P36.59
A hemisphere is too thick to be described as a thin lens.
The light is undeviated on entry into the ﬂat face. We next
consider the light’s exit from the second surface, for which
R
= −
6 00
.
cm
.
The incident rays are parallel, so
p
= ∞
Then,
n
p
n
q
n
n
R
1
2
2
1
+
=
−
becomes
0
1
1 00
1 56
6 00
+
=
−
−
q
.
.
.
cm
and
q
=
10 7
.
cm
P36.60
(a)
I
r
=
=
×
(
)
=
−
P
4
4 50
4
1 60
10
1 40
2
2
2
π
π
.
.
.
W
m
kW m
2
(b)
I
r
=
=
(
)
=
P
4
4 50
6 91
2
2
π
π
.
.
W
4
7.20 m
mW m
2
(c)
1
1
1
p
q
f
+
=
:
1
7 20
1
1
0 350
.
.
m
m
+
=
q
so
q
=
0 368
.
m
and
M
h
q
p
=
′
= −
= −
3 20
0 368
.
.
cm
m
7.20 m
′ =
h
0 164
.
cm
(d)
The lens intercepts power given by
P
=
=
×
(
)
(
)
⎡
⎣
⎢
⎤
⎦
⎥
−
IA
6 91
10
4
0 150
3
2
.
.
W m
m
2
π
and puts it all onto the image where
I
A
=
=
×
(
)
(
)
⎡
⎣
⎤
⎦
−
P
6 91
10
15 0
4
0 164
3
2
.
.
.
W m
cm
2
π
π
cm
(
)
2
4
I
=
58 1
. W m
2
FIG. P36.59

346
Chapter 36
P36.61
From the thin lens equation,
q
f p
p
f
1
1
1
1
1
6 00
12 0
12 0
6
=
−
=
−
(
)
(
)
− −
.
.
.
.
cm
cm
cm
00
4 00
cm
cm
(
)
= −
.
When we require that
q
2
→ ∞
,
the thin lens equation becomes
p
f
2
2
=
.
In this case,
p
d
2
4 00
=
− −
(
)
.
cm
Therefore,
d
f
+
=
=
4 00
12 0
2
.
.
cm
cm
and
d
=
8 00
.
cm
P36.62
(a)
For the light the mirror intercepts,
P
=
=
I A
I
R
a
0
0
2
π
350
1 000
2
W
W m
2
=
(
)
π
R
a
and
R
a
=
0 334
.
m or larger
(b)
In
1
1
1
2
p
q
f
R
+
=
=
we have
p
→ ∞
so
q
R
=
2
M
h
h
q
p
=
′
= −
so
′ = −
⎛
⎝
⎜
⎞
⎠
⎟
= −
⎛
⎝
⎞
⎠
°
°
⎛
⎝
⎞
⎠
⎡
⎣
⎢
h
q
h
p
R
2
0 533
.
π
rad
180
⎤
⎦
⎥
= −
⎛
⎝
⎞
⎠
(
)
R
2
9 30
.
m rad
where
h
p
is the angle the Sun subtends. The intensity at the image is
then
I
h
I
R
h
I R
R
a
a
=
′
=
′
=
(
)
×
−
P
π
π
π
2
0
2
2
0
2
2
3
4
4
4
2
9 30
10
.
r
ad
(
)
2
120
10
16 1 000
9 30
10
3
2
2
3
×
=
(
)
×
(
−
W m
W m
rad
2
2
R
R
a
.
)
2
so
R
R
a
=
0 025 5
.
or larger
FIG. P36.61

Image Formation
347
P36.63
For the mirror,
f
R
=
= +
2
1 50
.
m
. In addition, because the distance to the Sun is so much larger
than any other distances, we can take
p
= ∞
.
The mirror equation,
1
1
1
p
q
f
+
=
, then gives
q
f
=
=
1 50
.
m
Now, in
M
q
p
h
h
= −
=
′
the magnification is nearly zero, but we can be more precise:
h
p
is the angular diameter of the
object. Thus, the image diameter is
′ = −
=
−
(
)
°
⎛
⎝
⎞
⎠
(
)
= −
h
hq
p
0 533
1 50
0 0
.
.
.
o
π
rad
180
m
14 0
1 40
m
cm
=
−
.
P36.64
(a)
The lens makers’ equation,
1
1
1
1
1
2
f
n
R
R
=
−
(
)
+
⎛
⎝
⎜
⎞
⎠
⎟
becomes:
1
5 00
1
1
9 00
1
11 0
.
.
.
cm
cm
cm
=
−
(
)
−
−
(
)
⎡
⎣
⎢
⎤
⎦
⎥
n
giving
n
=
1 99
.
.
(b)
As the light passes through the lens for the first time, the thin lens equation
1
1
1
1
1
p
q
f
+
=
becomes:
1
8 00
1
1
5 00
1
.
.
cm
cm
+
=
q
or
q
1
13 3
=
.
cm
,
and
M
q
p
1
1
1
13 3
1 67
= −
= −
= −
.
.
cm
8.00 cm
This image becomes the object for the concave mirror with:
p
q
m
=
−
=
−
=
20 0
20 0
13 3
6 67
1
.
.
.
.
cm
cm
cm
cm
and
f
R
=
= +
2
4 00
.
cm
The mirror equation becomes:
1
6 67
1
1
4 00
.
.
cm
cm
+
=
q
m
giving
q
m
=
10 0
.
cm
and
M
q
p
m
m
2
10 0
1 50
= −
= −
= −
.
.
cm
6.67 cm
The image formed by the mirror serves as a real object for the lens on the second pass of
the light through the lens with:
p
q
m
3
20 0
10 0
=
−
= +
.
.
cm
cm
The thin lens equation yields:
1
10 0
1
1
5 00
3
.
.
cm
cm
+
=
q
or
q
3
10 0
=
.
cm
and
M
q
p
3
3
3
10 0
1 00
= −
= −
= −
.

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