# NCERT Solutions for Class 10 Science Chapter 10 Light- Reflection and Refraction

**NCERT Solutions for Class 10 Science Chapter 10 Light- Reflection and Refraction** are a comprehensive set of solutions to the problems present in NCERT textbooks. These play an important role in helping students improve their marks in examinations.

By using **Chapter 10 Class 10 Science NCERT Solutions**, students will be able to get an accurate gauge of their understanding of the concepts taught in class. It can help identify gaps in understanding, allowing students to focus their studies on areas that need improvement.

## Chapter 10 Light- Reflection and Refraction Class 10 Science NCERT Solutions

### In Text Questions

**1. Define the principal focus of a concave mirror.**

**Solution**

A point on the principal axis where the parallel rays of light meet after getting reflected from a concave mirror.

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

**Solution**

Given: Radius of curvature, R= 20 cm

We know that, f= R/2

f= 20/2 = 10 cm.

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

**Solution**

Concave mirror or converging mirror.

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

**Solution**

A convex mirror forms erect and diminished image of objects behind the vehicle due to which large traffic can be easily viewed.

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

**Solution**

Here,

R = 32 cm

f= R/2

f= 32/2

= 16 cm

**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?**

**Solution**

We have,

m = -3 (-ve sign as image is real)

u = -10 cm

but

m = -v/u

-3 = -v/(-10)

v = -30 cm

Image is formed 30 cm in front of the mirror.

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

**Solution**

The ray will be bend towards the normal as speed of light decreases when ray enters from air to water i.e. from rarer to denser medium.

**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 3×10 ^{8} ms^{-1}.**

**Solution**

Refractive index by glass, n =1.5

Speed of light in vacuum, c =3×10^{8} ms^{-1}

Speed of light in glass, v = ?

We know that

Refractive index of a medium- Speed of light in vacuum/Speed of light in medium

n = c/v

1.5 = 3×10^{8}/v

v = 2 × 10^{8} ms^{-1}

**3. Find out from the given table, the medium having highest optical density. Also, find the medium with lowest optical density.**

Material | R.I. | Material | R.I. |

1. Air | 1.0003 | 7. Kerosene | 1.44 |

2. Water | 1.33 | 8. Benzene | 1.50 |

3. Glass | 1.52 | 9. Ice | 1.31 |

4. Alcohol | 1.36 | 10. Sulphuric Acid | 1.43 |

5. Diamond | 2.42 | 11. Turpentine Oil | 1.47 |

6. Carbon disulphide | 1.63 | 12. Ruby | 1.71 |

**Solution**

The medium with highest optical density in diamond Refractive index of diamond = 2.42.

The medium with lowest optical density is air.

Refractive index of air = 1.0003.

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

**Solution**

We know that

n = c/V i.e., v ∝1/n

In the medium with least refractive index the speed of light is maximum. Therefore among kerosene, turpentine and water, speed of light is maximum in water.

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

**Solution**

It means that when light travels through diamond its speed decreases by a factor of 2.42.

**1. Define one dioptre of power of lens.**

**Solution**

A lens whose focal length is one metre has optical power of 1 dioptre.

**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 objects? Also find the power of the lens.**

**Solution**

We have, the image distance, v = 50 cm

Object distance, u = ?

Power of lens, P = ?

As size of the image is equal to that of object and the image is real and inverted, m = -1

m = v/u

-1 = 50 cm

but

1/f = 1/v -1/u

1/f = 1/50 – 1/-50

1/f = 2/50

⇒ f = 25 cm = 1/4 m

Power of lens, P = 1/¼ = 4 dioptre

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

**Solution**

Power, P = ?

Focal length, f = -2 m (concave lens)

So, P = 1/f = 1/-2

P = -0.5 dioptre

### Exercises

**1. Which one of the following material cannot be used to make a lens?(a) Water(b) Glass(c) Plastic(d) Clay**

**Solution**

(d) 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.**

**Solution**

(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.**

**Solution**

(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.**

**Solution**

(a) both concave

**5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be(a) plane(c) convex(b) concave(d) either plane or convex.**

**Solution**

(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.**

**Solution**

(c) A convex lens of focal length 5 cm.

**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.**

**Solution**

Given: Focal length of concave mirror = -15 cm

For erect image object must be placed between the pole of the mirror and its focus i.e. 0 cm <u < 15 cm.

Nature of image: erect, virtual and enlarged.

**8. 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 furnance.Support your answer with reason.**

**Solution**

(a) Concave mirror: So that beam of light covers large distance of the road in front of car.

(b) Convex mirror: So that view of large traffic behind can be seen ‘erect’ in a small mirror.

(c) Concave mirror: So that parallel rays of light coming from the Sun can be concentrated.

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

**Solution**

Yes, it will produce a complete image, but intensity of light in the image may be little less.

**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.**

**Solution**

Here, h = 5 cm, h/ = ?, u = -25 cm (Object distance is always negative)

v = ?, f = +10 cm (f of convex lens is positive)

Using,

1/f = 1/v – 1/u or 1/v = 1/u + 1/f

1/v = 1/-25 + 1/10 = -2+5/50 = 3/50

v = 16.7 cm

m = h’/h = v/u or h’ = h v/u

h’ = 5(50/3)/-25 = -10/3 = -3.3 cm

h’ = -3.3 cm

Negative sign shows image is inverted, real, diminished (3.3 cm) and at 16.7 cm on the right side of lens.

**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.**

**Solution**

Given f = −15 cm (concave lens)

v = -10 cm (in concave lens, image is formed on the same side as object)

u = ?

Using

1/f = 1/v – 1/u

⇒ 1/-15 = 1/-10 – 1/u

⇒ 1/u = 1/-10 + 1/15

⇒ 1/u = -15+10/150

⇒ u = -30 cm

For making a ray diagram, let’s take suitable scale of 5 cm = 1 cm.

So, according to the scale

f=-3 cm

u = -6 cm

v = -2 cm

Ray diagram is provided below.

**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.**

**Solution**

Given

u = -10 cm

f = +15 cm (convex mirror)

v = ?

Using mirror formula

1/f = 1/u +1/v

⇒ 1/15 = 1/-10 + 1/v

⇒ 1/v = 1/15 + 1/10

⇒ 1/v = 10 + 15/150

⇒ v = 6 cm

and magnification, m = -v/u = -6/-10 = 0.6

Image is formed 6 cm behind the mirror. It is virtual, erect and diminished.

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

**Solution**

Image is of the same size as object and image is virtual.

**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.**

**Solution**

Given

h_{o} = 5 cm

u = -20 cm

R = 30 cm

v = ?

hi = ?

Since

f = R/2

So,

f = 30/2 = 15 cm

Using mirror formula

1/f = 1/u +1/v

⇒ 1/15 = 1/-10 + 1/v

⇒ 1/v = 1/15 + 1/20

⇒ 1/v = 4+30/150

⇒ v = 8.57 cm

Magnification m =-v/u

⇒ m = -8.57/-20

⇒ m = 0.43

But, m = h^{i}/h_{0}

⇒ 0.43 = h_{i}/5

⇒ h_{i} = 2.15 cm

So image is formed behind the mirror at a distance of 8.57 cm; and its image is erect, virtual and of size 2.15 cm.

**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.**

**Solution**

Given

h_{0} = 5cm

u = -20 cm

R = 30 cm

v = ?

h_{i} = ?

Since

f = R/2

So,

f = 30/2 = 15 cm

Using mirror formula

1/f = 1/u +1/v

⇒ 1/15 = 1/-10 + 1/v

⇒ 1/v = 1/15 + 1/20

⇒ 1/v = 4+30/150

⇒ v = 8.57 cm

Magnification, m = -v/u

⇒ m = -8.57/-20

⇒ m = 0.43

But, m = h_{i}/h_{o}

⇒ 0.43 = h_{i}/5

⇒ h_{i} = 2.15 cm

So image is formed behind the mirror at a distance of 8.57 cm, and its image is erect, virtual and of size 2.15 cm.

**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.**

**Solution**

Given

ho = 7 cm

u = -27 cm

f = -18 cm (concave mirror)

v = ?

h_{i} = ?

Using mirror formula,

1/f = 1/u + 1/v

⇒ 1/-18 = 1/-27 + 1/v

⇒ 1/v = -1/18 + 1/27

⇒ 1/v = -3+2/54

⇒ v = -54 cm

Magnification, m = -v/u

⇒ m = -(-54/-27)

⇒ m = -2

Size of image

m = h_{i}/h_{o}

⇒ -2 = h_{i}/7

⇒ h_{i} = -14 cm

So, screen be placed at a distance of 54 cm in front of the mirror. Image formed is twice the size of object, real and inverted.

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

**Solution**

Given

P = -2.0 D

Using

P = 1/f

⇒ -2 = 1/f

⇒ f = -1/2 = -0.5 m

⇒ f = -50 cm

⇒ 1/v = 1/10 – 1/25

⇒ 1/v = 25-10/250

⇒ v = 250/15 = 16.67 cm

As v is +ve, image is formed on the other side of lens at a distance of 16.67 cm i.e. between F’ and 2F’, as shown in figure.

To find hi (height of image)

h_{i}/h_{o} = v/u

⇒ h_{i}/5 = 16.67/-25

⇒ h_{i} = 3.33 cm

Since, h_{i} is -ve, image is inverted and of size 3.33 cm, i.e. smaller than the size of object (diminished)

m = v/u = 16.67/-25 = -0.67

⇒ m = -0.67

As magnification is negative, image is real.

**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?**

**Solution**

P = +1.5 D

1/f = 1.5

⇒ f = 1/1.5m = 100/1.5 cm

f = +66.67 cm

The lens is converging as focal length is positive.