NCERT Solution Class 10 chapter Light” Reflection & Refraction of Light”

 

Page 168

1. Define the principal focus of a concave mirror.

Ans : The principal focus of a concave mirror is the point on its principal axis where parallel rays of light converge after reflection.

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

Ans : The focal length of a spherical mirror can be calculated using the formula: f = R/2, where R is the radius of curvature. Therefore, in this case, the focal length would be 10 cm.

 

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

Ans : A convex mirror can give an erect and enlarged image of an object.

 

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

Ans : Convex mirrors have a wider field of view and give a smaller image compared to the object, making them ideal as a rear-view mirror in vehicles. This allows the driver to see a larger area behind the vehicle and estimate distances more easily. Additionally, convex mirrors are able to provide a wider view of the surroundings, which is important in ensuring safe driving.

Page 171

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

Ans : The focal length of a convex mirror can be calculated using the formula: f = R/2, where R is the radius of curvature. Therefore, in this case, the focal length would be 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?

Ans :

1. The magnification (m) of an image formed by a concave mirror can be calculated using the formula: m = -v/u, where v is the image distance and u is the object distance. In this case, we know that the magnification is 3, so we can write: m = -v/u = 3

Solving for v, we get:

v = -3u

We also know that the object distance u is 10 cm. Substituting this into the equation above, we get:

v = -3 x 10 cm = -30 cm

Since the magnification is positive, the image is real and upright. Since the image distance is negative, the image is formed on the same side of the mirror as the object. Therefore, the real and enlarged image is located 30 cm behind the concave mirror.

Page  : 176

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 obliquely from air into water, it bends towards the normal. This is because water has a higher refractive index than air, which means that the speed of light is slower in water than in air. According to Snell’s law, when a light ray passes from a medium of lower refractive index to a medium of higher refractive index, it bends towards the normal.

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.

Ans : The speed of light in a medium is given by the formula: speed of light in vacuum / refractive index of the medium. Therefore, the speed of light in glass is:

Speed of light in glass = 3 × 108 m/s / 1.50 = 2 × 108 m/s

3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Ans : The medium with the highest optical density is diamond, with a refractive index of 2.42. The medium with the lowest optical density is air, with a refractive index of 1.00.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Ans : Light travels fastest in kerosene, as it has the lowest refractive index among the given options. According to Snell’s law, the speed of light in a medium is inversely proportional to its refractive index.

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

Ans : The refractive index of diamond is 2.42. This means that when light passes from air into diamond, it will bend towards the normal. The refractive index is a measure of how much a material can bend light, and a higher refractive index means that the material can bend light more. Therefore, diamond is a highly refractive material and is often used in jewellery to create the sparkling effect.

 

page : 184

1. Define 1 dioptre of power of a lens.

Ans :

The power of a lens is defined as the reciprocal of its focal length in meters. Therefore, a lens with a focal length of 1 meter has a 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 object? Also, find the power of the lens.

Ans : If the image formed by a convex lens is real and inverted, it means that the object is placed beyond the focal point of the lens. If the image is equal in size to the object, it means that the object is placed at twice the focal length of the lens. Therefore, in this case, the object (needle) is placed at a distance of 100 cm (twice the focal length) in front of the convex lens.

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

Ans :

To calculate the power of the lens, we can use the formula: P = 1/f, where P is the power of the lens in dioptres and f is the focal length in meters. Since the image distance is given as 50 cm (0.5 m), we can use the formula for thin lens equation:

1/f = 1/v – 1/u

Where v is the image distance and u is the object distance. Since the image is formed at a distance of 50 cm (0.5 m) from the lens, we have:

v = -0.5 m

Substituting this into the formula, we get:

1/f = 1/-0.5 – 1/1

Solving for f, we get:

f = -1.67 cm or -0.0167 m (rounded to 3 decimal places)

The negative sign indicates that the lens is a convex lens. Therefore, the power of the lens is:

P = 1/f = 1/-0.0167 = -59.88 dioptres (rounded to 2 decimal places)

3. The power of a concave lens can also be calculated using the formula: P = 1/f, where f is the focal length in meters. Therefore, in this case, the power of the concave lens is:

P = 1/(-2) = -0.5 dioptres.

Page 185

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

Ans : (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.

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 : (c) the mirror is concave and the lens is convex.

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.

Ans : (a) plane.

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.

Ans  : (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.

Ans : The object should be placed between the pole of the mirror and its principal focus (i.e., at a distance less than the focal length). The nature of the image will be virtual, erect and enlarged. The image will be larger than the object. The ray diagram is as follows:

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 furnace.
Support your answer with reason.

Ans :(a) Concave mirror, as it reflects parallel beams of light in a converging manner to give a real and inverted image of the source. (b) Convex mirror, as it reflects diverging beams of light to give a virtual and erect image of the source with a wider field of view. (c) Concave mirror, as it reflects parallel beams of light in a converging manner to produce high temperatures at the focus.

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.

Ans:

A convex lens, also known as a converging lens, can form a complete image of an object even if part of it is covered by black paper. This is because the convex lens refracts light rays that pass through it and converges them to a point on the other side of the lens, where the image is formed.

In case I, when the upper half of the lens is covered, the lower half of the lens still refracts the light rays coming from the object and converges them to form an image on the other side of the lens.

Similarly, in case II, when the lower half of the lens is covered, the upper half of the lens still refracts the light rays coming from the object and converges them to form an image on the other side of the lens.

This is because a convex lens is symmetrical, so the shape of the lens and the way it refracts light rays is the same whether the upper or lower half is covered

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.

Ans : Solution

Here : Object distance, u= -25 cm,
Object height, h = 5 cm,
Focal length, f = +10 cm
According to the lens formula, 1f=1ν−1u , we have
⇒ 1ν=1f−1u=110−125=15250orν=25015=16.66cm
The positive value of v shows that the image is formed at the other side of the lens.

The negative value of image height indicates that the image formed is inverted.
The position, size, and nature of image are shown alongside in the ray diagram.

 

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.

Ans :

f= -15 cm, v= -10 cm1/v -1/u = 1/f1/u = 1/15 – 1/10 = -1/30

u = -30 cm.

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.

Ans : f = +15 cm, u = -10 cm.1/f = 1/v +1/u1/v = 1/15 +1/10

1/v = 5/30

v = + 30 cm.

The image is formed 6 cm behind the mirror, it is a virtual and erect image.

The image is formed at a distance of 6 cm from the mirror on the same side as the object. The image is virtual, erect and of the same size as the object.

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

Ans :

It means that the image is of the same size as the object, and is formed at the same distance behind the mirror as the distance of the object in front of the mirror. The image is virtual and erect.

 

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.

Ans : Radius of curvature (R) = 30 cmf = R/2 = 30/2 = 15 cmu = –20 cm, h= 5 cm.

1/v +1/u = 1/f

1/v = 1/15+ 1/20 = 7/60

v = 60/7 = 8.6 cm.

image is virtual and erect and formed behind the mirror.

h_i/h₀= v/u

h_i/5= 8.6/20

h_i = 2.2 cm.

Size of image is 2.2 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.

Ans :  u = – 27 cm, f = – 18 cm. h_o= 7.0 cm1/v = 1/f- 1/u1/v = -1/18 + 1/27 = -1/54

V = – 54 cm.

Screen must be placed at a distance of 54 cm from the mirror in front of it.

h_i/h₀= v/u

h_i/h₀= v/u

h_i/7 = +54/-27

h_i = -2 x 7 = -14 cm.

Thus, the image is of 14 cm length and is inverted image.

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

Ans :Power of lens (P) = -2.0 DP = 1/f or f = 1/mf = 1/-2.0 = -0.5 m.

(-ve) sign of focal length means that the lens is concave lens.

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?

Ans : The power of the lens is given by P = 1/f, where f is the focal length of the lens. Therefore, f = 1/P = 1/1.5 D = 0.67 m. This is a converging lens.