Ch.14: Selected Solutions
14.1. (1) What is the speed of light in a) water? b) In crown glass?
We will use the definition of index of refraction:
c
m
n ≡ ; Where c is 3 x 10 8
v
s
c
Solving for v, we get v= ;
n
n w = The index of refraction of water = 1.33
n glass =The index of refraction of crown glass = 1.52
Ans. a) Inserting 1.33 for the index of refraction of water, the speed of
m
3 x 10 8
s = 2.26 x 10 8 m
light in water is: Vwater =
1.33
s
Ans. b) The speed of light in the glass is:
m
3 x 10 8
s = 1.97 x 10 8 m
Vglass =
1.52
s
14.2
Calculate the speed of light in diamond?
c
m
; Where c is 3 x 10 8
v
s
c
Solving for v, we get v= ;
n
n diamond = The index of refraction of diamond = 2.42
8 m
c 3 x 10 sec
m
v= =
= 1.24 x 10 8
n
2.42
s
m
Ans. 1.24x10 8
s
14.3 (I) Calculate the index of refraction of a liquid in which the speed of light is found
to be 2.21 x 108 m/sec. Identify the most likely substance using Table14.1
n ≡
Ch.14: Selected Solutions
m
3 x 108
c
sec = 1.36
=
Ans. n ≡
m
v
2.21 x 10 8
sec
Ethyl alcohol
14.4. (1) What is the index of refraction of a substance in which the speed of light is 2.0 x
108 m/sec? Which of the substances in Table 14.1 is this likely to be?
m
3 x 108
c
sec = 1.50 the index of refraction of glass.
=
Ans. n ≡
8 m
v
2 x 10
sec
14.5 (I) How long does it take light to reach the earth from the sun 1.5 x 108 km away?
We know that the speed of light in a vacuum is 3.00 x108m/s. Let’s first convert
1.5x108 km to meters.
1000m
1.5x10 8 km
= 1.5x1011 m
1km
Solve∆ X = Vt for time
∆ X 1.5x1011 m
=
= 500s That’s approximately 8.33 minutes.
Ans. t =
V
8 m
3.00x10
s
14.6 (I) How long does it take light to cross a room 5.0 m wide?
distance
5m
=
=1.67 x 10 -8 seconds
Ans. ∆X = Vt ; t=
m
V
3 x 10 8
s
14.7. (I) What is the strength in diopters of a camera lens of 50-mm focal length? Note
that the focal length f must be in meters in order for the strength to be measured in
Diopter.
1
1
Ans. Strength S ≡
=
= 20 Diopter .
f 50 x 10 -3 m
14.8. (I) A zoom lens on a camera has a focal length adjustable from 80 to 200 mm. What
are the largest and smallest strengths of this lens in diopters?
1
1
Ans. a. Strength S ≡
=
= 12.5 Dioter. The largest.
f
80 x 10-3 m
1
1
Ans. b. Strength S ≡ =
= 5 Diopter . The smallest.
f 200 x 10-3 m
Ch.14: Selected Solutions
14.9. (I) Calculate the focal length of an eyeglass lens of strength -3.5 D.
Strength
1
S ≡ ; we know that f is in meters because the strength S is in Diopters
f
Ans. Solving for f we find that f=
1
= -.286 meters = -286 millimeters
-3.5
The negative sign indicates that the lens was concave.
14.10. (I) How far from a lens of strength 25.0 D will sunlight be maximally converged?
Please read the question several times. What are we asked to find?
We are asked to determine the focal length f of the lens.
1
1 1
Ans. S ≡ ; therefore f = =
= .04 meters
f
S 25
How do we know that the focal length would be in meters?
Just for fun, convert the answer to millimeters____________.
14.11. (I) It is possible to observe the glowing coils of a light bulb filament by projecting
its image on a screen. Suppose a light bulb filament is 15.0 cm
from a lens of 12.0-cm focal length. (a) Where is its image located? (b) What is its
magnification?
f = 12 cm Do = 15 cm
Do
Ho
object
f
F
Hi
Di
image
a. Determine the distance between the image and the lens. Determine Di
1 1
1
1 1 1
1 1
using 60 as a common denominator we
=
+ ; therefore
= =
−
f Do Di
Di f Do 12 15
get:
1
5
4
1
1
1
We have found that
=
−
=
=
Di 60mm 60mm 60mm
Di 60mm
Ans. Solving for Di, we determine that Di = 60 centimeters.
b.
What is the magnification?
Hi 60 cm
Ans. M ≡
=
= 4
Ho 15 cm
This indicates that the image is four times larger than the object.
c.
What type of image do we get?
Ans. The image is real and inverted.
Ch.14: Selected Solutions
14.12. (I) A camera with a 50-mm focal length lens is being used to photograph a person
1. 75 m tall standing 2.00 m away. (a) How far from the lens must the film be in order for
the image to be in focus? (b) What is the height of the person's image on the film?
a. Determine the distance between the image and the lens. Determine Di
1 1 1
1 1 1
1
1
=
+ ; therefore
= =
−
f Do Di
Di f Do 50mm 2000mm
Use 2000 as a common denominator
1
40
1
39
=
−
=
Di 2000mm 2000mm 2000mm
Therefore,
2000mm
Ans. Di =
= 51.3mm
39
(b) What is the height of the person's image on the film?
Solve
H i Di
for H i
=
H o Do
Ans.
Hi =
Di
51.3mm
Ho =
⋅1.75m = 44.9mm
Do
2.00m