PHYSICS 1020
Optics Practice Problems
Solutions
1. What is the speed of light in (a) crown glass, (b) Lucite, and (c) ethyl alcohol?
Solution. The index of refraction n of a material is defined as
c
nD
v
where c is the speed of light in vacuum, and v is the speed of light in the material.
Therefore
vD
c
n
Referring to the table of indices of refraction (Table 23-1 in the text), and with
c D 299;792;458 m/s,
(a) For crown glass, n D 1:52, and so v D c=n D 197;231;880 m/s.
(b) For Lucite, n D 1:51, and so v D c=n D 198;538;052 m/s.
(c) For ethyl alcohol, n D 1:36, and so v D c=n D 220;435;631 m/s.
1
2. A diver shines a flashlight upward from beneath the water at a 42ı:5 angle to the
vertical. At what angle does light leave the water?
Solution. This is a problem in refraction, for which we use Snell’s law:
n1 sin 1 D n2 sin 2 :
Let’s label the air as medium 1 and the water as medium 2. Then n1 D 1:00 for air,
n2 D 1:33 for water, and 2 D 42ı:5. Solving Snell’s law for 1 ,
n2
sin 2
n1
1:33
D
sin 42ı:5
1:00
D 0:898535
sin 1 D
D sin1 0:898535 D 63ı:97
2
3. Rays of the Sun are seen to make a 31ı:0 angle to the vertical beneath the water. At
what angle above the horizon is the Sun?
Solution. This is a problem in refraction, for which we use Snell’s law:
n1 sin 1 D n2 sin 2 :
Let’s label the air as medium 1 and the water as medium 2. Then n1 D 1:00 for air,
n2 D 1:33 for water, and 2 D 31ı:0. Solving Snell’s law for 1 ,
n2
sin 2
n1
1:33
D
sin 31ı:0
1:00
D 0:685001
sin 1 D
D sin1 0:685001 D 43ı:24
Remember that this gives the angle from the normal; the problem asks for the angle
above the horizon, which is the complement of this angle. So the answer is
90ı 43ı:24 D 46ı:76
3
4. The critical angle (for total internal reflection) for a certain liquid-air surface is 47ı:7.
What is the index of refraction of the liquid?
Solution. The sine of the critical angle is equal to the ratio of the indices of refraction of the two media:
n2
sin c D
n1
We’re given c D 47ı:7 and n2 D 1:00 (which is always the lesser of the two indices),
and wish to find n1 (always the larger of the two indices):
n2
sin c
1:00
D
sin 47ı:7
D 1:352
n1 D
4
5. A stamp collector uses a converging lens with focal length 24 cm to view a stamp
18 cm in front of the lens. (a) Where is the image located? (b) What is the magnification?
Solution. We’re given the focal length f D C24 cm, and an object distance
d0 D C18 cm.
(a) The image distance di is found from the thin lens equation:
1
1
1
C
D ;
di
do
f
and so
Â
di D
Â
D
1
1
f
do
Ã1
1
1
24 cm 18 cm
Ã1
D 72 cm
So the image is virtual, and is located 72 cm in front of the lens.
(b) The magnification is found from the magnification equation,
di
hi
D
ho
do
72 cm
D
C18 cm
D C4
mD
5
6. Suppose you have a concave lens of focal length f D 23:0 cm. An object of height
5.0 cm is placed 33.0 cm from the lens. Using the algebraic method, determine: (a)
the image distance; (b) the image height; (c) the image magnification; (d) whether the
image is real or virtual; and (e) whether the image is upright or inverted.
Solution. We are given f D 23:0 cm, ho D C5:0 cm, and do D C33:0 cm.
(a) The image distance is given by the thin lens equation:
1
1
1
C
D ;
di
do
f
and so
Â
di D
Â
D
1
1
f
do
Ã1
1
1
23:0 cm C33:0 cm
Ã1
D 13:55 cm
(b) The image height is given by the magnification equation:
mD
di
hi
D ;
ho
do
and so
hi D ho
di
do
D .C5:0 cm/
13:55 cm
C33:0 cm
D C2:05 cm
(c) The magnification is given by the magnification equation:
hi
ho
C2:05 cm
D
C5:0 cm
D 0:411 cm
mD
(d) Since the image distance di is negative, the image is virtual.
(e) Since the image height hi is positive, the image is upright.
6
7. What is the illuminance due to a 1200-lumen light bulb at a distance of 1.4 meters
from the bulb?
Solution. Since a light bulb is isotropic, the illuminance E is given by
ˆ
ˆ
D
A
4 2
1200 lm
D
4.1:4 m/2
ED
D 48:72 lm
7