---
title: "NCERT Solutions Class 10 Science Light – Reflection and Refraction"
url: https://www.swavid.com/science/class/10/chapter/light-reflection-and-refraction/ncert-solutions
dateModified: 2026-10-07T16:46:48+00:00
---

# NCERT Solutions Class 10 Science Light – Reflection and Refraction

This chapter's questions cover concepts of reflection and refraction of light, including spherical mirrors, lenses, mirror and lens formulas, magnification, refractive index, and related numerical problems.

Free PDF (13 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-10/swavid-ncert-solutions-class-10-science-chapter-9-light-reflection-and-refraction-01879a4d9b.pdf

## QUESTIONS

### Question 1

*2 marks · Very short answer*

Define the principal focus of a concave mirror.

**Solution**

1. Light rays parallel to the principal axis falling on a concave mirror meet or intersect at a point on the principal axis after reflection.
2. This point on the principal axis is called the principal focus of the concave mirror.

**Answer:** The principal focus of a concave mirror is a point on its principal axis where all rays parallel to the principal axis converge after reflection from the mirror.

> Common mistake: Writing that rays diverge instead of converging for a concave mirror.

### Question 2

*3 marks · Numerical*

The radius of curvature of a spherical mirror is $20\text{ cm}$. What is its focal length?

**Solution**

1. Given: Radius of curvature, $R = 20\text{ cm}$
2. Formula: $f = \frac{R}{2}$
3. Substitution: $f = \frac{20\text{ cm}}{2}$
4. Result: $f = 10\text{ cm}$

**Answer:** $10\text{ cm}$

> Common mistake: Forgetting to include the unit or multiplying instead of dividing by 2.

### Question 3

*2 marks · Very short answer*

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

**Solution**

1. A concave mirror forms an erect and enlarged image of an object when the object is placed between the pole and the principal focus.
2. Thus, a concave mirror is the mirror that can give an erect and enlarged image of an object.

**Answer:** Concave mirror

> Common mistake: Naming a convex mirror, which always gives a diminished image.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Convex mirrors are preferred as rear-view mirrors in vehicles because they always give an erect, though diminished, image of the traffic behind.
2. They have a wider field of view as they are curved outwards, which enables the driver to view a much larger area than would be possible with a plane mirror.

**Answer:** Convex mirrors are preferred because they always provide erect images and have a wider field of view, enabling drivers to see a large area of traffic behind them.

> Common mistake: Stating that they form enlarged images instead of diminished images with a wider field of view.

## QUESTIONS

### Question 1

*3 marks · Numerical*

Find the focal length of a convex mirror whose radius of curvature is $32\text{ cm}$.

**Solution**

1. Given: Radius of curvature, $R = +32\text{ cm}$
2. Formula: $f = \frac{R}{2}$
3. Substitution: $f = \frac{32\text{ cm}}{2}$
4. Result: $f = +16\text{ cm}$

**Answer:** +16 cm

> Common mistake: Forgetting that the focus of a convex mirror is behind the mirror, hence its focal length is positive.

### Question 2

*3 marks · Numerical*

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

**Solution**

1. Given: Object-distance, $u = -10\text{ cm}$
2. Magnification, $m = -3$ (since the image is real and magnified)
3. Formula: $m = -\frac{v}{u}$
4. Substitution: $-3 = -\frac{v}{-10}$
5. Result: $v = -30\text{ cm}$

**Answer:** $30\text{ cm}$ in front of the mirror

> Common mistake: Forgetting the negative sign for real image magnification.

## QUESTIONS

### Question 1

*3 marks · Short answer*

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?

**Solution**

1. Light travels from air, which is an optically rarer medium, into water, which is an optically denser medium.
2. When a ray of light travels obliquely from a rarer medium to a denser medium, its speed decreases.
3. Therefore, the light ray bends towards the normal.

**Answer:** The light ray bends towards the normal because water is optically denser than air.

> Common mistake: Confusing mass density with optical density.

### Question 2

*3 marks · Numerical*

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 $3 \times 10^8\text{ m s}^{-1}$.

**Solution**

1. Given: Refractive index of glass $n_g = 1.50$, speed of light in vacuum $c = 3 \times 10^8\text{ m s}^{-1}$.
2. Formula: $n_g = \frac{c}{v}$, where $v$ is the speed of light in glass.
3. Substitution: $1.50 = \frac{3 \times 10^8\text{ m s}^{-1}}{v}$
4. Result: $v = \frac{3 \times 10^8}{1.50} = 2 \times 10^8\text{ m s}^{-1}$

**Answer:** $2 \times 10^8\text{ m s}^{-1}$

> Common mistake: Dividing the speed of light by the wrong factor.

### Question 3

*2 marks · Very short answer*

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

**Solution**

1. The medium with the highest optical density has the highest refractive index, which is diamond ($2.42$).
2. The medium with the lowest optical density has the lowest refractive index, which is air ($1.0003$).

**Answer:** Diamond has the highest optical density, and air has the lowest optical density.

> Common mistake: Looking at mass density instead of refractive index.

### Question 4

*2 marks · Very short answer*

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

**Solution**

1. From Table 9.3, the refractive indices are: water $= 1.33$, kerosene $= 1.44$, and turpentine oil $= 1.47$.
2. Since the speed of light is inversely proportional to the refractive index, light travels fastest in the medium with the lowest refractive index.

**Answer:** Light travels fastest in water because it has the lowest refractive index among the given media.

> Common mistake: Assuming light travels faster in higher refractive index media.

### Question 5

*2 marks · Very short answer*

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

**Solution**

1. The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium.
2. A refractive index of $2.42$ for diamond means that the speed of light in diamond is $\frac{1}{2.42}$ times the speed of light in vacuum.

**Answer:** The ratio of the speed of light in vacuum to the speed of light in diamond is equal to $2.42$.

> Common mistake: Stating it as speed in diamond is $2.42$ times that in vacuum.

## QUESTIONS

### Question 1

*2 marks · Very short answer*

Define $1\text{ dioptre}$ of power of a lens.

**Solution**

1. 1 dioptre is defined as the power of a lens whose focal length is 1 metre.
2. It is expressed as $1\text{ D} = 1\text{ m}^{-1}$.

**Answer:** 1 dioptre is the power of a lens having a focal length of 1 metre.

> Common mistake: Forgetting the unit relation or stating focal length in cm instead of metres.

### Question 2

*3 marks · Numerical*

A convex lens forms a real and inverted image of a needle at a distance of $50\text{ cm}$ 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.

**Solution**

1. Given: Image distance $v = +50\text{ cm} = +0.5\text{ m}$, magnification $m = -1$ (since the image is real, inverted, and equal in size to the object).
2. Formula: For a lens, magnification $m = \frac{v}{u}$, so $-1 = \frac{0.5}{u}$, which gives $u = -0.5\text{ m}$ (or $-50\text{ cm}$). Thus, the needle is placed at $2f$, so $2f = 50\text{ cm}$, giving focal length $f = +25\text{ cm} = +0.25\text{ m}$.
3. Formula: Power of a lens $P = \frac{1}{f(\text{in metres})}$.
4. Substitution: $P = \frac{1}{+0.25\text{ m}} = +4.0\text{ D}$.
5. Result: The needle is placed at $50\text{ cm}$ in front of the lens, and the power of the lens is $+4.0\text{ D}$.

**Answer:** The needle is placed at $50\text{ cm}$ in front of the lens, and the power of the lens is $+4.0\text{ D}$.

> Common mistake: Forgetting the negative sign for object distance or magnification for a real image.

### Question 3

*3 marks · Numerical*

Find the power of a concave lens of focal length $2\text{ m}$.

**Solution**

1. Given: Focal length of the concave lens $f = -2\text{ m}$ (focal length of a concave lens is negative).
2. Formula: Power $P = \frac{1}{f(\text{in metres})}$.
3. Substitution: $P = \frac{1}{-2\text{ m}}$.
4. Result: $P = -0.5\text{ D}$.

**Answer:** -0.5 D

> Common mistake: Omitting the negative sign for the focal length and power of a concave lens.

## EXERCISES

### Question 1

*1 mark · MCQ*

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

- Water
- Glass
- Plastic
- Clay

**Solution**

1. A lens requires a transparent medium for light to pass through and refract.
2. Water, glass, and plastic are transparent and can be used to make lenses, whereas clay is opaque.

**Answer:** (d) Clay

> Common mistake: Confusing transparent materials with opaque ones.

### Question 2

*1 mark · MCQ*

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?

- Between the principal focus and the centre of curvature
- At the centre of curvature
- Beyond the centre of curvature
- Between the pole of the mirror and its principal focus.

**Solution**

1. A concave mirror forms a virtual and erect image only when the object is placed very close to it.
2. Specifically, when the object is placed between the pole and the principal focus of the concave mirror, the image formed is virtual, erect, and enlarged.

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

> Common mistake: Choosing real image positions instead of virtual and erect ones.

### Question 3

*1 mark · MCQ*

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

- At the principal focus of the lens
- At twice the focal length
- At infinity
- Between the optical centre of the lens and its principal focus.

**Solution**

1. According to Table 9.4 in the textbook, a convex lens forms a real, inverted image of the same size as the object when the object is placed at $2F_1$.
2. In terms of focal length, this position corresponds to twice the focal length from the optical centre.

**Answer:** (b) At twice the focal length

> Common mistake: Confusing the condition for same-size image ($2F$) with the focus ($F$) or infinity.

### Question 4

*1 mark · MCQ*

A spherical mirror and a thin spherical lens have each a focal length of $-15\text{ cm}$. The mirror and the lens are likely to be

- both concave.
- both convex.
- the mirror is concave and the lens is convex.
- the mirror is convex, but the lens is concave.

**Solution**

1. A concave mirror has a negative focal length according to the New Cartesian Sign Convention.
2. A concave lens also has a negative focal length.
3. Therefore, both the spherical mirror and the thin spherical lens with a focal length of $-15\text{ cm}$ are concave.

**Answer:** (a) both concave.

> Common mistake: Confusing the sign of the focal length for convex and concave lenses or mirrors.

### Question 5

*1 mark · MCQ*

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

- only plane.
- only concave.
- only convex.
- either plane or convex.

**Solution**

1. A plane mirror always forms an erect image of the same size for any object position.
2. A convex mirror always forms a virtual, erect, and diminished image for all positions of the object.

**Answer:** (d) either plane or convex.

> Common mistake: Forgetting that convex mirrors also always form erect images.

### Question 6

*1 mark · MCQ*

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

- A convex lens of focal length $50\text{ cm}$.
- A concave lens of focal length $50\text{ cm}$.
- A convex lens of focal length $5\text{ cm}$.
- A concave lens of focal length $5\text{ cm}$.

**Solution**

1. To read small letters in a dictionary, we need a magnified erect image, which is obtained using a convex lens.
2. A lens with a shorter focal length provides a higher magnification ($m = 1 + \frac{D}{f}$), so a convex lens of $5\text{ cm}$ focal length is preferred over one of $50\text{ cm}$.

**Answer:** (c) A convex lens of focal length $5\text{ cm}$.

> Common mistake: Choosing a lens with a larger focal length thinking it magnifies more.

### Question 7

*3 marks · Short answer*

We wish to obtain an erect image of an object, using a concave mirror of focal length $15\text{ cm}$. 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.

**Solution**

1. To obtain an erect image using a concave mirror, the object must be placed between its pole and the principal focus.
2. Therefore, the range of distance of the object from the mirror should be between $0\text{ cm}$ and $15\text{ cm}$.
3. The nature of the image is virtual and erect, and it is larger (enlarged) than the object.

**Answer:** Object distance range: between $0\text{ cm}$ and $15\text{ cm}$; nature: virtual and erect; size: larger than the object.

> Common mistake: Giving the range beyond the focus, which produces real and inverted images.

### Question 8

*3 marks · Short answer*

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. Support your answer with reason.

**Part (a)**

1. A concave mirror is used in the headlights of a car.
2. When a light source is placed at the focus of a concave mirror, it produces a powerful parallel beam of light.

Answer (a): Concave mirror, because it produces a parallel beam of light when the source is at the focus.

**Part (b)**

1. A convex mirror is used as a side/rear-view mirror in vehicles.
2. It always gives an erect, diminished image and has a wider field of view, enabling the driver to see a large area behind.

Answer (b): Convex mirror, because it gives an erect and diminished image with a wide field of view.

**Part (c)**

1. A large concave mirror is used in a solar furnace.
2. It converges a parallel beam of sunlight to a sharp focal point, producing immense heat.

Answer (c): Concave mirror, because it converges sunlight to a point to produce heat.

**Answer:** Concave mirrors are used in headlights and solar furnaces, and convex mirrors are used as rear-view mirrors.

> Common mistake: Confusing the uses and properties of concave and convex mirrors.

### Question 9

*3 marks · Short answer*

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.

**Solution**

1. Yes, the lens will still produce a complete image of the object.
2. Light rays from all parts of the object pass through the uncovered half of the lens and form the image.
3. However, the intensity or brightness of the image will be reduced because fewer rays reach the screen.

**Answer:** Yes, a complete image is formed, but its brightness is reduced.

> Common mistake: Thinking that covering half the lens forms only half of the image.

### Question 10

*3 marks · Numerical*

An object $5\text{ cm}$ in length is held $25\text{ cm}$ away from a converging lens of focal length $10\text{ cm}$. Draw the ray diagram and find the position, size and the nature of the image formed.

**Solution**

1. Given: Height of object $h = +5\text{ cm}$, object distance $u = -25\text{ cm}$, focal length $f = +10\text{ cm}$.
2. Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$.
3. Substitution: $\frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{5 - 2}{50} = \frac{3}{50}$, giving $v = +16.67\text{ cm}$.
4. Result: Image is real, inverted, formed at $16.67\text{ cm}$ on the other side, with size $h' = -3.33\text{ cm}$.

**Answer:** Position: $v = +16.67\text{ cm}$; size: $-3.33\text{ cm}$; nature: real and inverted.

> Common mistake: Forgetting the negative sign for object distance $u$ according to the sign convention.

### Question 11

*3 marks · Numerical*

A concave lens of focal length $15\text{ cm}$ forms an image $10\text{ cm}$ from the lens. How far is the object placed from the lens? Draw the ray diagram.

**Solution**

1. Given: Focal length of concave lens $f = -15\text{ cm}$, image distance $v = -10\text{ cm}$.
2. Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, which rearranges to $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$.
3. Substitution: $\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}$.
4. Result: Object distance $u = -30\text{ cm}$, so the object is placed $30\text{ cm}$ in front of the lens.

**Answer:** The object is placed at a distance of $30\text{ cm}$ in front of the lens.

> Common mistake: Incorrect sign substitution for concave lens focal length or image distance.

### Question 12

*3 marks · Numerical*

An object is placed at a distance of $10\text{ cm}$ from a convex mirror of focal length $15\text{ cm}$. Find the position and nature of the image.

**Solution**

1. Given: Object distance $u = -10\text{ cm}$, focal length of convex mirror $f = +15\text{ cm}$.
2. Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$.
3. Substitution: $\frac{1}{v} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$.
4. Result: Image distance $v = +6\text{ cm}$, and the image is virtual, erect, and diminished behind the mirror.

**Answer:** Position: $v = +6\text{ cm}$ behind the mirror; nature: virtual and erect.

> Common mistake: Using the lens formula instead of the mirror formula.

### Question 13

*2 marks · Very short answer*

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

**Solution**

1. A magnification of $+1$ produced by a plane mirror means that the size of the image formed is exactly equal to the size of the object.
2. The positive sign indicates that a virtual and erect image is formed behind the mirror.

**Answer:** It means the image formed is virtual, erect, and of the same size as the object.

> Common mistake: Forgetting to mention that the positive sign indicates a virtual and erect image.

### Question 14

*3 marks · Numerical*

An object $5.0\text{ cm}$ in length is placed at a distance of $20\text{ cm}$ in front of a convex mirror of radius of curvature $30\text{ cm}$. Find the position of the image, its nature and size.

**Solution**

1. Given: Object-size, $h = +5.0\text{ cm}$; Object-distance, $u = -20\text{ cm}$; Radius of curvature, $R = +30\text{ cm}$.
2. Formula: Focal length $f = \frac{R}{2} = \frac{+30\text{ cm}}{2} = +15\text{ cm}$; Mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
3. Substitution: $\frac{1}{v} + \frac{1}{-20} = \frac{1}{15} \implies \frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$.
4. Result: $v = +8.57\text{ cm}$; Height of image $h' = -\frac{v}{u}h = -\frac{8.57\text{ cm}}{-20\text{ cm}} \times 5.0\text{ cm} = +2.14\text{ cm}$.
5. The image is virtual, erect, formed at a distance of $8.57\text{ cm}$ behind the mirror, and its size is $2.14\text{ cm}$. 

**Answer:** Image distance $v = +8.57\text{ cm}$, size $h' = +2.14\text{ cm}$, nature: virtual and erect.

> Common mistake: Incorrect sign convention used for object distance or radius of curvature.

### Question 15

*3 marks · Numerical*

An object of size $7.0\text{ cm}$ is placed at $27\text{ cm}$ in front of a concave mirror of focal length $18\text{ cm}$. 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.

**Solution**

1. Given: Object-size, $h = +7.0\text{ cm}$; Object-distance, $u = -27\text{ cm}$; Focal length, $f = -18\text{ cm}$.
2. Formula: Mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification $m = \frac{h'}{h} = -\frac{v}{u}$.
3. Substitution: $\frac{1}{v} + \frac{1}{-27} = \frac{1}{-18} \implies \frac{1}{v} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54}$.
4. Result: $v = -54\text{ cm}$; Height of image $h' = -\frac{v}{u}h = -\frac{-54\text{ cm}}{-27\text{ cm}} \times 7.0\text{ cm} = -14\text{ cm}$.
5. The screen should be placed at a distance of $54\text{ cm}$ in front of the mirror. The image is real, inverted, and of size $14\text{ cm}$.

**Answer:** Screen distance $54\text{ cm}$ in front of mirror, image size $-14\text{ cm}$, nature: real and inverted.

> Common mistake: Confusing the signs of focal length and object distance for a concave mirror.

### Question 16

*3 marks · Numerical*

Find the focal length of a lens of power $-2.0\text{ D}$. What type of lens is this?

**Solution**

1. Given: Power of the lens, $P = -2.0\text{ D}$.
2. Formula: Focal length $f = \frac{1}{P}$.
3. Substitution: $f = \frac{1}{-2.0\text{ D}} = -0.50\text{ m} = -50\text{ cm}$.
4. Result: Since the focal length is negative, the lens is a concave lens.

**Answer:** Focal length is $-0.50\text{ m}$ (or $-50\text{ cm}$), and the lens is a concave lens.

> Common mistake: Forgetting to specify the unit of focal length or misidentifying the type of lens from the sign of power.

### Question 17

*3 marks · Numerical*

A doctor has prescribed a corrective lens of power $+1.5\text{ D}$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

**Solution**

1. Given: Power of the lens, $P = +1.5\text{ D}$.
2. Formula: Focal length $f = \frac{1}{P}$.
3. Substitution: $f = \frac{1}{+1.5\text{ D}} = +\frac{2}{3}\text{ m} = +0.67\text{ m} = +67\text{ cm}$.
4. Result: Since the power and focal length are positive, the prescribed lens is a converging (convex) lens.

**Answer:** Focal length is $+0.67\text{ m}$ (or $+67\text{ cm}$), and the lens is converging.

> Common mistake: Stating the lens is diverging instead of converging when power is positive.

## Frequently asked questions

### How many questions and exercises are there in NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction on SwaVid?

This chapter page includes multiple questions broken down across textbook segments, along with 17 exercise questions. You can find SwaVid's free PDF and step-by-step solutions on this page only to help you practice.

### Which topics do the questions in this SwaVid chapter cover?

The questions cover important concepts like image formation by spherical mirrors, principal focus of a concave mirror, absolute refractive index, and the power of a lens. They also include comprehensive exercises on lens formulas and magnification.

### What are the hardest question types in Class 10 Science Chapter 9 and how should I approach them?

Numerical problems involving mirror and lens formulas or refractive index are usually considered the trickiest. To approach them, you should carefully note down the given values with proper sign conventions, use the correct formula such as $f = R/2$, and solve step by step.

### How can I write answers for full marks in the board exams for this Light chapter?

To secure full marks, always state the relevant laws or formulas clearly before starting your calculations. For ray diagrams, ensure you draw arrows, label all rays correctly, and mention units like diopters or centimeters in your final answers.

### Is the free PDF for Class 10 Science Chapter 9 Light Reflection and Refraction available on SwaVid?

Yes, SwaVid provides a free PDF and detailed step-by-step solutions on this page only. You can easily access these resources to revise concepts like the relation between focal length and radius of curvature before your exams.

## Related pages

- [Light – Reflection and Refraction: CBSE previous year questions](https://www.swavid.com/cbse/class-10/science/pyq/light-reflection-and-refraction)
- [Class 10 Science chapters](https://www.swavid.com/science/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student