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SEBA Class 8 Science Chapter 16 Light
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Light
Chapter: 16
PART – II
EXERCISE
1. Suppose you are in a dark room. Can you see objects in the room? Can you see objects outside the room? Explain.
Ans: In a dark room, you cannot see objects because light is required to reflect off them and reach your eyes. Since there is no light source in the room, no light is available for reflection, and hence, the objects remain invisible. For objects outside the room, if the room is entirely dark and there is no light coming from an external source like a window or streetlight, you would also not be able to see them. However, if some light does enter the room (for instance, through a window or a crack), you would be able to see the objects outside depending on the amount of light entering the room.
2. Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection?
Ans: Regular Reflection: Regular reflection occurs when all of the parallel rays reflected from a plane surface are parallel.
Diffused Reflection: Diffused or irregular reflection occurs when all of the parallel rays reflected from a plane surface are not parallel.
No, diffused reflection does not mean the failure of the laws of reflection. The laws are still followed (angle of incidence = angle of reflection), but the surface is rough, causing the rays to scatter.
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.
Ans: Regular Reflection occurs because the surface is smooth and shiny. When light strikes this smooth surface, it reflects in parallel rays, creating a clear reflection. This is what happens with polished or glossy surfaces.
(b) Chalk powder.
Ans: Diffused Reflection happens because the surface of chalk powder is rough and uneven. The roughness scatters the incoming light in various directions, meaning no clear image is formed, and the light diffuses over a wide area.
(c) Cardboard surface.
Ans: Diffused Reflection occurs because cardboard has a rough texture. When light hits this uneven surface, it scatters in different directions, so the reflection is not clear or sharp. You see the color of the cardboard but not a sharp image of objects.
(d) Marble floor with water spread over it.
Ans: Regular Reflection happens because water forms a smooth surface when spread over the marble floor. The smooth water surface reflects light in parallel rays, forming a clear, sharp reflection (similar to a mirror).
(e) Mirror.
Ans: Regular Reflection occurs in a mirror because the surface is extremely smooth and flat. When light strikes this perfectly smooth surface, it reflects at a precise angle, creating a clear image of the object in front of it.
(f) Piece of paper.
Ans: Diffused Reflection happens because the surface of the paper is rough, even though it may seem flat. The tiny fibers and texture of the paper scatter the incoming light in many directions, so you don’t get a sharp reflection. Instead, you see the general color or texture of the paper, but no clear image.
4. State the laws of reflection.
Ans: The law of reflection states that the angle of reflection equals the angle of incidence— θr = θi. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.
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.
Ans: Students, do yourself.
6. Fill in the blanks in the following.
(a) A person 1 m in front of a plane mirror seems to be ____meters away from his image.
Ans: 2.
(b) If you touch your _______ ear with the right hand in front of a plane mirror, it will be seen in the mirror that your right ear is touched with________ .
Ans: Right , left.
(c) The size of the pupil becomes________ when you see in dim light.
Ans: Larger.
(d) Night birds have_______ cones than rods in their eyes.
Ans: More.
7. Angle of incidence is equal to the angle of reflection.
(a) Always.
(b) Sometimes.
(c) Under special conditions.
(d) Never.
Ans: (a) Always.
8. 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.
Ans: (b) virtual, behind the mirror, and of the same size as the object.
9. Describe the construction of a kaleidoscope.
Ans: A kaleidoscope is made up of a hollow tube. Inside the tube, three mirrors are arranged in the shape of a triangular tube with their reflecting surfaces facing each other. One end of the tube is covered with a transparent sheet (usually containing small colored pieces of glass or beads), and the other end is covered with an opaque sheet that has an eyehole. When you look through the eyehole, light reflects off the mirrors, creating symmetrical and beautiful patterns from the small objects inside the transparent end.
10. Draw a labelled sketch of the human eye.
Ans:
11. Gurmit wanted to perform Activity 7.8 using a laser torch. Her teacher advised her not to do so. Can you explain the basis of the teacher’s advise?
Ans: The teacher likely advised against using a laser torch because laser beams are highly concentrated and intense. Direct exposure to a laser beam, especially to the eyes, can cause serious damage, as lasers can burn or permanently harm the retina. This is especially dangerous for students who may not be fully aware of the precautions required when working with lasers.
12. Explain how you can take care of your eyes.
Ans: To take care of your eyes, you should:
(i) Avoid staring directly at bright lights, including the sun, lasers, and strong artificial lights.
(ii) Maintain a proper distance from screens (phone, computer, TV) and take breaks using the 20-20-20 rule: look at something 20 feet away for 20 seconds every 20 minutes.
(iii) Eat a balanced diet rich in vitamins and minerals, such as vitamin A, to support eye health.
(iv) Wear protective glasses when working with harmful chemicals or bright lights.
(v) Ensure you get regular eye check-ups with an optometrist to monitor any changes in vision.
13. What is the angle of incidence of a ray if the reflected ray is at an angle of 90° to the incident ray?
Ans: The angle of incidence and the angle of reflection are always equal in accordance with the law of reflection. If the reflected ray is at an angle of 90° to the incident ray, then the angle of incidence must be 45°. This is because the sum of the angle of incidence and the angle of reflection must always equal 90°.
14. How many images of a candle will be formed if it is placed between two parallel plane mirrors separated by 40 cm?
Ans: When an object is placed between two parallel mirrors, the number of images formed is given by the formula:
For two parallel mirrors, the angle between them is 0°. However, the number of images formed is influenced by the number of reflections possible between the mirrors. When two parallel mirrors are placed facing each other, an infinite number of images are theoretically formed. In practice, the number of images may be limited by the size of the mirrors and the distance between the object and the mirrors.
15. Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30° as shown in Fig.7.19. Draw the reflected ray from the second mirror.
Ans: To find the reflected ray from the second mirror, we can follow these steps:
(i) Incident Ray: A ray of light is incident on the first mirror at an angle of 30°.
(ii) Reflection from the First Mirror: The ray reflects off the first mirror at an angle equal to the angle of incidence (30°).
(iii) Second Mirror Reflection: The reflected ray from the first mirror now strikes the second mirror. The angle of incidence on the second mirror will be 60° because the two mirrors meet at a right angle (90°), and the angle between the incident ray on the second mirror and the reflected ray from the first mirror will be complementary (i.e., 90°).
(iv) Final Reflected Ray: The ray will then reflect off the second mirror, following the law of reflection (angle of incidence = angle of reflection), and this reflected ray will form an angle of 60° with the second mirror.
16. Pragyan stands at A just on the side of a plane mirror as shown in Fig 7.20. Can he see himself in the mirror? Also can he see the image of objects situated at P, Q and R?
Ans: Based on the information provided, Pragyan stands at point A, just on the side of a plane mirror. Here’s an analysis of the situation:
(i) Can Pragyan see himself in the mirror?
= No, Pragyan cannot see his own reflection in the mirror if he is positioned on the side of the plane mirror, as the reflection can only be seen when the observer’s line of sight is perpendicular to the mirror surface.
(ii) Can Pragyan see the image of objects situated at P, Q, and R?
= If the objects (P, Q, R) are positioned such that their reflections can travel toward Pragyan’s line of sight, then he would be able to see their reflections. However, this depends on the positioning of these objects relative to the mirror and Pragyan’s position. The angles of incidence and reflection determine the visibility.
17. (a) Find out the position of the image of an object situated at A in the plane mirror (Fig. 7.21).
Ans: Position of the Image of Object A:
(i) The image of any object formed by a plane mirror is located at the same distance behind the mirror as the object is in front of it.
(ii) In this case, the image of Pragyan (at point A) will be formed at the same distance behind the mirror, at an imaginary point directly behind the mirror corresponding to A. It will be an upright and virtual image.
(b) Can Prahelika at Point B See the Image?
Ans: (i) Prahelika, being at point B, will be able to see the image of Pragyan at point A in the mirror if she is positioned such that the rays from the image (reflected from the mirror) can reach her eyes.
(ii) If B is along the line of sight for the reflection, Prahelika can indeed see the image.
(c) Can Pragyan at Point C See the Image?
Ans: Since Pragyan is at point A and the image is formed behind the mirror at an equal distance, Pragyan will not be able to see the image in the mirror directly from A. However, if Pragyan moves to point C, depending on the positioning and the angle, he may or may not be able to view the image from C.
(d) When Prahelika Moves from B to C, Where Does the Image of A Move?
Ans: As Prahelika moves from point B to point C, the position of the image of A (the object) does not change in the mirror; it remains fixed in the position directly behind the mirror. However, as Prahelika moves to a different location, the angle at which she views the image will change.
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