Chapter 10 – Light Reflection and Refraction Question Answers

Let’s make learning about light reflection and refraction fun! 🌈 Intext questions are like little quizzes that pop up while you’re reading to check your understanding. 🤔 Exercise questions are like homework where you get to practice what you’ve learned.

Intext Question Answers of Light Reflection and Refraction

1. Define the principal focus of a concave mirror. 

Ans: The principal focus (or focal point) of a concave mirror is the point where parallel rays of light, converging towards the mirror, meet after reflection. It lies on the principal axis of the mirror, in front of it.

2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length? 

Ans: The focal length fff of a spherical mirror is half the radius of curvature RRR. Given the radius of curvature is 20 cm:

► f = R/2 = 20/2 cm = 10 cm

3. Name a mirror that can give an erect and enlarged image of an object. 

Ans: A concave mirror can produce an erect and enlarged image when the object is placed between its pole and focus

4. Why do we prefer a convex mirror as a rear-view mirror in vehicles? 

Convex mirrors are preferred as rear-view mirrors in vehicles because they provide a wider field of view. They reflect light outward, which helps in viewing a larger area behind the vehicle, even though the images are smaller and appear further away.

1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answers: The focal length f of a convex mirror is related to its radius of curvature R by the formula:

F = −2R​

For a convex mirror with a radius of curvature of 32 cm:

F = −32/2 cm = −16 cm

(The negative sign indicates that the focal length is virtual and located behind the mirror.)

2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

To find the location of the image formed by a concave mirror, we can use the magnification formula and the mirror equation.

Given:

• Magnification, m=−3 (Since the image is real and magnified, the magnification is negative).
• Object distance, u=−10 cm (The object distance is negative because it is placed in front of the mirror).

1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why? 

Ans: When a ray of light travels from air (a less dense medium) into water (a denser medium) obliquely, the light ray bends towards the normal. This bending occurs due to the decrease in speed of light as it enters the denser medium (water). The refractive index of water is higher than that of air, causing the light to slow down and bend towards the normal, according to Snell’s law.

Snell’s Law is a rule that tells us how much the light will bend. It depends on the type of material the light is entering (like air, water, or glass) and the angle at which it hits the surface.

2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m s^-1 . 

3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density. 

The optical density of a medium is related to its refractive index; the higher the refractive index, the higher the optical density.

• Highest Optical Density: From the given table, Diamond has the highest refractive index of 2.42. Hence, diamond has the highest optical density.
• Lowest Optical Density: Air has the lowest refractive index of 1.0003. Thus, air has the lowest optical density.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3. 

The speed of light in a medium is inversely proportional to its refractive index. The lower the refractive index, the faster light travels through the medium.

• Refractive index of kerosene = 1.44
• Refractive index of turpentine oil = 1.47
• Refractive index of water = 1.33

Since water has the lowest refractive index (1.33) among kerosene, turpentine oil, and water, light travels fastest in water.

5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answers: The refractive index of diamond being 2.42 means that light travels 2.42 times slower in diamond than it does in a vacuum (or air). This also implies that diamond is highly optically dense, causing significant bending of light as it enters or exits the diamond from another medium.

1. Define 1 dioptre of power of a lens. 

The power of a lens is measured in dioptres (D). One dioptre is defined as the power of a lens with a focal length of 1 meter. In other words, if a lens has a focal length of 1 meter, its power is 1 dioptre. The formula for power P is:

P = 1/f 

2. 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. 

Understanding the Problem

• We have a convex lens.
• It forms a real and inverted image of a needle.
• The image distance (v) is 50 cm.
• The image size is equal to the object size.

Solution

1. Finding the Object Distance (u):

When the image size is equal to the object size, the object is placed at twice the focal length (2f) of the lens.

Since the image distance (v) is equal to 2f, we can find the focal length (f):

• v = 2f
• f = v/2 = 50 cm / 2 = 25 cm

So, the needle is placed at a distance of 25 cm in front of the convex lens.

2. Finding the Power of the Lens:

Power of a lens (P) is the reciprocal of its focal length (f) in meters.

• f = 25 cm = 0.25 m
• P = 1/f = 1 / 0.25 = 4 Diopters

Therefore, the needle is placed 25 cm in front of the convex lens, and the power of the lens is 4 Diopters.

3. Find the power of a concave lens of focal length 2 m.

The power of a concave lens is -1/f, where f is the focal length of the lens in meters.

Steps to solve:

► Substitute the focal length value:

The focal length of the concave lens is 2 m, so we substitute this value into the formula:

P = -1/2

► Evaluate the expression:

P = -1/2 = -0.5

Therefore, the power of the concave lens is -0.5 diopters.

Exercise Question of Light, Reflection, and Refraction 

1. Which one of the following materials cannot be used to make a lens?

(a) Water (b) Glass (c) Plastic (d) Clay 

Ans: Clay 

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 centre of curvature 
• (b) At the centre of curvature 
• (c) Beyond the centre of curvature 
• (d) Between the pole of the mirror and its principal focus. 

Ans: (d) Between the pole of the mirror and its principal focus.

3. Where should an object be placed in front of a convex lens to get a real image of the size of the object? 

• (a) At the principal focus of the lens 
• (b) At twice the focal length 
• (c) At infinity 
• (d) Between the optical centre of the lens and its principal focus. 

Ans: (b) At twice the focal length

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 is convex. 
• (d) the mirror is convex, but the lens is concave. 

Ans: (a) both concave.

Explanation:

• Concave mirrors have a negative focal length.
• Concave lenses also have a negative focal length.

5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be 

• (a) only plane. 
• (b) only concave. 
• (c) only convex. 
• (d) either plane or convex.

Ans: (d) either plane or convex.

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.

Explanation:

• Convex lenses are used as magnifying glasses.
• A shorter focal length results in a higher magnification.

Therefore, a convex lens with a focal length of 5 cm would provide the best magnification for reading small letters in a dictionary.

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. 

Ans: To obtain an erect image with a concave mirror, the object must be placed between the pole (P) and the focus (F) of the mirror. Since the focal length (f) is given as 15 cm, the object distance (u) should be less than 15 cm.

Nature of the Image: The image formed in this case is virtual, erect, and magnified.

8. Name the type of mirror used in the following situations. Support your answer with reason. 

(a) Headlights of a car. 

• Type of mirror: Concave mirror
• Reason: Concave mirrors are used to converge light rays. When the bulb is placed at the focus of a concave mirror, the light rays are reflected as a parallel beam, providing a long-range and intense beam of light, essential for car headlights.

(b) Side/rear-view mirror of a vehicle

• Type of mirror: Convex mirror
• Reason: Convex mirrors diverge light rays, providing a wider field of view. This allows the driver to see a larger area behind the vehicle, reducing blind spots and enhancing safety.

(c) Solar furnace

• Type of mirror: Concave mirror
• Reason: Concave mirrors concentrate sunlight at a focal point, creating high temperatures. This concentrated solar energy is used to heat substances in a solar furnace.

9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally.

Answer: Yes, the lens will produce a complete image of the object. When one-half of a convex lens is covered with black paper, it still produces a complete image of the object because the uncovered half of the lens still refracts light rays from all parts of the object. However, the intensity of the image will be reduced due to the decrease in the number of light rays passing through the lens.

Experimental Verification:

• Set up a convex lens and cover one-half with black paper.
• Place an object (like a candle) in front of the lens and adjust the screen until a clear image is formed.
• You will observe that the image is complete but dimmer compared to when the entire lens is uncovered.

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 the nature of the image formed. 

Ray Diagram and Image Formation

Given:

• Object height (h₀) = 5 cm
• Object distance (u) = -25 cm (negative due to convention)
• Focal length (f) = 10 cm

Using the lens formula: 

► 1/v – 1/u = 1/f 

► 1/v – 1/(-25) = 1/10 

► 1/v + 1/25 = 1/10 

► 1/v = 1/10 – 1/25 

► 1/v = 3/50 

► v = 50/3 ≈ 16.67 cm

Image distance (v): 16.67 cm (positive, so the image is real)

Magnification (m): 

► m = v/u 

► m = -(16.67)/(-25) 

► m ≈ – 0.67

Image height (h₁): 

► h₁ = – m * h₀ 

► h₁ = – 0.67 * 5 

► h₁ ≈ – 3.33 cm

Nature of the image:

• Since the magnification is positive, the image is upright.
• Since the image distance is positive, the image is real.
• Since the magnification is less than 1, the image is smaller than the object.

Concluded Answer: The image formed is a real, inverted (-sign of h₁ ), and smaller image located at a distance of approximately 16.67 cm from the lens. The image height is approximately 3.33 cm.

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. 

Given:

• Focal length (f) = -15 cm (negative due to concave lens)
• Image distance (v) = -10 cm (negative due to convention)

The object distance can be calculated using the lens formula:

 ► 1/v – 1/u = 1/f

where:

• v = image distance = -10 cm (negative sign indicates virtual image)
• u = object distance (to be found)
• f = focal length = -15 cm (negative sign indicates concave lens)

Substituting the values in the formula, we get:

• 1/(-10) – 1/u = 1/(-15)
• 1/u = 1/(-15) + 1/10
• 1/u = -1/30
• u = -30 cm

Therefore, the object is placed 30 cm from the lens.

Nature of the image:

• Since the image distance is negative, the image is virtual.
• Since the focal length is negative, the lens is concave.
• The image formed by a concave lens is always smaller than the object.

Concluded Answers: The object is placed at a distance of 6 cm from the concave lens. The image formed is a virtual, upright, and smaller image located 10 cm from the lens.

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. 

Given:

• Object distance (u) = -10 cm (negative due to convention)
• Focal length (f) = 15 cm

Using the mirror formula: 

► 1/v + 1/u = 1/f 

► 1/v + 1/(-10) = 1/15 

► 1/v – 1/10 = 1/15 

► 1/v = 1/15 + 1/10 

► 1/v = 1/6 

► v = 6 cm

Image distance (v): 6 cm (positive, so the image is virtual)

Nature of the image:

• Since the image distance is positive, the image is virtual.
• Convex mirrors always produce upright images.
• Since the image distance is smaller than the object distance, the image is smaller than the object.

Conclusion: The image formed is a virtual, upright, and smaller image located at a distance of 6 cm from the convex mirror.

13. The magnification produced by a plane mirror is +1. What does this mean? 

Magnification (m) = +1 for a plane mirror means:

• Upright image: The positive sign indicates that the image is oriented the same way as the object (upright).
• Same size: The magnitude of 1 means the image is the same size as the object.

In other words, a plane mirror produces a virtual, upright, and same-sized image of the object. This is why plane mirrors are commonly used for personal grooming and as a simple way to see oneself.

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. 

Solutions: Given:

• Object height (h₀) = 5.0 cm
• Object distance (u) = -20 cm (negative due to convention)
• Radius of curvature (R) = 30 cm

Step 1: Find the focal length (f)

• f = R/2 = 30 cm / 2 = 15 cm

Step 2: Use the mirror formula to find the image distance (v)

• 1/v + 1/u = 1/f
• 1/v + 1/(-20) = 1/15
• 1/v = 1/15 + 1/20
• 1/v = 7/60
• v = 60/7 ≈ 8.57 cm

Step 3: Find the magnification (m)

• m = -v/u
• m = -8.57/(-20)
• m ≈ 0.428

Step 4: Find the image height (h₁)

• h₁ = m * h₀
• h₁ = 0.428 * 5.0 cm
• h₁ ≈ 2.14 cm

Conclusion:

• Position of the image: 8.57 cm behind the convex mirror
• Nature of the image: Virtual, upright (due to the positive magnification)
• Size of the image: 2.14 cm (smaller than the object)

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 sharp focussed image can be obtained? Find the size and the nature of the image. 

Given:

• Object height (h₀) = 7.0 cm
• Object distance (u) = -27 cm (negative due to convention)
• Focal length (f) = -18 cm (negative for a concave mirror)

Step 1: Find the image distance (v) using the mirror formula:

• 1/v + 1/u = 1/f
• 1/v + 1/(-27) = 1/(-18)
• 1/v = 1/(-18) – 1/(-27)
• 1/v = -1/54
• v = -54 cm

Step 2: Find the magnification (m):

• m = -v/u
• m = -54/(-27)
• m = 2

Step 3: Find the image height (h₁):

• h₁ = m * h₀
• h₁ = 2 * 7.0 cm
• h₁ = 14 cm

Conclusion:

• Screen placement: The screen should be placed at a distance of 54 cm from the mirror.
• Image size: 14 cm
• Image nature: Real, inverted (due to the negative magnification)

16. Find the focal length of a lens of power – 2.0 D. What type of lens is this? 

Given:

• Power (P) = -2.0 D

Formula:

• Power (P) = 1/f (where f is the focal length in meters)

Calculation:

• f = 1/P
• f = 1/(-2.0)
• f = -0.5 m

Conclusion:

• The focal length of the lens is -0.5 meters.
• Since the focal length is negative, the lens is concave.

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?

Given:

• Power (P) = +1.5 D

Calculation:

• f = 1/P
• f = 1/(+1.5)
• f = 0.67 m

Conclusion:

• The focal length of the lens is 0.67 meters.
• Since the focal length is positive, the lens is convex.

Therefore, the doctor has prescribed a converging lens.