IMAGES AND OPTICAL INSTRUMENTS
31
E
XERCISES
Section 31.1 Images with Mirrors
17.
I
NTERPRET
This problem is about image formation in a plane mirror.
D
EVELOP
A small mirror (
M
) on the floor intercepts rays coming from a customer’s shoes (
O
), which are
traveling nearly parallel to the floor. The angle to the customer’s eye (
E
) from the mirror is twice the angle of
reflection, so
tan 2
h
d
α
=
E
VALUATE
Solving for the angle, we find
1
1
1
1
140 cm
tan
tan
35.2
2
2
50 cm
h
d
α
−
−
⎛
⎞
⎛
⎞
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
°
for the given distances. Therefore, the plane of the mirror should be tilted by
35
from the vertical to provide the
customer with a floor-level view of her shoes.
.2
°
A
SSESS
The angle tilted decreases if
d
is increased, and vice versa. This is what we expect from our experience at
shoe stores.
18.
(a)
Use
and
in the mirror equation (Equation 31.2) to get
s
s
36 cm
s
=
15 cm
f
=
/(
)
(15 cm)(36 cm)/
′
f
s
f
=
−
=
(36 cm
15 cm)
25.7 cm,
−
=
in front of the mirror.
(b)
The magnification (Equation 31.1) is
/
M
s s
′
= −
=
5
7
25.7/36
/(
)
71.4%
f
s
f
−
= −
−
= −
= −
(the image is reduced in size and inverted).
(c)
Since
s
′
is positive, the
image is real. Ray tracing, as in Table 31.1, confirms these conclusions.
19.
I
NTERPRET
We have an image formation problem involving a concave mirror. We want to know the orientation
of the image as well as its height compared to the object.
D
EVELOP
The magnification
M
, the ratio of the image height
h
′
to object height
h
, is given by Equation 31.1:
h
s
M
h
s
′
′
=
= −
where
s
and
s
′
are object and image distances to the mirror. The two quantities
s
and
s
′
are related by the mirror
equation (Equation 31.2):
1
1
1
s
s
f
+
=
′
31.1

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