Geometric Optics
3419
(c)
The principalray diagram is given in Figure 34.69.
Figure 34.69
E
VALUATE
:
For a real object, only virtual images are formed by a convex mirror. The virtual object considered
in this problem must have been produced by some other optical element, by another lens or mirror in addition to
the convex one we considered.
34.70.
I
DENTIFY
:
Apply
a
b
b
a
n
n
n
n
s
s
R
−
+
=
′
, with
R
→ ∞
since the surfaces are flat.
S
ET
U
P
:
The image formed by the first interface serves as the object for the second interface.
E
XECUTE
:
For the waterbenzene interface to get the apparent water depth:
1.33
1.50
0
0
7.33 cm.
6.50 cm
a
b
n
n
s
s
s
s
′
+
=
⇒
+
=
⇒
= −
′
′
For the benzeneair interface, to get the total apparent distance to the bottom:
1.50
1
0
0
6.62 cm.
(7.33 cm
2.60 cm)
a
b
n
n
s
s
s
s
′
+
=
⇒
+
=
⇒
= −
′
′
+
E
VALUATE
:
At the waterbenzene interface the light refracts into material of greater refractive index and the
overall effect is that the apparent depth is greater than the actual depth.
34.71.
I
DENTIFY
:
The focal length is given by
1
2
1
1
1
(
1)
n
f
R
R
⎛
⎞
=
−
−
⎜
⎟
⎝
⎠
.
S
ET
U
P
:
1
4.0 cm
R
= ±
or
8.0 cm
±
.
2
8.0 cm
R
= ±
or
4.0 cm
±
.
The signs are determined by the location of the
center of curvature for each surface.
E
XECUTE
:
1
1
1
(0.60)
4.00 cm
8.00 cm
f
⎛
⎞
=
−
⎜
⎟
±
±
⎝
⎠
, so
4.44 cm,
13.3 cm.
f
= ±
±
The possible lens shapes are
sketched in Figure 34.71.
1
2
3
4
5
6
7
8
13.3 cm;
4.44 cm;
4.44 cm;
13.3 cm;
13.3 cm;
13.3 cm;
4.44 cm;
4.44 cm.
f
f
f
f
f
f
f
f
= +
= +
=
= −
= −
= +
= −
= −
E
VALUATE
:
f
is the same whether the light travels through the lens from right to left or left to right, so for the
pairs (1,6), (4,5) and (7,8) the focal lengths are the same.
Figure 34.71
34.72.
I
DENTIFY
:
Apply
1
1
1
s
s
f
+
=
′
and the concept of principal rays.
S
ET
U
P
:
10.0 cm
s
=
. If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens
and the ray is parallel to the optic axis after it passes through the lens.
E
XECUTE
:
(a)
The ray is bent toward the optic axis by the lens so the lens is converging.
(b)
The ray is parallel to the optic axis after it passes through the lens so it comes from the focal point;
18.0 cm
f
=
.
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3420
Chapter 34
(c)
The principal ray diagram is drawn in Figure 34.72. The diagram shows that the image is 22.5 cm to the left of
the lens.
(d)
1
1
1
s
s
f
+
=
′
gives
(10.0 cm)(18.0 cm)
22.5 cm
10.0 cm
18.0 cm
sf
s
s
f
′ =
=
= −
−
−
. The calculated image position agrees with the
principal ray diagram.
E
VALUATE
:
The image is virtual. A converging lens produces a virtual image when the object is inside the focal point.
Figure 34.72
34.73.
I
DENTIFY
:
Since the truck is moving toward the mirror, its image will also be moving toward the mirror.
S
ET
U
P
:
The equation relating the object and image distances to the focal length of a spherical mirror is
1
1
1
s
s
f
+
=
′
, where
f = R
/2.
E
XECUTE
:
Since the mirror is convex,
f
=
R
/2 = (°1.50 m)/2 = °0.75 m. Applying the equation for a spherical
mirror gives
1
1
1
fs
s
s
s
f
s
f
′
+
=
⇒
=
′
−
.
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 Spring '06
 Buchler
 Physics, GEOMETRIC OPTICS

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