f
1
= 40 cm
f
2
=  40
60 cm
160 cm
2
Solutions to Practice Problems, Lecture 10 Review  Physics 7C Winter 2005
Slide 7: Double Lens Problem:
(a) Use ray traces and/or the thin lens equation to locate the image of the lens combination.
(b) Is the image real or virtual, upright or inverted?
( c) What is the magnitude of the final image?
Comments:
Solve the problem one lens at a time, starting with the left lens (converging)
then continuing with the right lens. The image from the left lens will be the object from the
right lens.
Lens 1: Converging—positive focal length! F = +40. You need three rays, typically.
1) Straight through middle. 2) Parallel, then toward far focal point. 3) Through near focal point,
then parallel.
1
f
1
=
1
o
1
+
1
i
1
1
40
=
1
60
+
1
i
1
i
1
= 120 cm
(real, inverted)
Now, since the first image is 120 cm from the first lens and is real (on the opposite side of the
lens as the object), and the two lenses are 160 cm apart, the first image is +40 cm away from the
second lens. So, o
2
is +40cm. What else do we need? We will need the height of the first image
to get the height of the final image. To find this, we use what we know about magnification:
M
lin
=
image size
object size
=
h
i
h
o
=
i
o
=
120
60
M
1
=2
h
i
h
o
=
h
i
2 cm
= 2
"
h
i
=  4 cm
For the second lens, use that o
2
is +40cm and the height of the object is – 4 cm (inverted).
1
f
2
=
1
o
2
+
1
i
2
1
40
=
1
40
+
1
i
2
i
2
= 20 cm
(virtual, inverted)
M
total
= M
1
•
M
2
=
1
2
•
"
2
=1
h
i
h
o
=M
2
=
1
2
=
h
i
4
"
h
i
= 2 cm
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View Full DocumentSlide 8: Pendulum Problem:
(a) A pendulum swings through an arc of one degree in one second from maximum height to
equilibrium. Another pendulum with the same length but with a bob of
twice
the mass is made
to swing through an arc of two degrees from maximum height to equilibrium. The time required
to swing through the twodegree arc is…
a)
½
second, one second, two seconds, or some other choice. Defend.
b) Write the appropriate harmonic oscillator equation for both scenarios and identify all
constants and variables.
Solution:
Since the first pendulum starts at maximum amplitude from rest and takes one second to go to
equilibrium, and since we know that the equilibrium position of a pendulum is the lowest point,
we know that it takes one second to go through
¼
of a period: T/4 = 1 sec. Why? Because in
order to return to the original position (and thus go through an entire period) it needs to swing
back up once to the opposite side, back down to equilibrium, then back up to the starting
position.
a) The second pendulum has a similar scenario, starting at maximum and down to equilibrium
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