Light

Exercises

Question 1

Suppose you are in a dark room. Can you see objects in the room? Can you see objects outside the room? Explain.

Answer

No, we cannot see objects in a dark room because there is no light to illuminate the objects in the room. When light falls on the eyes after reflecting from the object, it becomes visible. If the light is available outside the room, then the objects present would reflect it and hence we would be able to see objects.

Question 2

Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection?

Answer

Diffused reflection is not due to the failure of the laws of reflection. It is caused by the irregularities in the reflecting surface, like that of a cardboard.

Question 3

Mention against each of the following whether regular or diffused reflection will take place when a beam of light strikes. Justify your answer in each case.

(a) Polished wooden table

(b) Chalk powder

(c) Cardboard surface

(d) Marble floor with water spread over it

(e) Mirror

(f) Piece of paper

Answer

(a) Polished wooden table — Regular reflection
Reason — The polished wooden table has a smooth surface, hence, the reflections are regular.

(b) Chalk powder — Diffused reflection
Reason — Chalk powder spread on a surface will form an irregular surface, hence, the reflections are diffused.

(c) Cardboard surface — Diffused reflection
Reason — The surface of the cardboard is irregular, hence, the reflections are diffused.

(d) Marble floor with water spread over it — Regular reflection
Reason — Marble floor with water spread over it is a smooth surface, hence, the reflections are regular.

(e) Mirror — Regular reflection
Reason — A mirror has a very smooth surface, hence, the reflections are regular.

(f) Piece of paper — Diffused reflection
Reason — Piece of paper has many irregularities, hence, the reflections are diffused.

Question 4

State the laws of reflection.

Answer

Two laws of reflection are :

(i) The angle of incidence is equal to the angle of reflection.

(ii) Incident ray, reflected ray and the normal drawn at the point of incidence to the reflecting surface, lie in the same plane.

Question 5

Describe an activity to show that the incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.

Answer

Take a sheet of stiff paper or a chart paper. Let the sheet project a little beyond the edge of the table. Cut the projecting portion of the sheet in the middle. Look at the reflected ray. Make sure that the reflected ray extends to the projected portion of the paper. Bend that part of the projected portion on which the reflected ray falls.

When the whole sheet of paper is spread on the table, it represents one plane. The incident ray, the normal at the point of incidence and the reflected ray are all in this plane.

When we bend the paper we create a plane different from the plane in which the incident ray and the normal lie. Then we do not see the reflected ray.

Hence, the above activity indicates that the incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane.

Question 6

Fill in the blanks in the following.

(a) A person 1 m in front of a plane mirror seems to be ............... m away from his image.

(b) If you touch your ............... ear with your right hand in front of a plane mirror, it will be seen in the mirror that your right ear is touched with ...............

(c) The size of the pupil becomes ............... when you see in dim light.

(d) Night birds have ............... cones than rods in their eyes.

Answer

(a) A person 1 m in front of a plane mirror seems to be 2 m away from his image.

Reason — When a person stands in front of a plane mirror, their image appears to be as far behind the mirror as the person is in front of it.
So, when the person is 1 m away from the mirror. His image will be formed at 1 m from the mirror on the other side, hence, his image will be 1 + 1 = 2 m from the person himself.

(b) If you touch your left ear with your right hand in front of a plane mirror, it will be seen in the mirror that your right ear is touched with your left hand.

Reason — The image formed by a mirror is laterally inverted hence, when we touch our left ear with our right hand in front of a plane mirror, it will be seen in the mirror that our right ear is touched with our left hand.

(c) The size of the pupil becomes large when you see in dim light.

Reason — In dim light conditions, there's less available light for vision. To allow more light to enter the eye and improve visibility in low-light environments, the iris dilates the pupil. When the pupil dilates, it enlarges in size, allowing more light to pass through the eye and reach the retina at the back of the eye.

(d) Night birds have fewer cones than rods in their eyes.

Reason — Having fewer cones and more rods allows night birds to have better night vision. Rods are more sensitive to low levels of light, making them ideal for detecting prey and navigating in the dark. Additionally, having fewer cones means that night birds likely have reduced color vision or even monochromatic vision, which is less important for hunting in low-light conditions.

Question 7

The angle of incidence is equal to the angle of reflection.

1. Always
2. Sometimes
3. Under special conditions
4. Never

Answer

Always

Reason — According to the laws of reflection :

The angle of incidence is equal to the angle of reflection.

Question 8

Image formed by a plane mirror is

1. virtual, behind the mirror and enlarged.
2. virtual, behind the mirror and of the same size as the object.
3. real at the surface of the mirror and enlarged.
4. real, behind the mirror and of the same size as the object.

Answer

virtual, behind the mirror and of the same size as the object.

Reason — Image formed by a plane mirror is virtual, behind the mirror and of the same size as the object.

Question 9

Describe the construction of a kaleidoscope.

Answer

To make a kaleidoscope, we will take three rectangular mirror strips each about 15 cm long and 4 cm wide. We will join them together to form a prism as shown in the figure below:

After that, we will fix the arrangement of mirrors in a circular cardboard tube or tube of a thick chart paper. It should be slightly longer than the mirror strips. Close one end of the tube by a cardboard disc having a hole in the centre, through which we can see.

In order to make the disc durable, we will paste a piece of transparent plastic sheet under the cardboard disc. At the other end, touching the mirrors, we will fix a circular plane glass plate.

After that, we will place on this glass plate several small pieces of coloured glass (broken pieces of coloured bangles). Close this end of the tube by a ground glass plate. Allow enough space for the colour pieces to move around.

Our kaleidoscope is ready.

Question 10

Draw a labelled sketch of the human eye.

Answer

Labelled sketch of the human eye is shown below:

Question 11

Gurmit wanted to perform Activity 13.8 using a laser torch. Her teacher advised her not to do so. Can you explain the basis of the teacher's advice?

Answer

Intensity of laser light is very high. It can cause damage to the retina and lead to blindness. Hence, Gurmit's teacher advised her not to use a laser light for the activity as she was concerned about the risk of accidental exposure to the laser beam, which could potentially cause damage to Gurmit's eyes or the eyes of others nearby.

Question 12

Explain how you can take care of your eyes.

Answer

In order to take care of our eyes we should do the following:

1. We should always read at the normal distance for vision. We should not bring the book too close to our eyes or keep it too far.
2. We should not look at the Sun or a powerful light directly.
3. Too little or too much light is bad for eyes. Insufficient light causes eyestrain and headaches. Too much light, like that of the Sun, a powerful lamp or a laser torch can injure the retina.
4. We should never rub our eyes. If particles of dust go into our eyes, we should wash our eyes with clean water or visit the doctor if need be.
5. If advised, we should use suitable spectacles.

Question 13

What is the angle of incidence of a ray if the reflected ray is at an angle of 90° to the incident ray?

Answer

Given,

∠i + ∠r = 90° where

∠i = angle of incident.

∠r = angle of reflection

According to the law of reflection,

∠i = ∠r

Substituting the above law in the given condition we get,

2∠i = 90°

∠i = $\dfrac{90}{2}$ = 45°

Therefore, angle of incidence = 45°.

Question 14

How many images of a candle will be formed if it is placed between two parallel plane mirrors separated by 40 cm?

Answer

When a candle is placed between two parallel plane mirrors, infinite images are formed due to repeated reflections between the mirrors.

Question 15

Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30° as shown in Fig. 13.19. Draw the reflected ray from the second mirror.

Answer

Let MM' and M'M'' be the two plane mirrors at right angles.

The reflected ray from the second mirror is shown in the figure below:

Given that the ray is incident on mirror MM' at 30°,

∴ ∠AOX = 30°

According to the law of reflection,

∠i = ∠r

∴ ∠XOO' = ∠AOX = 30°

As OX ⊥ MM' and O'X ⊥ M'M''

∴ OX ⊥ O'X

∴ ∠OXO' = 90°

In ΔOXO', by angle sum property.

∠XOO' + ∠OXO' + ∠XO'O = 180°

⇒ 30° + 90° + ∠XO'O = 180°

⇒ ∠XO'O = 180° - 90° - 30°

⇒ ∠XO'O = 60°

∴ Angle of incidence of the ray on M'M'' = 60°

∴ Angle of reflection of the ray from M'M'' = 60° [∵ ∠i = ∠r]

Question 16

Boojho stands at A just on the side of a plane mirror, as shown in Fig. 13.20. Can he see himself in the mirror? Also, can he see the image of objects situated at P, Q and R?

Answer

Boojho cannot see himself in the mirror because light rays originating from him do not reach his eyes after reflection.

He can see the image of objects situated at P and Q because the light rays coming from P and Q get reflected by the mirror and reach his eyes.

Boojho can't see the image of object situated at R because the light rays from object R do not reach his eyes after reflection.

Question 17

(a) Find out the position of the image of an object situated at A in the plane mirror (Fig. 13.21).

(b) Can Paheli at B see this image?

(c) Can Boojho at C see this image?

(d) When Paheli moves from B to C, where does the image of A move?

Answer

(a) Image of the object placed at A is formed behind the mirror and distance of the image from the mirror is equal to the distance of A from the mirror.

(b) Yes, Paheli at B can see this image.

(c) Yes, Boojho at C can see this image.

(d) Image of the object at A will not move, hence, it will remain in the same position when Paheli moves from B to C.