Lesson: Light-Reflection and Refraction

# Question: 1

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

(a) Water

(b) Glass

(c) Plastic

(d) Clay

d

# 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.

d

# 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.

b

# 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

a

# 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

d

# 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

c

# 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.

## Solution

Concave mirror forms an erect image of an object when it is placed between the focus and the pole. Therefore, the range of the distance of the object = 0 to 15 cm from the pole of the mirror. The image will be virtual, erect and larger than the object. # Question: 8

Name the type of mirror used in the following situations:

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

(c) Solar furnace

## Solution

(a) Concave Mirror:

When the light source is placed at their principal focus of a concave mirror, it produces powerful parallel beam of light. This helps the driver to see considerable distance in the darkness.

(b) Convex Mirror:

The image formed by the convex mirror is highly diminished, virtual and erect. This provides a large field of view.

(c) Concave Mirror:

Concave mirrors can converge the parallel rays of sun at its principal focus.

The solar furnace when placed at the focus receives maximum amount of 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.

## Solution

Even when one-half of a convex lens is covered with a black paper, it will form complete image of an object.

Explanation

(i)     When the upper half of the lens is covered, it can be seen that a ray of light coming from the object (AB) is 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 (A’B’).

(ii)  When the lower half of the lens is covered, it can be seen that a ray of light coming from the object (AB) is 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 (A’B’).

# 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.

## Solution

Let the height of the object be ${h}_{o}$.

Given that,

Let the distance of the object from converging lens be u.

Given that,

The focal length of converging lens,

Using lens formula,

$\begin{array}{l}\frac{\text{1}}{v}-\frac{\text{1}}{u}\text{=}\frac{\text{1}}{f}\\ \frac{\text{1}}{v}\text{=}\frac{\text{1}}{f}\text{+}\frac{\text{1}}{u}\text{=}\frac{\text{1}}{\text{10}}\text{-}\frac{\text{1}}{\text{25}}\text{=}\frac{\text{15}}{\text{250}}\\ v\text{\hspace{0.17em}}\text{=}\frac{\text{250}}{\text{15}}\text{=16}\text{.66}\text{\hspace{0.17em}}\text{cm}\end{array}$

For a converging lens,

$\begin{array}{l}\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}\\ ⇒{h}_{i}=\frac{v}{u}×{h}_{\text{0}}\text{=}\frac{\text{50×5}}{\text{3×(-25)}}\text{=}\frac{\text{10}}{\text{-3}}\text{=}-\text{3}\text{.3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{cm}\end{array}$

Thus, the image measures 3.3 cm. It is inverted, formed at a distance of 16.7 cm and is behind the lens. The diagram is shown below. # 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.

## Solution

Let the focal length of concave lens be f.

Given that

Let the image distance be v.

Given that,

According to the lens formula,

$\begin{array}{l}\frac{\text{1}}{v}\text{-}\frac{\text{1}}{u}\text{=}\frac{\text{1}}{f}\\ ⇒\frac{\text{1}}{u}\text{=}\frac{\text{1}}{v}\text{-}\frac{1}{f}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-1}{10}+\frac{1}{15}=\frac{-5}{150}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\text{30}\text{\hspace{0.17em}}\text{cm}\end{array}$

The object is placed 30 cm in front of the lens. The ray diagram is shown below. # 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.

## Solution

Let the focal length of the convex mirror be f.

Given that the focal length,

Let the object distance be u.

Give that,

According to the mirror formula,

$\begin{array}{l}\frac{\text{1}}{v}\text{-}\frac{\text{1}}{u}\text{=}\frac{\text{1}}{f}\\ \frac{\text{1}}{v}\text{=}\frac{\text{1}}{f}\text{-}\frac{\text{1}}{u}\\ \frac{1}{15}+\frac{1}{10}=\frac{25}{150}\\ v=\text{6}\text{\hspace{0.17em}}\text{cm}\end{array}$

The image is formed behind the mirror.

Magnification,

$m=\frac{6}{10}=+0.6$

Thus, the image formed is virtual and erect.

# Question:13

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

## Solution

The positive sign means image formed is virtual and erect. The magnification by 1 means the image size is same as the object size.

# Question: 14

An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

## Solution

Given,

The object distance, ,

Object height, ,

$R=2\text{\hspace{0.17em}}f$

According to the mirror formula,

$\begin{array}{l}\frac{\text{1}}{v}\text{-}\frac{\text{1}}{u}\text{=}\frac{\text{1}}{f}\\ \frac{\text{1}}{v}\text{=}\frac{\text{1}}{f}\text{-}\frac{\text{1}}{u}\\ \text{=}\frac{\text{1}}{\text{15}}\text{+}\frac{\text{1}}{\text{20}}\text{=}\frac{\text{4+3}}{\text{60}}\text{=}\frac{\text{7}}{\text{60}}\\ v=\text{8}\text{.57cm}\end{array}$

$⇒$ The image is formed behind the mirror.

Magnification, m = image distance/object distance

$m=\frac{-8.75}{-20}=0.428$

$⇒$ The image formed is virtual.

Magnification (m) = height of the image/height of the object

$\begin{array}{l}⇒m=\frac{{h}^{\text{'}}}{h}\\ ⇒{h}^{\text{'}}=m×h=0.428×5=2.14\text{\hspace{0.17em}}\text{cm}\end{array}$

$⇒$ The image formed is erect.

$⇒$ The image formed is virtual, erect, and smaller in size.

# 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 focused image can be obtained? Find the size and the nature of the image.

## Solution

Given:

Object distance, ,

Object height, ,

Focal length,

According to the mirror formula,

$\begin{array}{l}\frac{\text{1}}{v}-\frac{\text{1}}{u}=\frac{\text{1}}{f}\\ \frac{\text{1}}{v}\text{=}\frac{\text{1}}{f}-\frac{\text{1}}{u}\\ =\frac{-1}{18}+\frac{1}{27}=\frac{-1}{54}\\ v=-\text{54}\text{\hspace{0.17em}}\text{cm}\end{array}$

The screen should be placed at a distance of 54 cm in front of the given mirror.

Magnification, m = image distance/object distance

$m=\frac{-54}{27}=-2$

$⇒$ The image is real.

Magnification, $m=$ height of the image/height of the object

$\begin{array}{l}m\text{=}\frac{{h}^{\text{'}}}{h}\\ ⇒{h}^{\text{'}}\text{=}m×h=7×\left(-2\right)=-14\text{\hspace{0.17em}}\text{cm}\end{array}$

$⇒$ The image formed is inverted.

# Question:16

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

## Solution

Power of lens,

Given,

Thus, it is a concave lens.

# 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?

## Solution

Power of lens, .

Given, P = 1.5 D

$⇒$ It is a convex lens or a converging lens.