3A Wave Motion I Chapter 2 Refraction of Light

3A Wave motion I Chapter 2 Refraction of Light

2 Refraction of Light

Practice 2.1 (p. 42) 1 C By Snell’s law,

sin θa sin 60 nX = = = 1.51 sin θX sin (90  55) 2 C 3 C c By n = , v speed of light in glass c 3108 = = = 1.82  108 m s1 n 1.65 4 B sin r sin 30 = sin16.5 sin19.5 r = 25.2 5 B 6 A 7 The light ray does not bend only if it enters glass along the normal, i.e. only if the angle of incidence is 0 . 8 Angle of incidence = 90  35 = 55 By Snell’s law,

sin θair nalcohol = sin θalcohol

sin θair sin 55 sin alcohol = = nalcohol 1.36

alcohol = 37.0 The angle of refraction is 37.0. 9 Refractive index sin sin 30 = a = = 1.46 sin sin 20 10 Light travels at different speeds in different media. Refraction results from the change in the speed of light when light crosses a boundary.

New Senior Secondary Physics at Work 1  Oxford University Press 2009 3A Wave motion I Chapter 2 Refraction of Light

11 By Snell’s law,

sin θa ng = sin θg

sin θa sin 45 sin g = = ng 1.50

g = 28.1 The angles that ray A makes with the normal at the air-glass interfaces are 28.1, 28.1, and 45. The angles that ray B makes with the normal at the air-glass interfaces are 0, 0, 0 and 0. 12 (a) By Snell’s law, 14 When he sees the fish above water surface, n sin  = n sin  A A B B the apparent depth of the fish is smaller than 1.6  sin 60 = 1.4  sin r its real depth. Therefore, the Indian should r = 81.8 aim his spear at a point below where the fish The angle of refraction is 81.8. appears to him. c (b) By n = , v 15 (a) c

v A n nB 1.4 0.875 = A = = = vB c n A 1.6 1

nB The ratio is 0.875 : 1. 13

(b) Stars appear slightly higher.

Practice 2.2 (p. 56) 1 D 2 B (3): Total internal reflection does not occur when light passes from an optically less dense medium to an optically denser medium, no matter how large the angle of incidence is.

3 B By Snell’s law, sin sin 45 n = a = = 1.414 sin sin 30  1   1  C = sin–1   = sin–1   = 45  n  1.414  The critical angle for the liquid-air interface is 45. 4 Critical angle  1   1  = sin–1   = sin–1   = 43.2  n  1.46  5

6 By Snell’s law,

nA sin A = nB sin B

If A equals the critical angle, B = 90.

1.7  sin 44.9 = nB  sin 90

nB = 1.20 The refractive index of medium B is 1.20. 7 (a) Critical angle  1   1  = sin–1   = sin–1   = 37.3  n  1.65  (b) Maximum value of  = 90 – 37.3 = 52.7 8

9 (a) By Snell’s law,

sin θa ng = sin θg sin 60 1.5 = sin w w = 35.3 x = 90  w = 54.7 On side BC, y = x = 54.7 On side CD, angle of incidence = 90  54.7 = 35.3 = w By Snell’s law,

sin θa ng = sin θg sin z 1.5 = sin 35.3 z = 60 (b) The angle of emergence of the ray is equal to the angle of incidence. 10 Critical angle for diamond-air interface  1   1  = sin–1   = sin–1   = 24.4  n   2.42  Critical angle for crystal-air interface  1   1  = sin–1   = sin–1   = 30  n   2.00  The critical angle for a crystal is larger, so a smaller amount of light going into a crystal is internally reflected back. Therefore, a crystal does not have the same brilliance as a diamond. 11

Critical angle C for water  1   1  = sin–1   = sin–1   = 48.75  n  1.33  Diameter of the diver’s view = 2  radius of the cone = 2  3  tan C ( = C) = 2  3  tan 48.75 = 6.84 m 12 Air near the ground is hotter and has a lower refractive index. With a continuous decrease of the refractive index, light from the sky is gradually refracted more towards the horizontal. When the light meets a layer of air at an angle beyond the critical angle, total internal reflection takes place. As a result, a person sees the image of the sky when he looks down, i.e. a mirage is formed.

Revision exercise 2 Multiple-choice (p. 59) 1 A By Snell’s law,

sin θa nm = sin θm sin 30 sin  = m 1.4

m = 20.9 2 C 3 C By Snell’s law, sin i sin 40 n = = = 1.62 sin r sin 90  66.6 4 B 5 B c 3108 n = = = 2 A 8 v A 1.510 c 3108 n = = = 1.5 B 8 vB 210

By Snell’s law, The angle of refraction in air is 41.7.

nA sin A = nB sin B

2  sin 20 = 1.5  sin B

B = 27.1 6 A By Snell’s law,

nP sin P = nQ sin Q

sin P nQ = = constant sin Q nP Then, sin C sin 35 =  C = 38 sin 90 sin 70 7 B 8 D Depth of eyes below water surface 0.8 =  98  = 0.695 m tan   2  9 A sin i Refractive index = = slope of the graph sin r Slope of Z > slope of Y > slope of X

 nZ > nY > nX Total internal reflection will not take place when light travels from medium X (an optically less dense medium) to medium Y (an optically denser medium). 10 C 11 C 12 (HKCEE 2006 Paper II Q32) 13 (HKCEE 2007 Paper II Q13)

Conventional (p. 61) 1 (a) By Snell’s law,

sin θa nw = (1M) sin θw

sin a = 1.33  sin 30

a = 41.7 (1A)

(b) Angle by which the ray is bent 4 To the goldfish, the girl appears to be smaller = 41.7 – 30 (1A) = 11.7 (1A) and further away from it. (1A) 2 (a) Refractive index 5 (a) 1 = (1M) sin i 0.174 0.342 0.5 0.643 sin C sin r 0.139 0.259 0.342 0.438 1 = sin 24.4 sin i 0.766 0.866 0.940 0.985 = 2.42 (1A) sin r 0.515 0.588 0.629 0.669 (b) By Snell’s law,

sin θa nd = (1M) sin θd sin r 2.42 = sin (90  80) r = 24.8 (1A) 3

(Correct path in oil) (1A) (Correct path in water) (1A) By Snell’s law,

sin θa no = sin θo (Correct labelled axes) (1A) sin 50 (Correct straight line) (1A) 1.47 = sin p Refractive index of Perspex p = 31.4 (1A) = slope of the graph (1M) q = p = 31.4 (1A) 0.95  0.44 = Then again by Snell’s law, 0.65  0.3

no sin o = nw sin w = 1.46 (1A) 1.47  sin 31.4 = 1.33  sin r r = 35.2 (1A)

(b) There may be error in measuring angles. (c) (1A)

If only one pair of data is used for red calculation, the percentage error will be significant. (1A) violet (c) The first point (0.139, 0.174) and the second point (0.259, 0.342) seem to be wrong. (1A) (Correct drawing of rays) (1A) The error may be due to inaccurate (Correct labels of coloured light) (1A) measurement (1A) 8 (a) Refractive index of water v or uneven texture within the Perspex. = a (1M) v (1A) w 3108 6 (a) = 2.26108 = 1.33 (1A) (b) By Snell’s law,

ng sin g = nw sin w (1M)

ng sin 50 = 1.33  sin 78.3

ng = 1.70 (1A) The refractive index of the glass is 1.70.

va (c) By ng = , vg (Correct drawing of rays) (1A) speed of light in the glass (Correct indication of the image va position) (1A) = (1M) ng (b) The letters will appear even higher. 3108 = (1A) 1.70 7 (a) Dispersion (1A) = 1.76  108 m s1 (1A) (b) Different colours of light travel at 9 (a) Direct a ray from the ray box to enter different speeds in the prism. (1A) the semicircular glass block from its Therefore, they are refracted by straight edge. The ray must hit the different amounts and separated from centre of the block. (1A) each other. (1A) Use the full-circle protractor to measure the angle of incidence i and the angle of refraction r. (1A)

Repeat with other angles of incidence. Record the (ii) results in a table. (1A) Plot a graph of sin i against sin r. The graph is a straight line passing through light ray from the origin. This means that sin i is an object directly proportional to sin r. (1A) (b) Direct a ray of light from the ray box to totally internally enter the semicircular glass block from reflected light ray its curved edge towards its centre. (1A) (Total internal reflection occurs on Slowly increase the angle of incidence the correct interface.) (1A) until the angle of refraction is 90 (1A) (Correct ray diagram) (1A) At this moment, the angle of incidence (iii) For each object, only one image is is equal to the critical angle C. (1A) formed by a prism. (1A) 10 Critical angle for diamond-air interface (c) The surface of a ground glass window is  1  = sin 1   not smooth. (1A)  n  When parallel light rays hit the ground  1  = sin 1   glass, they have different angles of  2.42  incidence. (1A) = 24.4 (1A) ground glass When there is a layer of oil, the critical angle parallel to an for the boundary changes. light rays observer outside the By Snell’s law, room

nd sin d = no sin o 2.42  sin C = 1.40  sin 90 C = 35.3 (1A) The critical angle increases if oil attaches on They would be refracted into the glass a diamond. (1A) and eventually leave the glass and reach As a result, fewer light rays are totally an observer outside the room at different reflected and the diamond loses brilliance. angles. (1A) (1A) Since the refracted light rays leave the 11 (a) Reflection occurs at X. (1A) glass in an irregular pattern, the image Refraction and (1A) formed is blurred. (1A) reflection occur at Y. (1A) On the other hand, most light from the (b) (i) Total internal reflection (1A) outside can pass through the glass and reach the room . The room is brighter when ground glass is used instead of curtain. (1A)

c 12 (a) (i) By ng = , vg speed of light in the glass c = ng 3108 = 1.45 = 2.07  108 m s1 (1A) (ii) By Snell’s law,

nl  sin l = ng  sin g

nl sin θ = g n g sin θl c

vl sin θg c = sin θl vg v sin θ g = g vl sin θl Speed of light in the liquid droplet

sin θl =  vg sin θg sin 29.2 =  2.07  108 sin 26.6 = 2.26  108 m s1 (1A) (Correct method) (1A) (b) Refractive index of the liquid c = (1M) vl 3108 = 2.26 108 = 1.33 (1A)

(Total internal reflection on the vertical surface) (1A) (Refraction at bottom surface with angle in air greater than that in glass) (1A) 13 (HKCEE 2004 Paper I Q1) 14 (HKCEE 2005 Paper I Q10)

sin1 15 (a) (i) By ng1 = , (1M) sin 2

sin 1 = 1.45  sin 15.5

1 = 22.8 (1A) (ii) When light travels from glass block 2 into the air, refraction occurs. By Snell’s law,

sin θa ng2 = (1M) sin θg 2 sin 90 = sin 38.7 = 1.60 The refractive index of the glass in block 2 is 1.60. (1A) (iii) By Snell’s law,

ng2  sin g2 = ng1  sin g1(1M)

1.60  sin 51.3 = 1.45  sin 3 (1M)

3 = 59.4 (1A)

(b) The speed of light is greater in block 1. (1A) c By v = , the smaller the refractive n index, the greater the speed of light in it. Since block 1 has a smaller refractive index than block 2, the speed of light will be greater in block 1. (1A) (c)

(Reflection at boundary with i = r) (1A) (Refraction at the bottom surface bending away from normal) (1A)

Physics in articles (p. 64) (a)

(Incident ray from emitter to the glass-air interface) (1A) (Reflected ray from the glass-air interface to the sensor) (1A) (No refracted ray at the glass-air interface) (1A) (b)

(Incident ray from emitter to the glass-air interface) (1A) (Reflected ray from the glass-air interface to the sensor) (1A) (Refracted ray in the water) (1A) (c) Critical angle of the glass-air interface  1  = sin 1  = 36.0 (1A) 1.7  By Snell’s law,

nglass  sin glass = nwater  sin water 1.70  sin C = 1.33  sin 90 C = 51.5 (1A) The critical angle of the glass-water interface is 51.5. The angle of incidence the infra-red light is 45. Therefore, total internal reflection would occur at a glass-air interface but not at a glass-water interface. (1A)

New Senior Secondary Physics at Work 13  Oxford University Press 2009