C3_Ch17_FS_e

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NSS Physics in Life Full Solution of Textbooks(Wave Motion)

Chapter 17 Refraction of Light

Checkpoint (p.132)

1. A. A light ray bends towards the normal upon entering an optically denser

medium.

2. P

3. Applying Snell’s law, we have

The angle of refraction is 25.4°.

4. nX = = 1.33.

Checkpoint (p.133)

1. C

2. Medium O. It is the optically least dense medium and thus light travels the fastest

in it.

3. C

Exercise (p.133)

1. C

2. C

3. C

4. (a) Tabulate the data as shown.

sin θa sin θg

0.000 0.000

0.174 0.115

0.342 0.225

0.500 0.326

0.643 0.423

0.766 0.515

The following graph is plotted:

1



(b) The refractive index is equal to the slope of the graph in (a), i.e.

≈ 1.51.


The angle of refraction is 25.9°.

6. From the diagram,

Applying Snell’s law, we have

The height of the building is 108 m.

2

0.05 0.1 015 0.2 0.25 0.3 0.350

0.1

0.4 0.45 0.5

0.2

0.3

0.4

0.5

0.6

0.7

0.8

sin θa

sin θg

0.55




β = α ≈ 32.1°


8. The refractive index of material X is or .

Thus we have

Checkpoint (p.137)

1. C

2.

3

observer

object

image

aquarium



3.

Checkpoint (p.140)

1. C

2. (a) Correct. Light ray a bends less upon entering the glass, showing that the

glass has a smaller refractive index for a.

(b) Incorrect. The glass has a smaller refractive index for a and thus the speed

of a is higher.

(c) Incorrect. All light waves travel at the same speed in vacuum.

Exercise (p.140)

1. A

2. Light rays from the girl’s legs bend away from the normal when they cross the

water-air boundary. The images of the legs are at higher positions than the actual

objects, causing the observer to think that the legs are shorter than they really

are.

3. Let n be the refractive index of the glass.

(a) Applying Snell’s law, we have

4

I



n = = 1.462 ≈ 1.46.

(b) Applying Snell’s law, we have

4. (a) Applying Snell’s law, we have

a = b = d = e ≈ 19.4°

Applying Snell’s law, c = f = 30°.

(b)

5. (a) (i) Light rays from the fish bend away from the normal when they cross

the water-air boundary. Thus the image of the fish is at a nearer

position to Michael than the actual object.

(ii) Light rays from Michael bend towards from the normal when they

cross the air-water boundary. Thus the image of Michael is at a farther

position to the fish than the actual object.

5

Michael’s eyes

fish

image of the fish

waterair



(b) (i) The image of the fish as observed by Michael becomes nearer to him.

(ii) The image of Michael as observed by the fish becomes farther from it.

6.

7. (a) Applying Snell’s law, the angle of refraction r for the red ray is

Thus the angle of refraction for the blue ray is 25.2120° − 0.27° = 24.9420°.

Applying Snell’s law, the refractive index for blue light is

≈ 1.524.

(b) The ratio is 1.524:1.509 for the travelling speed of red light in the glass to

that of blue light.

Checkpoint (p.145)

1. Table Q1:

medium refractive index critical angle

Perspex 1.50 41.8°

water 1.33 48.8°

diamond 2.42 24.4°

plastic block 1.54 40.5°

2. (a) Incorrect

(b) Incorrect

(c) Correct

6

Michaelfish’s eyes

waterair

image of Michael

image

image



Checkpoint (p.146)


Thus the vertex angle is 2θ ≈ 95.6°.

(b) From the diagram, we have

Exercise (p.153)

1. (a) the occurrence of fish-eye view (or any reasonable answer)

(b) light transmission in optical fibres (or any reasonable answer)


(b) The critical angle is ≈ 41.1°.

3. (a) The critical angle is of material X is ≈ 42.5°.

(b) Applying Snell’s law, we have

4. Consider the diagram below.

7



Applying Snell’s law, the refractive index is ≈ 1.49.

5. (a), (b)

(c) Using prismatic periscope can avoid the formation of multiple images and

hence a sharper image is obtained.

(d)

6. (a) (i) The critical angle of the water is = 47.79 ≈ 47.8°.

(ii)

(b) Yes. The diagram is shown below.

7. Let n be the refractive index of liquid X.


Thus the speed of light in liquid X is ≈ 2.25 × 108 m s−1.

Chapter Exercise (p.157)

1. D

2. B

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3. D

4. B

5. A

6. C

7. D

8. A

9. B

10. D

11. A

12. (a) (1A for larger degree of bending by the violet light)

(b) The critical angle of the glass for red light is larger than that for violet light.

(c) The violet light disappears first. (1A)

The angle of incidence increases as the block is moved to the right. (1A)

From (b), we know that he violet light undergoes total internal reflection

first and thus it disappears from the screen first. (1A)

13. (a) (1A for the image, 1A for correct bending of light)

(b) The key appears nearer to the swimmer. (1A)

(c) Consider the following diagram.

9

violet ray

image



(1M)


(1M)

Therefore,

(1A)

14. (a) Direct a light ray to the middle of the straight edge of the semi-circular

block (1A). Measure the angle of incidence i and the corresponding angle of

refraction r using the protractor (1A). Plot a graph of sin i against sin r and a

straight line will be obtained (1A). This shows that sin i is directly

proportional to sin r.

(b) The slope of the graph in (a) is the refractive index n of the material the

semi-circular block is made of (1A). The critical angle c of the material can

be calculated using the equation (1A).

15. (a) (i) reflection (1A) and refraction (1A)

(ii) (1A for refracted light rays parallel to the ones incident from the air)

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key

d

θa1.5 m

2.5 m

1 m

swimmer’s eyes

θw



(b) (i) (1A for correct paths in each case)

(ii) The phenomenon is known as total internal reflection (1A). Using

prisms can avoid the formation of multiple images and hence a clearer

image can be obtained (1A).

16. (a) (1A for rays and 1A for the images)

(b) No (1A). The image is virtual, i.e. light rays do not come from the image

directly, and it cannot be caught by a screen (1A).

17. (a) (i) Applying Snell’s law we have

(1M+1A)

(ii) (1A for correct label of c)

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image

image



(b) (i) total internal reflection (1A)

(ii) α = = 40° (1A)

Applying Snell’s law,

(1A)

(iii) (1A for extended rays, 1A for correct position of image)

(iv) The image is virtual. (1A)

(v) (1A)

18. (a) By the law of reflection, a = 30°. (½A)

Applying Snell’s law,

(1A)

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c

image



b = c = d = e ≈ 18.9° (1A)

Applying Snell’s law, f = 30° (½A)

(b) reflection (1A) and refraction (1A)

(c) (2A for correct paths in the bead and leaving the bead)

19. (a) (2A for correct paths of light rays and 1A for the image formation)

(b) (2A for correct paths of light rays and 1A for the image formation)

13

image

coin

image

image

observer



20. (a) The critical angle for the violet light is = 40.18°. (1M)

(1A for correct path in the prism and 1A for correct path leaving the prism)

(b) The critical angle for the red light is = 41.14°. (1M)

(1A for correct path in the prism and 1A for correct path leaving the prism)

21. (LE-GCE O-level May 2002 P1 Q11)

22. (LE-GCE O-level May 2005 P2 Q4)

23. (WJEC/ A/AS-level Jan 2006 Paper 541/01 Q5)

24. (UC-IGCE May/June 2006 P3 Q6)

25. (HKCEE 1999 P1 Q5)

26. (HKCEE 2001 P1 Q7)

27. (HKCEE 2003 P1 Q2)

28. (a) A plane mirror consists of a glass with a silver coating at the back. Light is

14

image

coin

image

observer

40°85.1°

40°77.7°



reflected by both sides and multiple images are formed (1A). In contrast, a

prism does not form multiple images and hence sharper images are obtained

(1A).

(b) (1A for correct paths of rays and 1A for correct orientation of the prism)

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C3_Ch17_FS_e

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