Question 1. Suppose you are in a dark room. Can you see objects in the room? Can you see objects outside the room? Explain.
Solution 1. In a dark room nothing will be visible because any object is visible only when it gets reflected by light hence, in a dark room you cannot see an object. Whereas, objects outside the dark room will be clearly visible because of the presence of light.
Question 2. Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection?
Solution 2.
| S. No | Regular Reflection | Diffused Reflection |
| 1. | It occurs on a smooth surface. | It occurs on a rough surface. |
| 2. | Light gets reflected in a particular direction. | Light gets reflected in random directions. |
| Example: Reflection by the plane mirror | Example: Reflection by the road surface. |
No, diffused reflection does not mean failure of the laws of reflection although, in diffused reflection, the rays aren’t parallel to the incident rays.
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
Solution:
a) Polished wooden table- Regular reflection will take place because this is a plane and polished surface.
b) Chalk powder- As the surface of the chalk powder is rough hence diffused reflection will take place.
c) Cardboard surface- As the surface of the cardboard is rough hence, the diffused reflection will take place.
d) Marble floor – Regular reflection will take place because the marble floor is a plane and smooth surface.
e) Mirror- Regular reflection will take place because the mirror is a plane and very smooth surface.
f) Piece of paper- As the surface of the paper is considered to be rough surface hence, the diffused reflection will take place.
Question 4. State the laws of reflection.
Solution 4.
The law of reflection states that
a) The angle of reflection is equal to the angle of incidence.
b) The reflected ray, the incident ray, and the normal to the reflective surface at the point of incidence all lie on 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.
Solution 5. Place a plane mirror on a table perpendicular to the plane of the table. Place a sheet of paper on the table in such a way that it makes a contact with the bottom of the mirror. Now, take another piece of paper make a hole in that paper sheet and pass the light through it on the bottom of the mirror. Now you will notice that the incident ray, normal line and the reflected ray all lie on 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 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.
Solution 6.
(a) A person 1 m in front of a plane mirror seems to be 2m away from his image.
(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.
(c) The size of the pupil becomes large when you see in dim light.
(d) Night birds have fewer cones than rods in their eyes.
Choose the correct option in Questions 7 – 8
Question 7. The angle of incidence is equal to the angle of reflection.
(a) Always
(b) Sometimes
(c) Under special conditions
(d) Never
Solution 7.
(a) Always
Question 8. The image formed by a plane mirror is
(a) virtual, behind the mirror and enlarged.
(b) virtual, behind the mirror and of the same size as the object.
(c) real at the surface of the mirror and enlarged.
(d) real, behind the mirror and of the same size as the object.
Solution 8.
(b) virtual, behind the mirror and of the same size as the object.
Question 10. Draw a labelled sketch of the human eye.
Solution 10
Question 11. Gurmit wanted to perform Activity 16.8 using a laser torch. Her teacher advised her not to do so. Can you explain the basis of the teacher’s advice?
Solution 11. The intensity of the laser light is very high, it can be harmful to the human eyes. It can lead to blindness causing damage to the retina which is why her teacher advised her not to perform that activity.
Question 12. Explain how you can take care of your eyes.
Solution 12. We can take care of your eyes in the following ways:
- a) we should avoid reading in bright light as well as in dim light.
- b) He should go for an eye check-up on regular basis.
- c) we should not rub our eyes in case of any irritation instead we should immediately wash our eyes with cold water.
- d) we should avoid direct exposure of sunlight to eye as it can be harmful.
- e) We should maintain a distance of at least 25 cm between the eyes and the book while reading.
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?
Solution 13. the angle of incidence is 45o because According to the law of reflection, the angle of incidence is equal to the angle of reflection. Therefore, both the angle of incidence and the angle of reflection are 90/2=45o.
Question 14. How many images of a candle will be formed if it is placed between two parallel plane mirrors separated by 40 cm?
Solution 14. multiple and infinite images of the candle will be formed if it is placed between two parallel plane mirrors separated by 40 cm, due to the multiple reflections between the mirrors. When two mirrors are placed parallel to each other then, infinite numbers of images are formed.
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. 16.19. Draw the reflected ray from the second mirror.
Solution 15.
The first law of reflection is used to obtain the path of reflected light.
It can be observed that the given ray of light will reflect from the second mirror at an angle of 60°.
The incident ray OA reflects at point O
Since Angle of Reflection = Angle of Incidence
OO’ makes an angle 300 with the normal of the first mirror
Now, drawing normal at O’
The two normals intersect at 900 angle
And applying angle sum property in △ OXO’
We get
The angle of Incidence in 2nd mirror = 600
Applying the law of reflection in 2nd mirror,
We get the below figure
mirror,
We get the below figure
Question 16. Boojho stands at A just on the side of a plane mirror as shown in Fig. 16.20. Can he see himself in the mirror? Also, can he see the image of objects situated at P, Q and R?
Solution 16. The reflected ray won’t reach Boojho’s eyes hence, he won’t be able to see his image. He can see the image of points P and Q but not the image of point R because the reflected rays from P and Q will reach his eyes but point R will not get reflected.
Question 17. (a) Find out the position of the image of an object situated at A in the plane mirror (Fig. 16.21).
(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?
Solution 17.
a) Image A’ of an object placed at A will be formed behind the mirror. The distance of image A’ from the mirror and the distance of A from the mirror will be equal.
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. When Paheli moves from B to C It will remain in the same position.