New Century Senior Physics

Second Edition

Worked solutions

Chapter 18 - Light

from Richard Walding

Q1

Q2

Q3

Q4

Q5

Q6

Q7

Q8

Q9

Q10

Q11

Q12

Q13

Q14

Q15

Q16

Q17

Q18

Q19

Q20

Q21

Q22

Q23

Q24

Q25

ngd = nd/ng

Q26

(a) (i) nm = va/vm vm = va/nm = 3 x 108/1.78 = 1.68 x 108 m s-1

(ii) vm = = 3 x 108/1.2 = 2.5 x 108 m s-1

(iii) vm = = 3 x 108/2.1 = 1.43 x 108 m s-1

(iv) vm = = 3 x 108/1.42 = 2.11 x 108 m s-1

(b) The more optically dense is the one with the greatest refractive index (medium iii)

Q27

Q28

nm = a/m g = a/ng = 633 nm/1.5 = 422 nm

Q29

n = true depth/apparent depth

1.33 = true depth/50 cm

true depth = 1.33 x 50 = 66.5 cm

Q30

Q31

Q32

Q33

Q34

Q35

(a)

i 20 30 40 50 60 70 80r 10 15 20 24 27 30 31

sin r 0.17 0.26 0.34 0.41 0.45 0.50 0.52sin i 0.34 0.50 0.64 0.77 0.87 0.94 0.98

(b)

(c) linear (c-intercept = 0)

(d) sin i sin r

(e) sin i/sin r = n = 1.8988 (gradient)

(f) zircon n = 1.9

Q36

Q37

Q38

i 10 15 20 25 30 35 40r 13 22 30 38 47 57 70sin r 0.22 0.37 0.50 0.62 0.73 0.84 0.94sin i 0.17 0.26 0.34 0.42 0.50 0.57 0.64

Gradient = sin i/sin r = nkero/air

Gradient = y/x = (0.643 - 0.174)/(0.940 - 0.225) = 0.6861

Therefore nkero/air = 0.6861; or nair/kero = 1/0.6861 = 1.457

sin c = 1/nkero = 1/1.457 = 0.6861

c = sin-1 0.6861 = 43.3

Alternatively, solve equation for when sin r = 1 (that is, when r = 90)

y = 0.6861 x 1

y = sin i = 0.6861

c = sin-1 0.6861 = 43.3

Q39

Q40

Since the incident ray on the plastic block is 90o it passes into the block undeflected. 

At the plastic /water interface the angle of incidence = 45o

n1 sin 1 = n2 Sin 2

1.62 sin 45 = 1.33 sin 2

2 = 59.5o

At the water/ air interface: the angle of incidence = 180 - 135 - 30.5 = 14.5o

n1 sin 1 = n2 sin 2

1.33 sin 14.5 = 1 sin 2

2 = 19.5o

Q41

Q42

Q43

It would appear transparent and wouldn’t be able to be seen as it doesn’t refract the light. You

could add a dye to make it less transparent, or you could use another plastic that had a

different refractive index to water, or use cardboard so that it would disintegrate when

exposed to water after a short time.