Rajasthan Board RBSE Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Exercise Questions and Answers.

## RBSE Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

### RBSE Class 10 Science Chapter 10 Light Reflection and Refraction InText Questions and Answers

Page 168.

Question 1.

Define the principal focus of a concave mirror.

Answer:

The rays of light parallel and closed to the principal axis of the mirror after reflection passes through a point on the principal axis is called principal focus of the concave mirror.

Question 2.

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

Answer:

Focal length, f = \(\frac { R }{ 2 }\)

Here R = 20cm.

∴ f = \(\frac { 20}{ 2 }\) = 10 cm.

Question 3.

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

Answer:

Concave mirror.

Question 4.

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

Answer:

We prefer rear view mirror in vehicles due to the following causes:

- It always forms an erect image of an object, although diminished image.
- They have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area.

Page 171.

Question 1.

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

Answer:

Here R = 32 cm.

As f = \(\frac { R }{ 2 }\) = \(\frac { 32 }{ 2 }\) = 16 cm.

Question 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?

Answer:

Magnification produce by a spherical mirror (m)

Given that the image is 3 times magnified than the object.

∴ \(h^{\prime}= -3 h\)

(negative sign for real image) Putting the values:

\(\frac { 3h }{ h }\) = \(\frac { -v }{ u }\) ∴ \(\frac { v }{ u }\) = 3

Distance of object (u) = -10 cm

∴ \(\frac { v }{ -10 }\) = 3

v = 3 x (-10) = -30 cm.

Hence the image is formed 30 cm. in front of mirror.

Page 176.

Question 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?

Answer:

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

Question 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 x 10^{8 }m s^{-1}.

Answer:

Given, n = 1.5

c = 3 x 10^{8 }m s^{-1}.

v = ?

n = \(\frac { c }{ v }\) ⇒ 1.5 = \(\frac{3 \times 10^{8}}{v}\)

v = 2 x 10^{8 }m s^{-1}

Question 3.

Find out, from Table 10.3 (See textbook), the medium having highest optical density. Also find the medium with lowest optical density.

Answer:

The medium with highest optical density is diamond (refractive index = 2.42). The medium with lowest optical density is air [refractive index = 1.0003]

Question 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 (See textbook).

Answer:

We know that,

n = \(\frac { c }{ v }\) ⇒ v α \(\frac { 1 }{ n }\)

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.

Question 5.

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

Answer:

It means when light travels through diamond its speed decreases by a factor of 2,42.

Page 184.

Question 1.

Define 1 dioptre of power of a lens.

Answer:

When the focal length of a lens is 1 m the power of lens is called 1 Dioptre i.e. 1 D = 1 m^{–}^{1}.

Question 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.

Answer:

Given,

Image distance v = 50 cm. Object distance u = ?

Power of lens P = ?

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

∵ m = \(\frac { v }{ u }\) ⇒ -1 = \(\frac { 50 }{ u }\)

⇒ u = -50 cm.

\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)

⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ 50 }\) \(\frac { 1 }{ -50 }\)

⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ 20 }\)

⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ 25 }\)

⇒ f = 25 cm = \(\frac { 1 }{ 4 }\) m

power of lens,

\(\frac { 1 }{ 1/4 }\) = 4 dipotre.

Question 3.

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

Answer:

Given

Power, P = ?

Focal length, f = -2 m [concave lens]

∴ p = \(\frac { 1 }{ f }\) = \(\frac { 1 }{ -2 }\)

p = -0.5 dipotre.

### RBSE Class 10 Science Chapter 10 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

(b) Glass

(c) Plastic

(d) Glay

Answer:

(d) Clay

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 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.

Answer:

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

Question 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.

Answer:

(b) At twice the focal length

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 is convex.

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

Answer:

(a) Both concave.

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.

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:

(a) A convex lens of focal length 50 cm.

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:

(i) To obtain erect image of an object in a concave mirror, the object should be placed between pole and principal focus of die mirror, therefore the range of distance of the object from the mirror must be less than 15 cm.

(ii) The image will be erect and virtual.

(iii) The size of the image would be larger

Question 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.

Answer:

(a) Concave mirror: When light source is placed at the focus of the mirror then on reflection a strong parallel beam of light emerges.

(b) Convex mirror: Field of view of convex mirror is larger and it forms virtual, erect and diminished images of objects behind.

(c) Concave mirror: Light rays from the sun on reflection from the mirror is cocentrated at the focus of the mirror and produce heat.

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:

The convex lens will form complete image of an object, even its one half is covered with black paper. It can be understood by following two cases:

- CASE I : When the upper half of the lens is covered a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.
- CASE II : When the lower half of the lens is covered, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of object.

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 the nature of the image formed.

Answer:

Given,

Object distance u = -25 cm

Object height h = 5 cm

Focal length f = +10 cm

∴ Image formed on the other side at 16.66 cm and will be real (as v is positive) and inverted W is -ve).

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:

Given,

Focal length f = +15 cm,

Object distance v = -10 cm

According to lens formula

⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)

⇒ \(\frac { 1 }{ -15 }\) = \(\frac { 1 }{ -10 }\) – \(\frac { 1 }{ 4 }\)

⇒ \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 15 }\) \(\frac { 1 }{ 10 }\)

u = -30 cm

The negative value of u indicates that object placed 30 cm infront of the lens.

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:

Given,

Focal length f = +15 cm,

Object distance u = -10 cm

According to mirror formula,

⇒ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)

⇒ \(\frac { 1 }{ 15 }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ -10 }\)

⇒ \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 15 }\) ⇒ v = 6 cm.

The positive value of v indicates that the image is formed behind the mirror.

Magnification, m = \(\frac { -v }{ u }\) = \(\frac { -6 }{ -10 }\) = +3

The positive value of magnification indicates that the image formed is virtual and erect.

Question 13.

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

Answer:

m = \(\frac{h^{\prime}}{h}\) = +1

⇒ \( h^{\prime}\) = h

i.e size of image is equal to the size of object. Positive sign of m indicates that the image is erect and hence virtual.

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:

Given

Object size h = 5.0 cm

Object distance u = -2 cm

Radius of curvature R = 30 cm

Image distance v = ?

Image size h’ = ?

⇒ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) = \(\frac { 2 }{ R }\)

\(\frac { 1 }{ v }\) = \(\frac { 2 }{ R }\) – \(\frac { 1 }{ u }\)

\(\frac { 2 }{ 30 }\) + \(\frac { 1 }{ 20 }\) = \(\frac { 7 }{ 60 }\)

⇒ v = \(\frac { 60 }{ 7 }\) = 8.5 cm

Positive sign of v indicates that the image is at the back of the mirror. It is virtual and erect.

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

Answer:

Given,

Objet size h = 7.0 cm

Object distance u = -27 cm

Focal length f = -18 cm

Image distance v = ?

Image size h’ = ?

By mirror formula

\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)

\(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ -18 }\) + \(\frac { 1 }{ 27 }\)

⇒ v = -54 cm

Hence, the screen should be placed in front of the mirror at a distance of 54 cm from the mirror. The image obtained on screen will be real. Now, magnification

m = \(\frac{h^{\dagger}}{h} = \frac{-v}{u}

\frac{h^{\prime}}{7.0} = \frac{(-54)}{(-27)}

h^{\prime} = -14.0 \mathrm{~cm}\)

The negative sign of h! shows that image is inverted

Question 16.

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

Answer:

Given,

Focal length f = ?

Power P = – 2.0 D

As f = \(\frac { 100 }{ D }\) = \(\frac { 100 }{ -2.0 }\)

= -50cm

As the power of lens is negative, the lens must be concave.

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:

Here, P = +1.5 D

f = ?

f = \(\frac { 1 }{ p }\)

f = \(\frac { 100 }{ 1.5 }\) = 6.67 cm

As power is positive, the prescribed lens is converging or convex.

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