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Solution for Homework 23 Optical Systems
Solution to Homework Problem 23.1(How long to accelerate to light speed?)
Problem:
A problem you should have easily been able to do in UPI, but I get asked it all the time when I ask
for a good question on electricity and magnetism the Frst day of class.
Suppose a human can confortably live in a spaceship accelerating at
1
g
= 9
.
81
m
s
2
, ignoring relativistic e±ects,
how long does it take the spaceship to reach the speed of light? Report your answer in years. You may use the
approximate conversion
1yr =
π
×
10
7
s
.
Select One of the ²ollowing:
(a)
15
.
7yr
(b)
0
.
19yr
(cAnswer)
0
.
97yr
(d)
15
,
000yr
(e) Toby.
Solution
²rom UPI,
v
=
at
=
gt
, so
t
=
c
g
=
3
×
10
8 m
s
9
.
81
m
s
2
= 3
.
05
×
10
7
s = 0
.
97yr
Total Points for Problem: 3 Points
Solution to Homework Problem 23.2(Two Lens Problem)
Problem:
A converging lens of focal length
12cm
and a diverging lens of focal length

15cm
are separated by
55cm
. An object is placed
22cm
to the left of the converging lens. Compute the distance (with correct sign) the
Fnal image forms from the diverging lens.
Select One of the ²ollowing:
(a)
+8
.
1cm
(bAnswer)

9
.
8cm
(c)
+1
.
5cm
(d)

15cm
(e) Toby.
Solution
(a)
The ray diagram is drawn below.
1
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View Full Document Converging
Lens
5cm
15cm
25cm
5cm
15cm
25cm
object
image
Ray Diagram Must be Drawn to Scale
F’
F
0
P
CC
FF
f
1
s
1
s
1
’
(b)
Use the Thin Lens Equation for the Frst lens:
1
f
1
=
1
s
1
+
1
s
′
1
,
where
f
1
= 12cm
is the focal length of the Frst lens,
s
1
= 22cm
is the object distance, and
s
′
1
is the image
distance of the Frst lens. Solve for
s
′
1
s
′
1
=
1
1
f
1

1
s
1
.
s
′
1
=
1
1
12cm

1
22cm
= 26
.
4cm
(c)
The object distance for the second lens is the separation of the lenses,
d
, minus the image distance of the
Frst lens,
s
2
=
d

s
′
1
= 55cm

26
.
4cm = 28
.
6cm
. Use the Thin Lens Equation to compute the Fnal image
distance,
s
′
2
.
s
′
2
=
1
1
f
2

1
s
2
,
where
f
2
=

15cm
is the focal length of the second lens. Solve for
s
′
2
:
s
′
2
=
1
1
f
2

1
s
2
=
1
1
−
15cm

1
28
.
6cm
=

9
.
8cm
Total Points for Problem: 3 Points
Solution to Homework Problem 23.3(Crossed Polarizers)
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This note was uploaded on 05/04/2008 for the course PHYS 2074 taught by Professor Stewart during the Winter '08 term at Arkansas.
 Winter '08
 Stewart
 Work, Light

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