Name _________________
Solutions to Final Exam
May 2, 2014
This test consists of five parts. Please note that in parts II through V, you can skip one
question of those offered. Some possibly useful formulas appear below
Constants, etc.
e  1.602  1019 C
Magnification
M  h h   q p
m   0
eV  1.602  1019 J
ke  8.99  109 N  m 2 /C2
Telescope
Magnification
m  fo fe
0  4 107 T  m/A
h  6.626  1034 J  s
h  4.136  10
15
Focal Length
f  12 R
Diffraction
sin  dark  m d
Compton Scattering
h
   
1  cos  
mc
Diffraction Grating
sin   m d
X-Ray Scattering
2d cos   m
eV  s
 sin  a sin    
I  I max 

  a sin   
Position on a screen
x  L tan   L sin 
2
1  n2   1 1 
   1   
f  n1   R1 R2 
Brewster’s
Angle
tan  P  n2 n1
LC Circuits
  1 LC
Thin Film
2tWeak   m
2-Slit Interference
sin  bright  m d
  d sin  
I  I max cos 2 
 

Electromagnetic Waves
E0  cB0
S  cB02 2 0
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. Can you use the formulas relating objects and images when working with a flat
mirror?
A) Yes, but you have to use the size of the mirror as the focal length
B) Yes, but you have to use one-half the size of the mirror as the focal length
C) Yes, but you have to use zero as the focal length
D) Yes, but you have to use infinity as the focal length
E) No
2. Which of the following must be true about any charge you put on a conductor?
A) It must always be at the center of the conductor
B) It must be uniformly spread throughout the conductor
C) It must be concentrated at one point on the surface of the conductor
D) It must be on the surface, but not at a point
E) Conductors can never have any charge on them
3. How can a transformer change the voltage from high voltage to low voltage?
A) There are resistors that reduce the voltage, acting as a voltage divider
B) A capacitor stores the charge, then releases it at a lower voltage
C) A diode only lets current through one way, but it is a lower voltage afterwards
D) Charge is stored which creates an electric field, and a second wire feels the
voltage and produces the lesser voltage
E) A coil produces a changing magnetic field, a second coil with a different
number of terms experiences a voltage due to the changing magnetic flux
4. One reason astronomical telescopes are so large in diameter is so they can gather lots
of light. What is a second reason for this?
A) Only large diameter telescopes can let in long wavelength light
B) There is a physical limit to how small of angles can be resolved based on the
size of the opening
C) There is a maximum magnification that can be implemented for any given
diameter
D) It’s just a way to pad their budgets, bigger telescopes are more expensive
E) Only with a large diameter can they view a large portion of the sky
5. A light ray coming in parallel to the optic axis for a concave mirror will, after
reflection, pass through the ______ of the mirror
A) Focus
B) Center
C) Vertex
D) Radius
E) None of the above
6. Which of the following is approximately a correct statement about the uncertainty of
position and/or momentum based on quantum mechanics?
A) There is a minimum position uncertainty for any particle
B) There is a minimum momentum uncertainty for any particle
C) There is a minimum for the sum of the position and momentum uncertainty
D) There is a minimum for the product of the position and momentum
uncertainty
E) There is a minimum for the ratio of the position and momentum uncertainty
7. Electric field lines
A) Start at positive charges and end at negative charges
B) Start at negative charges and end at positive charges
C) Always form closed circles around currents
D) Are always parallel on the surface of conductors
E) None of the above
8. What does it mean that there is a phase shift between the voltage and the current in an
AC circuit?
A) The two are not of the same magnitude
B) The peaks and valleys of the current and the voltage do not occur at the same
time
C) The frequencies of the two are slightly different
D) The voltage oscillates, but the current does not
E) The current oscillates, but the voltage does not
9. When can we trust the results of geometric optics, also known as the ray
approximation?
A) When all apertures/holes/barriers, etc., are large compared to the wavelength
B) When all apertures/holes/barriers, etc., are small compared to the wavelength
C) Whenever we are detecting individual photons
D) Whenever we are not detecting individual photons
E) Whenever multiple wavelengths are being used simultaneously
10. Which of the following imaging system produces an image with no distortion or
imperfect focus?
A) A single thin lens
B) A pair of matched thin lenses
C) A single curved mirror
D) A pair of matched curved mirrors
E) A single flat mirror
11. Suppose light is moving in air, and hits water, but reflects off. How does the
wavelength or frequency of the reflected wave compare to the incident wave?
A) Frequency is the same but wavelength decreases
B) Frequency is the same but wavelength increases
C) Frequency decreases and wavelength increases
D) Frequency increases and wavelength decreases
E) Both stay the same
12. What happens if you charge up a capacitor, then connect the ends to a resistor?
A) The capacitor will instantly discharge
B) The capacitor will discharge gradually through the resistor
C) The capacitor and resistor will trade energy back and forth, oscillating
D) The capacitor will give its energy into the resistor, where it will stay
E) Nothing, because the resistor doesn’t allow current through it
13. Ampere’s Law relates the integral of the magnetic field around a loop to what?
A) The voltage through the loop
B) The electric charge inside the loop
C) The electric flux through the loop
D) The current going through the loop
E) The magnetic charge inside the loop
14. How do corner reflectors, such as you have in reflectors on cars and so on, work?
A) They have detectors that sense the incoming light, then LED’s that send a signal
back the way it came
B) They have a single mirror that pivots in different directions based on the incoming
light
C) They have sets of three mirrors that always reflect the light back the
direction it came from
D) They have numerous tiny mirrors reflecting light in all directions, and one of them
will be correct to reflect the light back
E) They use optical fibers to take the light, loop it around, then send it back from
where it came
15. A Michelson interferometer compares the distance between two long paths by
A) Measuring the amount of time it takes light to go through each path
B) Adding together the light that follows two paths, and seeing how much they
combine constructively or destructively
C) Forcing waves to fit exactly between mirrors on two different paths, and
measuring how the wavelengths of the light compare
D) Counting the exact number of waves along each of two paths and subtracting
them
E) Measuring the blurring of the image caused by the motion of one of the mirrors
16. The force on a moving charge in a magnetic field is ______ to the direction of motion
and _____ to the direction of the magnetic field
A) Perpendicular, perpendicular
B) Perpendicular, parallel
C) Parallel, perpendicular
D) Parallel, parallel
E) None of the above
17. Which type of electronic component is best at blocking low frequency AC sources
while allowing high frequencies through?
A) Capacitor B) Resistor C) Inductor D) Diode
E) Battery
18. When light passes through a very small hole shaped like the one at
right, what will the pattern of light look on a screen far away?
A) It will tend to maintain this rectangular shape and size indefinitely
B) It will tend to spread equally in both the vertical and horizontal directions
C) It will spread out in both directions, but mostly in the vertical direction
D) It will spread out in both directions, but mostly in the horizontal direction
E) No significant light can make it through small holes
19. A capacitor with capacitance 2 F has a voltage 6 V on it. What is the charge on the
capacitor?
A) 121 C
B) 13 C
C) 3 C
D) 12 C
E) 24 C
20. In a velocity selector, a charged particle is in a strong magnetic field. Why does the
charge move in a straight line anyway?
A) It is moving parallel to the magnetic field, and hence feels no force
B) It is moving perpendicular to the magnetic field, and hence feels no force
C) It is so massive that the tiny force of the magnetic field has no effect
D) It is moving so quickly that the force has little effect on it
E) There is an electric field that cancels out the force from the magnetic field
21. How does placing a material in a capacitor increase the capacitance?
A) It effectively increases the area of the capacitor
B) It effectively increases the distance between the plates
C) It adds extra charge to the capacitor plates
D) It acts as a battery, changing the voltage
E) Charges in the material shift, cancelling out part of the electric field
22. Inductors resist changes in
A) Charge
B) Voltage
C) Current
D) Power
E) Inductance
23. The chromatic dispersion of a lens refers to the fact that
A) The small angle approximation is not always accurate, and hence things don’t
focus perfectly
B) Lenses absorb a certain fraction of the light, ruining the image
C) Lenses do not bend light of all frequencies the same, ruining the focus
D) The shape of lenses is not perfect, making the image imperfect
E) Because of the delay from the slowdown of different colors, the light doesn’t all
arrive at the same time
24. According to Fermat’s principle, the path light takes from one point to another is
always the one that takes the
A) Most time
B) Least time
C) Shortest path
D) Longest Path
E) None of the above
25. How does the direction the charge carriers in a conductor move compare to the
direction of current flow?
A) They are the same, always
B) They are always opposite directions
C) They are the same if charge carriers are positively charged, and opposite if
they are negatively charged
D) They are the same if charge carriers are negatively charged, and opposite if they
are positively charged
E) None of the above
Part II: Short answer, old material: [20 points]
Choose two of the following three questions and give a short answer (2-3
sentences) (10 points each)
26. Suppose you want to know the total charge inside a box. Explain how you could,
in principle, determine this without looking inside the box. An equation is
advised.
According to Gauss’s law, the total charge inside an arbitrary box is related to the
total electric flux coming out of the box. The charge inside can be determined from

qin   0  E   0  E  nˆ da .
27.Describe in words or equations Kirchoff’s two laws for circuits. You do not need
to give a detailed description of how you apply it.
Kirchoff’s first law says that charge is conserved, which basically says that the
current into any vertex is zero. Kirchoff’s second law says that the voltage drop around
any loop has to be zero. If written as equations, these equations would look something
like
I
in
  I out
0   V
loop
28. Explain qualitatively how an electric generator can create electrical power. A
diagram is not necessary, nor do you have to explain what is rotating the
relevant parts.
An electric generator operates by rotating a loop of wire in a constant magnetic field,
or by rotating magnets in the presence of a loop of wire. The magnetic flux through the
wire is changing, and this creates an electromotive force according to the Faradays’ Law:
 
dB
.
dt
Part III: Short answer, new material: [20 points]
Choose three of the following four questions and give a short answer (2-3
sentences, or comparable) (10 points each)
29. As a beam of white light moves from air to glass, will its direction change? Give
a relevant equation. Will all colors bend the same amount? Explain which
colors are more likely to bend more.
Because air and water have different indices of refraction the direction of the
wave compared to the normal will change according to Snell’s law, n1 sin 1  n 2 sin  2 .
However, the index of refraction (especially in water) depends on frequency. In general,
long wavelengths tend to bend less than short wavelengths (red refracts rotten, blue bends
better).
30. Explain what a birefringent crystal is, and why a single beam of light can be
divided into two beams by it
Light normally consists of two different directions the electric field points. For
most materials, the index of refraction is the same for both polarizations, but for
birefringent crystals, the index of refraction is different, and hence when a beam of mixed
polarizations passes into it, the two polarizations will bend by different amounts,
allowing them to be separated.
31. A curved mirror has a radius of R. What is its focal length? Draw a sketch or
otherwise indicate how a mirror could have a positive or negative focal length
The focal length is half the radius of the mirror, f  12 R . The
radius (or focal length) is considered positive if it is concave on the side
the light is entering, and negative if it is convex on the side the light is
entering, as sketched at right.
f >0
f <0
32 Explain qualitatively why it is that soap bubbles and thin films of oil sometimes
appear to have detailed colors
When light bounces off of a thin surface, like a soap bubble, there are two
reflections, one when the light enters the water, and one when it exits back into air.
These two reflected waves can interfere, constructively or destructively, enhancing or
suppressing the reflection. Whether it interferes constructively or destructively depends
on wavelength; for example, reflection is destructive if the thickness is 2tWeak   m , and
strong if 2tStrong    m  12  , where m is an integer, so the same thickness will reflect
some colors very well and others very poorly, so that white light gains distinct colors
upon reflection from a thin film.
Part IV: Calculation, old material: [60 points]
Choose two of the following three questions and perform the indicated
calculations (20 points each)
33. A charge of +4.00 C is placed at the origin. A point P is 3.00
cm to the right of it.
(a) What is the electric field at P?
3.00 cm
+ 4.00 C
P
The electric field is computed using
 ke q
8.99  109 N  m 2 /C2  4.00  106 C 

E  2 rˆ 
rˆ  3.996 107 rˆ N/C .
2
r
 0.03 m 
The r̂ simply denotes the direction of the force, which is away from the charge.
(b) What is the electric potential at P?
The potential is given by
9
2
2
6
ke q  8.99 10 N  m /C  4.00 10 C 
V
rˆ  1.199  106 N  m/C  1.199  106 V .

r
0.03
m


(c) A proton (m = 1.67210-27 kg, q = 1.60210-19 C) at rest is now placed at the
point P. What is its initial acceleration at P?




We use the formula F  ma , and use the formula F  qE to obtain the
acceleration, which is

 F q  1.602 1019 C
a  E
 3.996 107 rˆ N/C   3.829 1015 rˆ m/s2 .
m m
1.672 1027 kg
(d) Eventually the proton gets very far away from the point charge. At this
point, what is its velocity?
We use conservation of energy to solve the problem. The initial energy is all
potential energy, which is U  qV , so
U  qV  1.602  1019 C 1.199 106 V   1.9211013 J .
Now, as the proton heads to infinity, the potential energy falls to zero, and it is all
converted to kinetic energy. So U  12 mv 2 . We then solve for the velocity:
13
2U 2 1.921 10 J 
v 

 2.23 1014 m 2 /s 2 ,
27
m
1.672  10 kg
2
v  2.23 1014 m 2 /s 2  1.516 107 m/s .
34. An electric teapot is basically a resistor connected to a power
source, which we’ll treat as a 120 V battery. Water has a specific
heat capacity of 4.18 J/g/C
(a) How much energy is required to heat 100.0 g of water from 20C
to 100C?
–
+
b
120 V
R
Since the specific heat capacity of water is 4.18 J/g/C, and we have 100.0 g of
water, the heat capacity is
C   4.18 J/g/ C 100.0 g   418.0 J/ C
Since the total temperature change is 80C, the total energy is therefore
E  C T   418.0 J/ C  80C   3.34  104 J.
(b) Suppose it takes 3.00 minutes to heat it up this much. What is the rate of
power consumption for the resistor? What is the current I?
Power is energy over time, so since it takes 180 seconds, the power is
E 3.34 104 J

186 W.

t
180 s
We then use the formula   I V to solve for the current I¸ which yields
I
 186 W

 1.548 A .
V 120 V
(c) Find the resistance of the resistor.
We use the formula V  IR and solve for the resistance R to yield
R
V
120 V

 77.5  .
I
1.548 A
35. Flipping an electron’s spin in a hydrogen atom emits microwaves with a
frequency of f  1.420 109 Hz .
(a) Find the angular frequency , the period T, the wavelength  , and the wave
number k for this wave.
Frequency, angular frequency, and period are related by
  2 f  2 1.420 109 s 1   8.92 109 s 1 ,
T
1
1

 7.04  1010 s  0.704 ns .
9 1
f 1.420  10 s
The wavelength can be found from f   c and the wave number from k   2 :
c 2.998 108 m/s

 0.2111 m  21.11 cm ,
f
1.420  109 s 1
2
2
k

 29.8 m 1 .
 0.2111 m

In fact, this emission feature is usually called the 21 cm line.
(b) An astronomical source produces waves with an intensity
S  1.08 1013 W/m 2 . Find the magnitude of the electric field E0 and the
magnetic field B0 of this wave.
We use the formula relating the intensity to the magnetic field, S  cB02 2 0 , so
7
13
2
2 0 S 2  4 10 T  m/A 1.08  10 W/m 
B 

 9.05  1024 T 2 ,
3.00  108 m/s
c
B0  3.011014 T .
2
0
The electric field is then found from E0  cB0 , so
E0  cB0   2.998  108 m/s
1.64 10
13
T   4.92  105 V/m  49.2  V/m .
(c) An astronomer is trying to make an LC circuit that vibrates at this exact
frequency. He has already purchased an inductor with L  1.20  H . What
capacitance should he use to get the right frequency?
This is straightforward. We use the formula   1
and solve for C. We have
C
1
 L
2

1.20 10
1
6
H  8.92  10 s
9

1 2
LC , and simply square it
 1.05 1014 F .
Part V: Calculation, new material: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
air
water
36. A light beam passing through a layer of flint glass
(n = 1.62) moving at an angle of 40 compared to

the perpendiular, passes into a layer of water (n =
1.33), and then continues towards a layer of air, as
sketched at right
(a) Assuming at least some of the light makes it
40
into the water, calculate the angle  compared
to the perpendicular at which it enters the water.

’ 
glass
We use Snell’s Law, n1 sin 1  n2 sin  2 , to find the resulting angle. We have
1.62sin 40  1.33sin  ,
1.62
1.62
sin  
sin 40 
0.643  0.783
1.33
1.33
  sin 1  0.783  51.5
(b) Assume now that some of the light is reflected at the water-air interface,
calculate the angle  at which it is reflected, compared to the perpendicular.
By alternate interior angles, the angle ’ is equal to . Then, by the law of
reflection, we have        51.5 . That was easy.
(c) For the second step, the water-air interface, is any of the light actually
reflected? Is any of it transmitted? Explain why you believe your answer.
In principle, we could calculate the angle of the transmitted angle , so we would
have
1.33sin    1.00sin  ,
sin   1.33  0.783  1.041
But this result is greater than one, which is impossible. This is a sign that, in fact, all of
the light is reflected, so we have total internal reflection.
37. An object of size 1.00 cm is being examined through a thin lens. When the object
is placed 20.0 cm in front of the lens, and examined through the lens, it appears
to be 3.00 cm high, and upright
(a) What is the apparent location of the image compared to the lens? What is
the focal length of the lens?
The magnification is
M
q h 3.00 cm
 
3,
p
h 1.00 cm
q  3 p  3  20.0 cm   60.0 cm .
So q  60.0 cm . The image appears to be 60.0 cm on the side the light is entering the
lens. The focal length can then be determined from
1 1 1
1
1
3 1
1
,
  



f
p q 20.0 cm 60.0 cm 60.0 cm 30.0 cm
f  30.0 cm .
(b) The object is now moved to a distance such that it appears to be infinitely far
away as viewed through the lens. What distance is it at?
The image is at q   . We therefore have
1 1 1 1 1 1
1
     0  .
p f q f  f
f
It follows that p  f  30.0 cm , so the object is 30.0 cm in front of the lens.
(c) Which of the four lenses sketched at right has the
right shape for this lens? If it is made from cubic
zirconium with index of refraction n = 2.16, and the
two radii are equal in magnitude, what are those
radii?
Lenses which are thicker in the middle are converging,
and hence the circled lens is the correct one. If the light is
coming from the left, then the first surface has radius positive
R, and the second surface has radius negative R. We then use
the lensmaker’s equation to obtain
 1 1 
1
1 n
 1 1  2 1.16
   2  1      2.16  1    
,
R
30.0 cm f  n1   R1 R2 
R R
R   2 1.16  30.0 cm   69.6 cm .
38. Helium produces a yellow spectral line with
  588 nm and a violet spectral line with a
wavelength you determine below. It is then
passed through diffraction grating
(a) The yellow light’s first order diffraction line
is at an angle of 35.0. Determine the spacing
of the diffraction grating.
2nd order
2nd order
yellow
violet
violet
yellow
The angle by which the light is diffracted for the m’th order diffraction line is
given by sin   m d . Solving for d, we have
d
m 1 588 nm 

 1025 nm  1.025  m .
sin  sin  35.0 
(b) The violet one is diffracted by an angle of 25.8. What is the wavelength 
for the violet line?
The spacing is, of course, the same, so we use the same equation sin   m d :

d sin  1025 nm  sin  25.8 

 446 nm .
m
1
(c) Find the angles for the second order diffraction for these two waves.
Interestingly, only one of these two lines has a second order spectral line.
Which one?
The angles of the second order spectral lines can be solved for by rearranging the
equation again. The angles can be found using
m 2  588 nm 

 1.1473 ,   ???
d
1025 nm
m 2  446 nm 
violet: sin  

 0.870 ,   sin 1  0.870   60.5 .
d
1025 nm
yellow: sin  
The sine of an angle can never exceed one, so in fact the yellow line has no second order
diffraction.
39. The work function for Gallium is 4.32 eV.
(a) When light with frequency  = 157.0 nm impacts on Gallium, what is the
kinetic energy of the electrons that come off of it?
We need to work out the frequency for this light, which can be found from
f   c . We can then get the energy from E  hf . So we have
E  hf 
hc

 4.136 10

15
eV  s  2.997 108 m/s 
157.0 109 m
 7.90 eV .
From this we subtract the work function, 4.32 eV, to get
K  hf    7.90 eV  4.32 eV  3.58 eV .
(b) Waves of unknown frequency impact on Gallium, causing the emission of
electrons with kinetic energy 2.01 eV. What is the frequency of the light?
Turning the equation around, we have
hf  K    2.01 eV  4.32 eV  6.33 eV ,
f 
6.33 eV
6.33 eV

1  1.53 1015 s 1  1.53 1015 Hz .
h
4.136  1015 eV  s
(c) What is the longest wavelength that can eject any electrons from Gallium?
A photon can eject an electron only if it has sufficient energy to overcome the
work function, so hf   . Hence the frequency must exceed
f 

h

4.32 eV
 1.044 1015 s 1 .
15
4.136 10 eV  s
Since f   c , it follows that high frequency corresponds to short wavelength, and hence
we must have

c
f min

2.998 108 m/s
 2.87 107 m  287 nm .
15
1.044 10 /s