Physics 54!
Summer 2012
Exam III
Solutions
Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.
1.!
A series circuit has an AC generator, a light bulb (resistor), an inductor and a
capacitor. Measurement shows that the rms voltage across the capacitor is larger
than that across the inductor. To make the light bulb glow brighter, one can:
!
Decrease the capacitance.
!
Decrease the frequency.
√ !
!
2.!
Increase the inductance. [Initially XC > XL .]
All of these make the bulb glow less brightly.!
Unpolarized light is
incident at angle θ1
R
from medium 1 on
the interface with
medium 2. Shown is
a reflectivity plot for
these two media.
!
The plot is for
Einc
perpendicular
to the plane of
incidence.
√ !
!
θ1 (radians)
The indices
obey n1 < n2 .
The plot for the same media but the other orientation of Einc relative to
the plane of incidence differs from this one in that the angle at which
R = 0 is smaller.
!
π /2
The angle at which R = 0 is given by sin θ1 = n2 /n1 . !
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Physics 54!
3.!
Summer 2012
The capacitor shown has charge Q as indicated, which is
decreasing. Point P is between the capacitor plates, near the
top edge. Which of the following is wrong?
!
At P the E-field is to the right.
!
At P the B-field is into the page.
!
At P the Poynting vector is up the page.
√ !
4.!
+Q
P
•
The current in the wires is to the right. [Left. Q is decreasing.]
Which of the following devices, when used as intended, produces a virtual
enlarged image to be viewed?
!
A camera.
!
A slide projector.
√ !
A microscope.
!
None of the above.
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.!
When the frequency in a series LCR AC circuit is lower than the resonant
frequency, the rms voltage across the capacitor is lower than that across the
inductor.
T
6.!
F [Higher. XC > XL .]
On a clear afternoon the sky light from a point directly above an observer is
partially polarized with E perpendicular to the plane made by rays from the sun
to the point and from the point down to the observer.
√
7.!
√
T
F
To correct the far vision of a person who cannot see clearly beyond distance D
(meters), a lens should be prescribed with power −1/D (diopters).
√
T
F
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Physics 54!
Summer 2012
Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method
clearly. Problems carry the point values shown.
1.!
A lecture demonstration uses a circuit with a capacitor, an inductor, and a light
bulb. Assume the only resistance is that of the bulb. If the bulb alone is in the
circuit, across the (rms) emf E of the wall outlet, it consumes (average) power
P0 . The capacitance is such that XC = 3R , where R is the bulb’s resistance. With
all three elements in the circuit, the power consumed by the bulb is P0 /5 . When
L is increased the bulb glows brighter. [All answers are numbers.]
a.!
What is XL /R initially?
b.!
What is the phase angle φ initially? [Give tan φ .]
c.!
When L is increased until resonance is reached, what is VC /E ?
[15 points]!
a.!
With the bulb alone P0 = E 2 /R . With the whole circuit P0 /5 = (E /Z)2 R , so
Z 2 = 5R 2 . Thus R 2 + (XL − XC )2 = 5R 2 , or XL − XC = 2R . Since increasing L makes
Z smaller, we see that XC > XL . So XL = XC − 2R = R and XL /R = 1 .
XL XC
−
= −2 .
R
R
b.!
We find tan φ =
c.!
The current is now I = E /R so VC = IXC = (E /R) ⋅ 3R , so VC /E = 3 .
!
[This is roughly like the demonstration done in lecture.]
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Physics 54!
2a.!
Summer 2012
The E-field of a harmonic e-m wave is given by Ez (x,t) = E0 cos(kx − ω t) .
a.!
Write the formula for the B-field, in terms of the given quantities.
b.!
Write the formula for the Poynting vector.
[Be sure to use subscripts to indicate which component you are describing.]
[10 points]
a.!
With E in the z-direction and the wave moving in the x-direction, B must be in the
negative y-direction, so By = −(E0 /c)cos(kx − ω t) .
b.!
The Poynting vector is in the x-direction, so Sx = (E02 /c µ0 )cos2 (kx − ω t) .
2b.!
!
a.!
Shown is a resistor in a circuit, with current as
indicated. The resistor has length  , crosssection radius a, and resistance R.
I
P
•
S
E
B
a.!
Find E, B, and S at the point on the
surface shown, indicating directions on the diagram. [Treat E as uniform.
What is the potential difference from one end of the resistor to the other?]
b.!
Use the flux of the Poynting vector through the surface of the resistor to
find the total power flowing into it.
[10 points]
E is to the right (toward lower potential) so E = IR / . B is out of the page, and is
like the field of a long wire, so B = µ0 I / 2π a . This gives S down, with magnitude
S = EB/ µ0 = I 2 R / 2π a .
b.!
The Poynting vector is everywhere perpendicular to the surface and has the same
magnitude everywhere, so the flux is just S times the surface area 2π a . The power
going into the resistor is I 2 R as expected.!
!
!
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Physics 54!
Summer 2012
3a.!
A small object is near a plate glass window. Draw
two rays (one at normal incidence to the glass) to
show where the object appears to be to a viewer
on the other side of the window. [No
calculations.]
!
[An assignment problem, to calculate how far the
apparent location is from the actual one.]
• •
!
[5 points]
Appparent location
3b.!
In an assignment problem, light traveling out of the page is polarized vertically.
By passing it through N polaroid filters, each with axis making angle π / 2N with
the previous one, the light emerges polarized horizontally. Here we consider the
energy absorbed by the filters.
a. !
What fraction of the intensity that enters it is absorbed by each filter?
[Which component of E is absorbed?]
b.!
Show that for large N this fraction decreases as 1/ N 2 . [Use small angle
approximations.]
c.!
!
a.!
This happens N times, once for each filter. What does this mean about the
total energy absorbed as N → ∞ ?
[15 points]
The component perpendicular to the polaroid axis is absorbed, which is Esin θ , so
the fraction of intensity absorbed by one filter is sin 2 θ = sin 2 (π / 2N) .
b.!
Small angles: sin θ ≈ θ , so we have approximately (π / 2N)2 .
c.!
We find (π / 2N)2N , which clearly goes to zero as N → ∞ . All the energy is
transmitted, as the assignment problem showed.
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Physics 54!
4a.!
Summer 2012
An object is placed directly in front in front of a mirror of focal length +10 cm. At
distance 30 cm from this mirror is another mirror of focal length +10 cm.
a.!
Use the formulas to locate the image formed by the right mirror, and find
its magnification. Then use this as object for the left mirror, finding the
location and overall magnification of the final image.
b.!
Draw principal rays on the diagram to verify your calculations. The dots
are at 5 cm intervals.
Final image
•
•
•
•
•
•
First image
!
a.!
[10 points]
For the right mirror: p = 30 , f = 10 ; this gives q = 15 and m = −1/ 2. For the left
mirror: p = 15 , f = 10 ; this gives q = 30 and m = −2 . The final image is at the right
mirror, and m = +1 . [An assignment problem and demo.]
b.!
4b.!
See drawing. Red for right mirror, blue for left mirror.
The diagrams indicate eyeballs for two people, one nearsighted, one farsighted.
a.!
Draw two rays coming from a great distance and entering the eye,
showing where the image will be formed by the eye’s focusing system.
b.!
In each case, indicate what type of lens (with positive or negative power)
would be prescribed to correct the far vision of these people.
[10 points]
Negative power
Positive power
Nearsighted
Farsighted
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Physics 54!
Summer 2012
Median = 72
SD = 13.6
Median = 66
SD = 11.8
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