Making Connections
4. (a) critical angle in zircon:
na = 1.00 (assuming in air)
θa = 90.0º
nz = 1.92
θz = θc = ?
nz sin θ z = n a sin θ a
sin θ c =
n a sin θ a
nz
(1.00)sin 90.0°
1.92
sin θ c = 0.521
=
critical angle in diamond:
na = 1.00 (assuming in air)
θa = 90.0º
nd = 2.42
θd = θc = ?
nd sin θ d = n a sin θ a
sin θ c =
n a sin θ a
nd
(1.00)sin 90.0°
2.42
sin θ c = 0.413
=
θ c = 31.4°
θ c = 24.4°
The critical angle for zircon is 31.4º, and the critical angle for diamond is 24.4º.
(b) A real diamond will have more light totally internally reflect and exit from the face of the diamond. It also gives the
light inside the diamond more reflections to be dispersed into the colours.
(c) A zircon gem is much cheaper to produce because zircon can be made, not mined, and is much easier to cut than
diamond.
5. The important issue is to maintain balance. New and expanding industry with better technology is always beneficial to
any country. If Canada does not help lead the communications industry, it will be taken over by others. Copper is an
important metal for much more than just communications. It is used in most houses for wiring. The government could
help look for other uses for copper (roofing, cooking pots, etc.) and also develop the use of fibre optic technology in
communications.
Reflecting
6.
Some other dispersion effects:
• “colours” on a soap bubble or oil spill on water
• coloured edges around a magnifying lens
CHAPTER 9 REVIEW
(Pages 352–353)
Understanding Concepts
1. (a) Medium A has a higher index of refraction because the light is bending away from the normal.
(b) Light is travelling slower in medium A because it has a higher index of refraction.
2. You could measure the angle of incidence and the angle of refraction and use Snell’s law to calculate the index of
refraction. You could also measure the critical angle and then use Snell’s law to calculate the index of refraction.
3. (a) If the index of refraction for blue light is higher than red light, then the blue light would travel more slowly in a
substance.
(b) If they are both incident on the same surface with the same angle, blue light will bend closer to the normal, and
therefore have a smaller angle of refraction than red light.
4. The more concentrated the salt solution, the higher the index of refraction. By conducting an experiment to determine the
index of refraction, you could calculate the concentration of salt in a solution by comparing it to a chart.
5. As the sun’s light enters the earth’s atmosphere, it bends. At sunrise and sunset, the light of a sun just below the physical
horizon is bent enough for an observer to see, so the sun still appears to be “up in the sky”. This provides us with light for
a longer period than without an atmosphere.
236 Unit 4 Light and Geometric Optics
Copyright © 2002 Nelson Thomson Learning
6.
The light is refracted at the air-water boundary. When we see the fish, the light from the fish is bent. If we aim at where
the fish appears, the spear will not ‘refract’ at the water surface, and so will pass through where the fish’s image is, not the
fish.
7.
c = 3.00 × 108 m/s
v = 1.24 × 108 m/s
n=?
8.
n=
=
1.24 ×10 8 m / s
n = 2.42
This substance is most likely diamond.
for violet light:
for red light:
c = 3.00 × 108 m/s
c = 3.00 × 108 m/s
n = 1.53
n = 1.51
v=?
c
c
n=
n=
v
v
c
c
v=
v=
n
n
=
3.00 × 108 m / s
1.53
=
8
9.
c
v
3.00 × 108 m / s
3.00 × 108 m / s
1.51
8
v = 1.96 × 10 m / s
v = 1.99 × 10 m / s
The speed of violet light in crown glass is 1.96 × 108 m/s, and the speed of red light in crown glass is
1.99 × 108 m/s.
angle of refraction in diamond:
angle of refraction for fused quartz:
na = 1.00
na = 1.00
θa = 40.0º
nq = 1.46
θd = ?
θq = ?
na sin θ a = nq sin θ q
na sin θ a = nd sin θ d
n sin θ a
n sin θ a
sin θ q = a
sin θ d = a
nq
nd
(1.00)sin 40.0°
2.42
sin θ d = 0.266
=
(1.00)sin 40.0°
1.46
sin θ q = 0.440
=
θ d = 15.4°
θ q = 26.1°
The difference between the angles of refraction is 26.1º – 15.4º = 10.7º.
Copyright © 2002 Nelson Thomson Learning
Chapter 9 Light Rays, Reflection, and Refraction
237
10.(a) na = 1.00
θa = 0º
np = 1.50
θp = ?
na sin θ a = n p sin θ p
sin θ p =
n a sin θ a
np
(1.00)sin 0°
1.50
sin θ p = 0
=
θ p = 0°
The angle of refraction in polyethylene is 0º when the incident angle is 0º.
na sin θ a = n p sin θ p
(b) na = 1.00
θa = 30.0º
n sin θ a
sin θ p = a
np = 1.50
np
θp = ?
(1.00)sin 30.0°
=
1.50
sin θ p = 0.333
θ p = 19.5°
The angle of refraction in polyethylene is 19.5º when the incident angle is 30.0º.
na sin θ a = n p sin θ p
(c) na = 1.00
θa = 60.0º
n sin θ a
sin θ p = a
np = 1.50
np
θp = ?
(1.00)sin 60.0°
=
1.50
sin θ p = 0.577
θ p = 35.3°
The angle of refraction in polyethylene is 35.3º when the incident angle is 60.0º.
na sin θ a = n w sin θ w
11. na = 1.00
θa = ?
n sin θ w
sin θ a = w
nw = 1.33
na
θw = 25.0º
(1.33) sin 25.0°
=
1.00
sin θ a = 0.562
θ w = 34.2°
The angle of incidence in air is 34.2º.
ng sin θ g = n a sin θ a
12.(a) na = 1.00
θa = 90.0º
n sin θ a
sin θ c = a
ng = 1.68
ng
θg = θc = ?
(1.00)sin 90.0°
=
1.68
sin θ c = 0.595
θ c = 36.5°
The critical angle of the medium is 36.5º.
238 Unit 4 Light and Geometric Optics
Copyright © 2002 Nelson Thomson Learning
(b) na = 1.00
θa = 90.0º
nm = ?
θm = θc = 40.0º
nm sin θ m = n a sin θ a
nm =
n a sin θ a
sin θ m
(1.00)sin 90.0°
sin 40.0°
n m = 1.56
The index of refraction of the medium is 1.56.
13. If we consider the PRL, then the amount of refraction would be larger for light travelling in air into the substance with a
critical angle of 27.0º. This means the light has slowed down more. Therefore, the light will be travelling faster in the
substance whose critical angle is 32.0º
=
Applying Inquiry Skills
14.
15. θa = 48.0º
θs = 15.5º
na = 1.00
ns = ?
na sin θ a = ns sin θ s
ns =
n a sin θ a
sinθ s
(1.00)sin 48.0°
sin 15.5°
ns = 2.78
The index of refraction of the substance is 2.78.
=
16.
Copyright © 2002 Nelson Thomson Learning
Chapter 9 Light Rays, Reflection, and Refraction
239