between the emergent ray and this normal, which is the angle of emergence.
(d) To draw the angle of incidence at \( 30 \) degrees on side AB, first draw a normal at the point of incidence on surface AB as described in part (a). Then, using a protractor, draw an incident ray such that the angle between this ray and the normal is \( 30 \) degrees. Ensure the ray enters the slab at this angle.
In simple words:
(a) Draw a straight line at 90 degrees to surface AB where the light hits it.
(b) Connect \( P_3 \) and \( P_4 \) to show light coming out. Then, draw a 90-degree line at that exit point on CD and measure the angle the light makes with it.
(d) Draw the incoming light ray so it makes a 30-degree angle with the 90-degree line you drew on side AB.

Exam Tip: For ray diagrams, ensure normals are drawn perpendicular to the surface at the point of incidence/emergence. Use a protractor for accurate angle measurements, and label all rays and angles clearly.