PHYSICS 6C
Ch.24 Worksheet Solutions
1) The reflecting surfaces of two plane mirrors form an angle of 120°. If a
ray of light strikes mirror 1 with an angle of 55°, find the angle of reflection
when the light leaves mirror 2.
55
55
angle of incidence = angle of reflection
the angles in a triangle must add up to 180°
enough said.
120
65
65
2) Sunlight reflects from a concave piece of broken glass, converging to a
point 15 cm from the glass. What is the radius of curvature of the glass?
This is a concave mirror, so the radius is positive, and double the focal length. The sunlight
is coming from a distant object (S=∞), so the image is at the focal point. So R=+30cm.
3) An object with a height of 42cm is placed 2.0m in front of a convex
mirror with a focal length of -0.5m. Find the location and size of the
resulting image. Is this image real, or is it virtual?
Use the standard thin lens formula with S=2m and f=-0.5m.
1 1
1
1
1
1
S
0.4m This is a virtual image, 40cm behind the mirror.
f
S S
S
0.5 2
y
S
0.4
y
(42cm)(
) 8.4cm
For size, use m
S
y
2
4) A magician wishes to create the illusion of a 2.74m tall elephant. He
plans to do this by forming a virtual image of a 50cm tall model elephant
with the help of a spherical mirror.
a) Should the mirror be concave or convex?
b) If the model must be placed 3.0m from the mirror, what radius of
curvature is needed?
c) How far from the mirror will the image be formed?
part a): A concave mirror can produced an enlarged virtual image.
2.74
parts b) and c): We are given the magnification – m
5.48
0.5
S
with that we can find the image postion:
5.48
S
16.4m
S
1 1
1
Now we can find the focal length:
f
3.67m
R
f
3
16.4
clas.ucsb.edu/staff/vince/
7.34m
PHYSICS 6C
Ch.24 Worksheet Solutions
5) To a person swimming 0.8m below the surface of the water in a pool, the
diving board directly overhead appears to be a height of 5.2m above the
swimmer. What is the actual height of the diving board above the surface of
the water?
na
Sa
1.33
5.2
nb
Sb
0 (R=∞ for a flat surface)
1
Sb
0
Sb
3.9 (this is the actual height above the swimmer)
Subtract 0.8m to get the height above the water: 3.1m.
6) A lens produces a real image that is twice as large as the object and is
located 15cm from the lens. Find the focal length of the lens and the object
distance. Draw a ray diagram.
Since the image is twice as large as the object, then it is also twice as far from the lens.
1
1
1
Thus S=7.5cm. Using our basic formula, focal length is:
f
5cm
f
7.5 15
f
S
5cm
S
15cm
7.5cm
7) An arrow 2cm long is located 75cm from a lens with focal length f=30cm.
a) If the arrow is vertical, what is the magnification of the image?
b) If the arrow is placed horizontally, so that its tip is 74cm from the lens
and its tail is 76cm from the lens, what is the magnification of the image
(i.e. how long does the image of the arrow appear to be from tip to tail)?
part a):
1
30
part b):
1
1
30 74
1
75
1
S
1
S
S tip
S
50cm
50.45cm;
1
30
50
75
m
1
76
1
S
2
3
S ta il
49.565cm
image
length
0.89cm
m
0.44
8) Repeat problem 7 for the case of a lens with focal length f=-30cm.
part a):
part b):
1
1
30 74
1
30
1
S
1
75
S tip
1
S
S
21.4cm
m
1
30
1
76
21.35cm;
clas.ucsb.edu/staff/vince/
21.4
75
1
S
S ta il
0.29
21.51cm
image
length
0.16cm
m
0.08