Lesson: Light-Reflection and RefractioDo you want to know click on

Question: 1

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

(a) Water

(b) Glass

(d) Clay

Solution

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

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

Solution

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

(d) Between the optical centre of the lens and its principal focus.

Solution

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

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

Solution

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

(d) either plane or convex

Solution

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

(d) A concave lens of focal length 5 cm

Solution

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:

(a) Headlights of a car

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

Support your answer with reason.

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.

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,

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, $h_{0} = 5 cm$

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

Given that, $u = 25 cm$

The focal length of converging lens, $f = 10 c m$

Using lens formula,

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

For a converging lens,

$\begin{array}{l} \frac{h_{i}}{h_{0}} = \frac{v}{u} \\ \Rightarrow h_{i} = \frac{v}{u} \times h_{0} = \frac{50×5}{3×(-25)} = \frac{10}{-3} = - 3 .3 cm \end{array}$

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 $f = - 15 c m$

Let the image distance be v.

Given that, $v = - 10 c m$

According to the lens formula,

$\begin{array}{l} \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \\ = \frac{- 1}{10} + \frac{1}{15} = \frac{- 5}{150} \\ = - 30 cm \end{array}$

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, $f = + 15 cm$

Let the object distance be u.

Give that, $u = - 10 c m$

According to the mirror formula,

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

The image is formed behind the mirror.

Magnification, $m = \frac{image distance}{object distance}$

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

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, $u = - 20 cm$ ,

Object height, $h = 5 cm$ ,

Radius of curvature, $R = 30 cm$

$R = 2 \times f (f = focal length)$

$R = 2 f$

$f = 15 cm$

According to the mirror formula,

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

$\Rightarrow$ The image is formed behind the mirror.

Magnification, m = image distance/object distance

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

$\Rightarrow$ The image formed is virtual.

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

$\begin{array}{l} \Rightarrow m = \frac{h^{'}}{h} \\ \Rightarrow h^{'} = m \times h = 0.428 \times 5 = 2.14 cm \end{array}$

$\Rightarrow$ The image formed is erect.

$\Rightarrow$ 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, $u = - 27 cm$ ,

Object height, $h = 7 cm$ ,

Focal length, $f = - 18 cm$

According to the mirror formula,

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

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

Magnification, m = image distance/object distance

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

$\Rightarrow$ The image is real.

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

$\begin{array}{l} m = \frac{h^{'}}{h} \\ \Rightarrow h^{'} = m \times h = 7 \times (- 2) = - 14 cm \end{array}$

$\Rightarrow$ 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, $P = \frac{1}{f}$

Given,

$\begin{array}{l} P = - 2 D \\ \Rightarrow f = - \frac{1}{2} = - 0.5 m \end{array}$

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, $P = \frac{1}{f}$ .

Given, P = 1.5 D

$\Rightarrow f = \frac{1}{1.5} = \frac{10}{15} = 0.66 m$
$\Rightarrow$ It is a convex lens or a converging lens.

Lesson: Light $-$ Reflection and Refraction

EXEMPLAR

Question: 1

Which of the following can make a parallel beam of light when light from a point source is incident on it?

(a) Concave mirror as well as convex lens

(b) Convex mirror as well as concave lens

(d) Concave mirror as well as concave lens

Solution

Question: 2

A 10-mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is:

(a) 30 cm

(b) 20 cm

(d) 60 cm

Solution

Question: 3

Under which of the following conditions a concave mirror can form an image larger than the actual object?

(a) When the object is kept at a distance equal to its radius of curvature

(b) When object is kept at a distance less than its focal length

(d) When object is kept at a distance greater than its radius of curvature

Solution

Question: 4

In given below figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is:

(a) $\frac{\sqrt{3}}{\sqrt{2}}$

(b) $\frac{\sqrt{2}}{\sqrt{3}}$

(d) $\sqrt[]{2}$

Solution

Question: 5

A light ray enters from medium A to medium B as shown in the following figure. The refractive index of medium B relative to A will be

(a) Greater than unity

(b) less than unity

(d) Zero

Solution

Question: 6

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the figure. Which of the following could be inside the box?

(a) A rectangular glass slab

(b) A convex lens

(d) A prism

Solution

Question: 7

A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the given below figure. Which of the following could be inside the box?

(a) Concave lens

(b) Rectangular glass slab

(d) Convex lens

Solution

Question: 8

Which of the following statements is true?

(a) A convex lens has 4 dioptre powers having a focal length 0.25 m

(b) A convex lens has $-$ 4 dioptre powers having a focal length 0.25 m

(d) A concave lens has $-$ 4 dioptre powers having a focal length 0.25 m

Solution

Question: 9

Magnification produced by a rear-view mirror fitted in vehicles:

(a) is less than one

(b) is more than one

(d) Can be more than or less than one depending upon the position of the object in front of it.

Solution

Question: 10

Rays from Sun converge at a point 15 cm in front of a concave mirror. Where an object should be placed so that size of its image is equal to the size of the object?

(a) 15 cm in front of the mirror

(b) 30 cm in front of the mirror

(d) More than 30 cm in front of the mirror

Solution

Question: 11

A full-length image of a distant tall building can definitely be seen by using:

(a) A concave mirror

(b) A convex mirror

(d) Both concave as well as plane mirror

Solution

Question: 12

In torches, search lights and headlights of vehicles the bulb is placed:

(a) Between the pole and the focus of the reflector

(b) Very near to the focus of the reflector

(d) At the centre of curvature of the reflector

Solution

Question: 13

The laws of reflection hold good for:

(a) Plane mirror only

(b) Concave mirror only

(d) All mirrors irrespective of their shape

Solution

Question: 14

The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in the given figure. Which one of them is correct?

(a) A

(b) B

(d) D

Solution

Question: 15

You are given water, mustard oil, glycerin and kerosene. In which of these media a ray of light incident obliquely at same angle would bend the most?

(a) Kerosene

(b) Water

(d) Glycerin

Solution

Question: 16

Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure X?

(a) Fig. A

(b) Fig. B

(d) Fig. D

Solution

Question: 17

Which of the following ray diagrams is correct for the ray of light incident on a lens shown in the following figure?

(a) Fig. A

(b) Fig. B

(d) Fig. D

Solution

Question: 18

A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.

(a) Plane, convex and concave

(b) Convex, concave and plane

(d) Convex, plane and concave

Solution

Question: 19

In which of the following, the image of an object placed at infinity will be highly diminished and point sized?

(a) Concave mirror only

(b) Convex mirror only

(d) Concave mirror, convex mirror, concave lens and convex lens

Solution

Short answer Questions

Question: 20

Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.

(a) Object is placed between device and its focus, image formed is enlarged and behind it.

(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.

(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.

(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

Solution

(a) Concave mirror

(b) Convex lens

(d) Convex mirror

Question: 21

Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

Solution

The refraction of light in a glass slab can be shown as given below.

On passing through a rectangular glass slab, a ray of light bends first at the air-glass interface and then again at the glass- air interface.

The angle of refraction by first interface is equal to the angle of incidence at the second interface, i.e. $∠ r 1 = ∠ r 2$ .

The final emergent ray is always parallel to the incident ray and is displaced through a distance.

Thus $∠ e = ∠ i$

Question: 22

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine? Support your answer with reason.

Solution

No, the bending of light will be different in different mediums since each medium has a different refractive index. The amount of light refracted or bent depends on the refractive index of the second medium. Since the refractive index of kerosene or turpentine is more than that of water, the pencil will appear to be more bent.

Question: 23

How is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?

Solution

Refractive index of medium is the ratio of the speed of light in vacuum (or air) to the speed of light in the medium. So, the refractive index,

$n = \frac{c}{v}$

Let one of the medium be $m_{1}$ . Its refractive index is $n_{1} = \frac{c}{v_{1}}$

Let the other medium be $m_{2}$ . Its refractive index is $n_{2} = \frac{c}{v_{2}}$

$∴ n_{21} = \frac{v_{1}}{v_{2}}$

Question: 24

Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

Solution

Given that, the refractive index of diamond with respect to glass is,

$\begin{array}{l} n_{d g} = \frac{v_{g}}{v_{d}} = 1.6 \end{array}$

The absolute refractive index of glass is,

$n_{g} = \frac{C}{v_{g}} = 1.5$

The absolute refractive index of diamond is,

$n_{d} = \frac{C}{v_{d}}$

Therefore, $\frac{v_{g}}{v_{d}} \times \frac{c}{v_{g}} = n_{d} = 1.6 \times 1.5 = 2.40$

Question: 25

A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

Solution

The given statement is correct.

Magnified virtual image

For the statement to be correct, the object has to be placed within 20 cm from the lens in the first case.

Magnified real image

For the statement to be correct, the object has to be placed between 20 cm and 40 cm.

Question: 26

Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

Solution

Sudha should move the screen towards the lens to obtain a clear image of the building because as the object distance increases, the image distance decreases.

The approximate focal length of this lens will be 15 cm.

Question: 27

How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

Solution

The power of a lens is inversely proportional to its focal length.

$\begin{array}{l} P = \frac{1}{f} \end{array}$

The lens of focal length 20 cm will have more power than the lens of focal length of 40 cm.

Therefore, the lens having focal length of 20 cm will provide more convergence.

Question: 28

Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.

Solution

In two plane mirrors, the incident ray and reflected ray will always be parallel to each other for any angle of incidence when the two plane mirrors are placed at right angle to each other. The following diagram illustrates this property.

Question: 29

Draw a ray diagram showing the path of rays of light when it enters with oblique incidence:

(i) From air into water

(ii) From water into air

Solution

Long answer Questions

Question: 30

Draw ray diagrams showing the image formation by a concave mirror when an object is placed:

(a) Between pole and focus of the mirror

(b) Between focus and centre of curvature of the mirror

(d) A little beyond centre of curvature of the mirror

(e) At infinity

Solution

Question: 31

Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) Between optical centre and focus of the lens

(b) Between focus and twice the focal length of the lens

(d) At infinity

(e) At the focus of the lens

Solution

Question: 32

Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Solution

The Laws of Refraction

(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media.

Let us consider the refraction of light through the following glass slab.

Here AO is the incident ray. OB is the refracted ray and BC is the emergent ray.

As per the law of refraction, the incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence (here O), all lie in the same plane.

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for light of a given colour and for a given pair of media. This law is also known as Snell’s law of refraction and is given as:

$\frac{\sin i}{\sin r} = cons \tan t (refractive index)$

Question: 33

Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) At the focus of the lens

(b) Between focus and twice the focal length of the lens

Solution

Question: 34

Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) At infinity

(b) At finite distance from the mirror

Solution

Question: 35

The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of80 cm and the lens?

Solution

The magnification,

$\begin{array}{l} m = - \frac{υ}{u} = - 3, \\ \Rightarrow u = \frac{80}{- 3} = - 26.67 cm \end{array}$

The object should be placed at a distance of 26.67 cm from the lens.

The image is real and inverted. The lens is convex.

Question: 36

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Solution

As image could be real or virtual, there are two possible situations.

Case I: In case of concave mirror.

$\begin{array}{l} M = - \frac{1}{3} \\ f = - 20 cm \\ m = - \frac{1}{3} = \frac{- v}{u} \\ v = \frac{u}{3} \\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ - \frac{1}{20} = \frac{3}{u} + \frac{1}{u} = \frac{4}{u} \\ - \frac{1}{20} = \frac{4}{u} \\ u = - 80 cm \end{array}$

Case II: In case of a convex mirror

$\begin{array}{l} M = + \frac{1}{3} \\ f = + 20 cm \\ m = + \frac{1}{3} = \frac{- v}{u} \\ v = \frac{- u}{3} \\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{20} = \frac{- 3}{u} + \frac{1}{u} = \frac{- 2}{u} \end{array}$

Therefore, $u = - 40 cm$

Question: 37

Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of $-$ 50 cm. What is the nature of the lens and its power used by each of them?

Solution

Power of a lens is the degree of its convergence or divergence. It is the reciprocal of its focal length in metres. Its unit is dioptre.

Case 1

The power of the lens in the first case,

$\begin{array}{l} P_{1} = \frac{1}{f} = \frac{1}{\frac{50}{100}} = 2 D \end{array}$

It is a convex lens.

Case 2

The power of the lens in the second case,

$\begin{array}{l} P_{2} = \frac{1}{f} = \frac{1}{\frac{- 50}{100}} = - 2 D \end{array}$

It is a concave lens.

Question: 38

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under:

Position of candle $= 12.0 cm$

Position of convex lens $= 50.0 cm$

Position of the screen $= 88.0 cm$

(i) What is the focal length of the convex lens?

(ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?

(iii) What will be the nature of the image formed if he further shifts the candle towards the lens?

(iv) Draw a ray diagram to show the formation of the image in case(iii) as said above.

Solution

(i) Object distance, $u = - (50 - 12) = - 38 cm$

Image distance $= (88 - 50) cm = 38 cm$

Therefore, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (f is the focal length)

$\Rightarrow \frac{1}{f} = \frac{1}{19} cm$

$\Rightarrow f = 19 cm$

(ii) Object distance on shifting is $(50 - 31) = 19$ cm.

So, $u = - 19$ cm.

Since the object is placed at the focus, the image will be formed at infinity.

(iii) When the student further shifts the candle towards the lens, the lens forms enlarged, virtual and erect image of the candle.

(iv) The ray diagram showing the formation of the image is given below.

Lesson: Light-Reflection and Refraction

Question: 1

Define the principal focus of a concave mirror.

Solution

It is a point on the principal axis where the light rays that are parallel to the principal axis converge, after reflecting from the mirror.

Question: 2

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

Solution

Radius of curvature, $R = 20 cm$

Radius of curvature of the spherical mirror $= 2 \times focal length (f)$

i.e.,

$\begin{array}{l} R = 2 f \\ f = \frac{R}{2} = \frac{20}{2} = 10 cm \end{array}$

Question:3

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

Solution

Concave mirror.

Question:4

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

Solution

Convex mirrors always form a virtual, erect, and diminished image of the objects placed in front of it. This helps in getting a wider view of objects, traffic and people behind the vehicle.

Question: 5

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

Solution

Given:

The radius of curvature, $R = 32 cm$ .

Radius of curvature,

$\begin{array}{l} R = 2 \times focal length (f) \\ \Rightarrow R = 2, f = \frac{R}{2} = \frac{32}{2} = 16 cm \end{array}$

Question: 6

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

Solution

Magnification produced by a spherical mirror, m = height of the image/height of the object.

Let the height of the object be $h_{0}$ and the height of the image be h₁.

$\Rightarrow m = \frac{h_{1}}{h_{0}} = - \frac{u}{v}$

The height of the image, $h_{1} = - 3 h_{0}$ (The image formed is real.)

$\begin{array}{l} \frac{-3 h}{h} = \frac{- v}{u} \\ \frac{v}{u} =3 \end{array}$

Object distance, $u = - 10 cm$
$v = 3 \times (- 10) = - 30 cm$
$\Rightarrow$ An inverted image is formed at a distance of 30 cm in front of the given concave mirror.

Question: 7

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

When a ray of light enters from an optically rarer medium to an optically denser medium, it bends towards the normal. Since water is optically denser than air, a ray of light travelling in water from air will bend towards the normal.

Question: 8

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 $3 \times 10^{8}$ ms^-1.

Solution

Let the refractive index of the glass be n_g.

$\Rightarrow n_{g} =$ speed of light in vacuum/speed of light in glass
Speed of light in vacuum, $c = 3 \times 10^{8} m s^{- 1}$
Refractive index of glass, $n_{g} = 1.50$
Therefore, the speed of light in the glass

= speed of light in vacuum/refractive index of glass
$= \frac{c}{n_{g}} = \frac{3 \times 10^{8}}{1.50} = 2 x 10^{8} {ms}^{-1}$

Question: 9

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

Material medium	Refractive index	Material medium	Refractive index
Air	1.0003	Canada Balsam	1.53
Ice	1.31	-	-
Water	1.33	Rock salt	1.54
Alcohol	1.36	-	-
Kerosene	1.44	Carbon disulphide	1.63
Fused quartz	1.46	Dense flint glass	1.65
Turpentine oil	1.47	Ruby	1.71
Benzene	1.50	Sapphire	1.77
Crown glass	1.52	Diamond	2.42

Solution

Diamond has the highest optical density.

Air has the lowest optical density.

Question: 10

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

Material medium	Refractive index	Material medium	Refractive index
Air	1.0003	Canada Balsam	1.53
Ice	1.31	-	-
Water	1.33	Rock salt	1.54
Alcohol	1.36	-	-
Kerosene	1.44	Carbon disulphide	1.63
Fused quartz	1.46	Dense flint glass	1.65
Turpentine oil	1.47	Ruby	1.71
Benzene	1.50	Sapphire	1.77
Crown glass	1.52	Diamond	2.42

Solution

Light travels faster when the refractive index of a medium is less. The speed of light is inversely proportional to the refractive index. Of all three given media, water has the lowest refractive index. So light will travel the fastest in water.

Question: 11

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

Solution

Light travels faster when the refractive index of a medium is less. The speed of light is inversely proportional to the refractive index. The refractive index of air is approximately 1.The refractive index of diamond is 2.42.

This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air.

Question:12

Define 1 dioptre of power of a lens.

Solution

1 dioptre is defined as the power of a lens having a focal length of 1 metre.

Question: 13

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.

Solution

Given:

i. Image is real and of same size. It implies that the position of image is at 2F.

ii. Object distance, $u = - 50 cm$
Image distance, $v = 50 cm$
Focal length $= f$

According to the lens formula,

$\begin{array}{l} \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \frac{1}{f} = \frac{1}{50} - \frac{1}{(-50)} \\ = \frac{1}{50} + \frac{1}{50} = \frac{1}{25} \\ \Rightarrow f = 25 cm \\ = 0 .25 m \end{array}$

Power of the lens, $P = \frac{1}{f (in metres)} = \frac{1}{0 .25} =+4D$

Question: 14

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

Solution

Let the power of lens be P and the focal length be f

The power of the lens (P) of focal length $f = \frac{1}{f}$

$= \frac{1}{- 2} = - 0.5 D$

Questions

Q1	What are the laws of reflection of light?

Q2	List the properties of an image formed by a plane mirror.

Q3	Describe refraction by double convex lens with figure.

Q4	Illustrate the nature, position and relative size of the image formed by a concave lens.

Q5	Derive the power of lens.

Q6	Differentiate between object distance and image distance.

Q7	Describe the relationship between the radius of curvature, R and focal length, f of a spherical mirror.

Q8	Explain the mirror formula for spherical mirrors.

Q9	What is meant by the magnification produced by a lens?

Q10	Describe diverging lenses.

Q11	Illustrate the new Cartesian sign convention for spherical mirrors using a diagram.

Q12	Draw and describe the two types of spherical mirrors

Q13	Tabulate the image formation by a concave mirror for different positions of the object.

Q14	Illustrate and explain the images formed by a convex mirror.

Q15	Illustrate the ray diagrams for the formation of image by a concave mirror for various positions of the object.

Q16	What are the sign conventions for reflection by spherical mirrors?

Q17	A convex mirror used for rear-view on an automobile has a radius of curvature of 30.00 m. If a bus is located at 10.00 m from this mirror, find the position, nature and magnification of the image.

Q18	Explain image formation in lenses using ray diagrams.

Q19	A concave lens has focal length of 10 cm. The object should be placed at what distance from the lens, so that it forms an image at 5 cm from the lens? Also, find the magnification produced by the lens.

Q20

The ray of the sun was reflected on a sheet of paper using a concave lens.

This produced fire. Explain with an illustration why fire was produced.

Questions

Q1	What are the laws of reflection of light?

A1	(i) The angle of incidence is equal to the angle of reflection and (ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane.

Q2	List the properties of an image formed by a plane mirror.

A2	(a) The image formed by a plane mirror is always virtual and erect. (b) The size of the image is equal to that of the object. (c) The image formed is as far behind the mirror as the object is in front of it. (d) The image is laterally inverted.

Q3	Describe refraction by double convex lens with figure.

A3	A lens may have two spherical surfaces, bulging outwards. Such a lens is called a double convex lens. It is simply called a convex lens. It is thicker at the middle as compared to the edges. Convex lens causes convergence of light rays as shown in the figure. Hence convex lenses are called converging lenses.

Q4	Illustrate the nature, position and relative size of the image formed by a concave lens.

A4	In the first case, the object is placed at infinity. The corresponding image is: a) Highly diminished. b) Virtual and erect. c) Formed at the focus (F). In the second case, the object is placed anywhere between the optical centre (O) and infinity. The corresponding image is: a)Diminished. b)Virtual and erect. c)Formed between the optical centre (O) and the focus (F).

Q5	Derive the power of lens.

A5	The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter P. The power P of a lens of focal length f is given by P = 1/ f.

Q6	Differentiate between object distance and image distance.

A6	Object distance: The distance of an object from the pole of the mirror is called the object distance (u). Image distance: The distance of an image from the pole of the mirror is called the image distance (v).

Q7	Describe the relationship between the radius of curvature, R and focal length, f of a spherical mirror.

A7	For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. We put this as R = 2f. This implies that the principal focus of a spherical mirror lies midway between the pole and the centre of curvature.

Q8	Explain the mirror formula for spherical mirrors.

A8

The mirror formula for the spherical mirrors is given by:

(1/v) + (1/u) = (1/f)

where, v = distance of the image from the pole of the mirror,

u = distance of the object from the pole of the mirror,

f = distance of the focus point from the pole of the mirror.

This formula is valid in all situations for all spherical mirrors, for all the positions of the object.

Q9	What is meant by the magnification produced by a lens?

A9

It is represented by the letter m. If h is the height of an object and h’ is the height of the image given by a lens, then the magnification produced by the lens is given by, m = height of the image/ height of the object = h/ h′.

Magnification produced by a lens is also related to the object-distance u, and the image-distance v.

This relationship is given by magnification, m = h′/h = -v/u.

Q10	Describe diverging lenses.

A10	A double concave lens is a transparent material bound by two spherical surfaces, curved inwards. It leads to the divergence of light rays and thus is also called diverging lens. It is thin in the middle and thick at the edges.

Q11	Illustrate the new Cartesian sign convention for spherical mirrors using a diagram.

A11

1. The object is always placed to the left of the mirror, which means that the light from the object falls on the mirror from the left-hand side.

2. All distances parallel to the principal axis are measured from the pole of the mirror.

3. All the distances measured to the right of the origin (along +ve x-axis) are taken as positive while those measured to the left of the origin (along –ve x-axis) are taken as negative.

4. Distances measured perpendicular to and above the principal axis (along +ve y-axis) are taken as positive.

5. Distances measured perpendicular to and below the principal axis (along –ve y-axis) are taken as negative.

Q12	Draw and describe the two types of spherical mirrors

A12	A spherical mirror is a mirror that has a curved reflecting surface. A spherical mirror, whose reflecting surface faces towards the centre of the sphere, is called a concave mirror. A spherical mirror, whose reflecting surface is curved outwards, is called a convex mirror.

Q13	Tabulate the image formation by a concave mirror for different positions of the object.

A13

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus, F	Highly diminished, point-sized	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Virtual and erect

Q14	Illustrate and explain the images formed by a convex mirror.

A14

The nature, position and relative size of the image formed by a convex mirror is tabulated below.

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus, F, behind the mirror	Highly diminished, point sized	Virtual and erect
Between infinity and the pole, P of the mirror	Between P and F, behind the mirror	Diminished	Virtual and erect

The ray diagrams for the formation of image by a convex mirror for these two positions of the objects are shown in figure (a) and (b).

Q15	Illustrate the ray diagrams for the formation of image by a concave mirror for various positions of the object.

A15

Position of the object	Position of the image	Size of the image	Nature of the image
At infinity	At the focus, F	Highly diminished, point sized	Real and inverted
Beyond C	Between F and C	Diminished	Real and inverted
At C	At C	Same Size	Real and inverted
Between C and F	Beyond C	Enlarged	Real and inverted
At F	At infinity	Highly enlarged	Real and inverted
Between P and F	Behind the mirror	Enlarged	Real and inverted

Q16	What are the sign conventions for reflection by spherical mirrors?

A16

In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system. The conventions are as follows –

(i) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.

(ii) All distances parallel to the principal axis are measured from the pole of the mirror.

(iii) All the distances measured to the right of the origin (along +ve x-axis)are taken as positive while those measured to the left of the origin (along –ve x-axis) are taken as negative.

(iv) Distances measured perpendicular to and above the principal axis (along +ve y-axis) are taken as positive.

(v) Distances measured perpendicular to and below the principal axis (along –ve y-axis) are taken as negative.

Q17	A convex mirror used for rear-view on an automobile has a radius of curvature of 30.00 m. If a bus is located at 10.00 m from this mirror, find the position, nature and magnification of the image.

A17

Radius of curvature, R = + 30.00 m

Object-distance, u = – 10.00 m

Image-distance, v =?

Height of the image, h ′ =?

∴ Focal length, f = R/2 = + 30.00 m/ 2 = + 15 m.

Since,

1/v + 1/u = 1/f

1/v = 1/f -1/u = + 1/ 15 + 1
(10) = (2+3)/30 = 5/30 =1/6

v = + 6 m

The image is 6 m at the back of the mirror.

Magnification, m = −v/u = (–6 m)
(- 10.00) m = + 0.6 m.

The image is virtual, erect and smaller in size by a factor of 0.6.

Q18	Explain image formation in lenses using ray diagrams.

A18

For drawing ray diagrams in lenses, alike of spherical mirrors, we consider any two of the following rays –

(i) A ray of light from the object, parallel to the principal axis, after refraction from a convex lens passes through the principal focus on the other side of the lens, as shown in figure (a). In case of a concave lens, the ray appears to diverge from the principal focus located on the same side of the lens, as shown in figure (b).

(ii) A ray of light passing through a principal focus, after refraction from a convex lens, will emerge parallel to the principal axis. This is shown in figure (a). A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis. This is shown in figure (b).

(iii) A ray of light passing through the optical centre of a lens will emerge without any deviation. This is illustrated in figure (a) and figure (b).

Q19	A concave lens has focal length of 10 cm. The object should be placed at what distance from the lens, so that it forms an image at 5 cm from the lens? Also, find the magnification produced by the lens.

A19

A concave lens always forms a virtual, erect image on the same side of the object. Image distance, v = –5 cm

Focal length, f = –10 cm

Object distance u =?

Since 1/v – 1/u = 1/f

or,

1/u = 1/v – 1/f

⇒1/u= -1/5 – 1/(-10) = -1/5 +1/10

⇒1/u = (-2+1)/ 10 = -1/10

or, u = – 10 cm.

Thus, the object-distance is 10 cm.

Magnification, m = v/u

m = - 5cm/ - 10 cm = 1/2 = + 0.50

The positive sign shows that the image is erect and virtual. The image is half of the size of the object.

Q20	The ray of the sun was reflected on a sheet of paper using a concave lens. This produced fire. Explain with an illustration why fire was produced.

A20

The given phenomenon is explained through the illustration given below:

The figure illustrates that a number of rays parallel to the principal axis fall on a concave mirror. The reflected rays meet/intersect at the same point on the principal axis of the mirror. This point is called the principal focus of the concave mirror.

Since the light from the sun is converged at the focus, the image of the object appears as a spot of light. The paper is placed at the focus. The fire is produced due to the concentration of sunlight at the focus.