Physics 142!
Summer 2013
Exam III
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.!
In as class demonstration the AC circuit consists of the 120 V (rms) voltage from
the wall outlet as generator, a light bulb of 60 Ω resistance, and a capacitor and
inductor, each of which has 300 V (rms) across it. Which of the following is
wrong?
!
The circuit is in resonance.
!
The capacitor and inductor each consume zero power in a cycle.
√ !
[The current is 2 A so XL = XC = 150 Ω .]
!
!
2.!
The reactances of the capacitor and inductor are XL = XC = 300 Ω .
One of the above is wrong.
Which of the following devices, when used as intended, produces a (final) real
image to be viewed by an observer?
!
A magnifying glass.
!
An astronomical telescope.
√ !
!
A slide projector.
An optical microscope.
!
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Physics 142!
3.!
Summer 2013
Unpolarized light is incident at angle θ1 from a medium with refractive index n1
onto the interface with a medium of refractive index n2 . The light is totally
reflected. Which of the following is wrong?
!
If θ1 < 90° then n2 < n1 .
!
If n2 > n1 then θ1 = 90° .
√ !
The reflected light is partially polarized with E perpendicular to the plane
of incidence. [No polarization for total reflection.]
If n2 < n1 then θ1 is at least as large as the critical angle.
!
4.!
A person cannot focus effectively on objects closer than 50 cm or farther away
than 5 m. The bifocal lenses to correct the vision (with effective near point at 25
cm) should have focal lengths:
!
–50 cm and +5 m.
√ !
+50 cm and –5 m.
!
–50 cm and –5 m.
!
+50 cm and + 5 m.
Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each
question carries a value of 3 points.
5.!
If a surface absorbs energy E from a beam of light, it also absorbs momentum
E/c from the beam.
√
6.!
T
Total reflection for incident angles less than 90° can occur when light is reflected
from a medium with a higher index of refraction.
T
7.!
F
√
F [Must be lower.]
The transmission axis of polarizing sunglasses is horizontal to absorb reflected
glare from horizontal surfaces.
T
√
F [Vertical.]
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Physics 142!
Summer 2013
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method
clearly. Problems carry the point values shown.
1.!
The input voltage (rms) to the filter is Vin at frequency ω .
a.!
Find the current through the resistor in terms of Vin
and the circuit elements shown.
b.!
Find Vout /Vin in terms of ω and the circuit
What is this ratio for ω → 0 ? For ω → ∞ ?
d.!
For what value of ω is the ratio equal to 1?
Vin
R
L
elements.
c.!
C
[20 points]
a.!
The impedance is Z = R 2 + (XL − XC )2 so the current is
I = Vin / R 2 + (XL − XC )2 .
Vout
=
Vin
R
b.!
Since Vout = IR we have
c.!
For both limits the ratio vanishes.
d.!
When ω L − 1/ω C = 0 the ratio is 1. This means ω 2 = 1/LC , i.e., resonance.
R 2 + (ω L − 1/ω C)2
3
.
Vout
Physics 142!
2a.!
Summer 2013
Light is incident from air onto a block of a transparent material
in the shape of an equilateral triangle.
a.!
The angle of refraction at the top surface is θ 2 and the
angle of incidence of this ray on the side of the triangle
is θ 3 . Show that θ 2 + θ 3 = 60° . [Draw the ray.]
b.!
θ1
60°
Show that if n ≥ 2 then reflection is total at the side of
the triangle, no matter what θ1 is. [Show it for θ1 = 90°
and argue it must be true for smaller values of θ1 . Use sin 30° = 1/ 2 .]
[10 points]
a.!
See drawing. The angles of the dotted triangle are θ 2 , θ 3 , and 120°, so
θ 2 + θ 3 = 60° .
b.!
For θ1 = 90° , nsin θ 2 = 1 , or sin θ 2 = 1/n ≤ 1/ 2 . This means θ 2 ≤ 30° , so θ 3 ≥ 30°
and sin θ 3 ≥ 1/n . Thus θ 3 is at least equal to the critical angle for total reflection. If
θ1 < 90° then θ 3 is even larger.
2b.!
The magnifying powers of telescopes and microscopes are inversely proportional
to the focal length of the eyepiece. Why is it not possible to obtain arbitrarily
large magnifying powers by making that focal length extremely small?
[5 points]
!
!
Reducing the focal length requires reducing radii of curvature of the lens surfaces. This
increases the problem of spherical aberration, making it necessary to use only the very
center of the lens, which reduces the brightness of the image.
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Physics 142!
3a.!
!
Summer 2013
Shown is a flashlight battery connected in
a circuit with a current I as shown. The
radius of the battery is a and its length is
L. The potential difference between its
ends is ΔV.
P
I
+
a.!
At point P shown, what are the
magnitudes and directions of the electric and magnetic fields (assuming
the E-field is uniform along the length of the battery)?
b.!
What is the Poynting vector S (direction and magnitude) at P? Explain the
physical significance of the direction.
c.!
Show that the outward flux of S through the battery’s surface is equal to
the power it supplies.
[15 points]
a.!
E is to the left (toward lower potential), B is out of the page (right hand rule). The
magnitudes are E = ΔV /L , B = µ0 I / 2π a .
b.!
S is up the page, with magnitude S =
EB IΔV
=
. The direction shows power is
µ0 2π aL
leaving the battery through its sides.
c.!
The surface area of the sides is A = 2π aL and the total power is P = SA = IΔV , as
expected. (There is no power outward through the ends of the battery because of
the direction of S.)
3b.!
Explain why swimmers and scuba divers wear goggles when under water.
[5 points]
The material of the eye has almost the same index of refraction as water, so the focusing
power of the eye in water is very small. Placing a layer of air in front of the eye restores
normal vision. This is what the goggles do.
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Physics 142!
4.!
Summer 2013
An object is 15 cm if front of a mirror of focal length 10 cm as shown. The dots
are spaced 5 cm apart. At 50 cm from the mirror is a lens of focal length –10 cm.
1st image
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2nd image
•
•
•
•
•
a.!
Use the formulas to locate the image formed by the mirror and find its
magnification.
b.!
Use this image as the object for the lens, locate the final image and find the
overall magnification.
c.!
Draw careful principal ray diagrams to show how these images are
formed. The vertical lines represent the location of the mirror and lens in
paraxial ray approximation.
d.!
Suppose the lens is only 25 cm from the mirror. Use the formulas to locate
the final image and find the overall magnification. [What is the object
distance for the lens?]
[20 points]
a.!
Here p1 = 15 so we have
b.!
Now p2 = 20 so
1
1
1
1
=
−
=
, so q1 = 30 . Also m1 = −30/15 = −2 .
q1 10 15 30
1
1
1
3
−20/3
=
−
=−
so q2 = −20/3 . Also m2 = −
= 1/3 .
q2 −10 20
20
20
The overall magnification is m = m1m2 = −2/3 .
c.!
Drawing shown. Red rays for mirror, blue for lens.
d.!
Now the object for the lens is virtual. We have p2 = −5 and find s2 = 10 , so it is a
real image. The magnification is m2 = +2 and the overall magnification is −4 . [Ray
diagram on next page for those curious.]
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Physics 142!
Summer 2013
1st image
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2nd image
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Physics 142!
Summer 2013
Exam III
25
Median = 81.5
St Dev = 13
Students
20
15
10
5
0
<40
40
50
60
70
80
90
80
90
Exam Average
25
Median = 74.3
St Dev = 13.4
Students
20
15
10
5
0
<40
40
50
60
8
70