1. Young’s double-slit experiment underlies the instrument landing system at most
airports and is used to guide aircraft to safe landings when the visibility is poor. Suppose
that a pilot is trying to align her plane with a runway as suggested by figure (a) below.
Two radio antennas A1 and A2 separated by 40.0m are positioned at the end of the
runway. The antennas broadcast coherent radio waves at 30 MHz.
A1
A2
a) Find the wavelength of the radio waves.
Since c = f, then  = c/f = (3x108 m/s) / (30x106 Hz) = 10m.
b) The pilot “locks onto” the strong signal radiated along an interference maximum and
steers the plane to keep the signal strong. Suppose that the pilot is flying along the first
side maximum (figure (b)). Calculate the angle  between the plane’s trajectory and the
centerline of the runway.
 10m
d sin   m  sin   
 0.25    14.5o
d 40m
c) How far to the side of the runway is the plane if it is 2 km from the end of the runway
and is flying along the first side maximum?
x
sin  
 x  (2km)(0.25)  0.5km  500m
2km
2. Consider the mass spectrometer shown below where a proton of charge +e and mass
m traveling with velocity v enters a region of electric and magnetic fields whose
magnitudes are 720 N/C and 0.2 T respectively.
+
+
+
-
+
-
-
E
v
-
+
-
a) If the magnetic field, B is directed into the page everywhere, what is the direction of
the magnetic force FB on the proton? (Explain your reasoning.)
up since the direction is given by the rh rule and the charge is +
b) Draw the electric field lines on the diagram above and then find the magnitude and
direction of the electric force on the proton.
Electric force = eE = 1.15 x 10-16 N
c) Based on your answers to parts a and b above, what is the velocity of the proton such
that it travels in a straight line?
evB = eE so v = E/B = 3600 m/s
d) If the proton leaves the electric field and is only under the influence of the magnetic
field, what is the charge to mass ratio of the proton if it travels in a circle of radius 0.188
mm? (In order to receive credit, do not simply take the proton’s charge divided by the
proton’s mass!)
F = evB = mv2/r so e/m = v/(rB) = 9.57 x 107 C/kg as it should
3. Consider the following circuit where the resistors are 150 and the battery is 45V.
B
6
5
4
A
3
2
1
a) What is the equivalent resistance of the circuit?
Labeling the resistors as shown:
R6 in parallel with R5 so R56 = 75
R4 in series with R56 so R456 = 225
R3 in parallel with R456 so R3456 = 90
R1 and R2 in series with R3456 so R123456 = 390.
Thus Req = 390.
b) What is the current produced by the battery?
Since V = IReq, I = V/Req = 45V/390 = 0.115A = 115mA.
c) What is the voltage difference between the points A and B?
Vbattery = 45V and V1 = V2 = IR1 = 17.3V, so V3456 = 45V – 34.6V = 10.4V.
Since R3 in parallel with R456, V3 = V456 = 10.4V.
By Ohm’s law, I3 = 0.069A and I456 = 0.046A (which sum to 0.115A)
This gives V4 = 6.9V and V65 = 3.5V (which sum to 10.4V since V4 and
V65 are in series.)
The potential difference across points A and B is given by V65 = 3.5V.
(For completeness, I5 = I6 = 0.023A = 23mA)
4.
A proton sits at the origin with an electron to either side of it along an axis, each
at a distance of 0.53 x 10-10 m from the proton.
a. Draw a sketch and find the electric field (magnitude and direction) at one of the
electrons due to the proton and other electron.
-e
+e
-e
E
ke
ke
3 ke


 3.84 x1011 N / c
2
2
r (2r )
4 r2
b. Find the external work required to have brought the second electron from a large
distance away to its distance from the proton and first electron in part (a). Is it
positive or negative work? Explain why.
ke ke ke
but Wext = eV = - 2.2x10-18J; negative since you need to apply a force
V 

r 2r 2r
away from the charges as the electron is held back while moving in towards its final
position.
5. a. In the circuit shown below, find the current through each resistor, using the
labels shown.
1 K
I2
2.5 K
I3
I1
4 K
10 V
1 K
Can find equivalent resistance directly – top branch has 5K which is in parallel with
middle branch – so these together give 1.67 K which is then in series with 1K – so the
total Req = 2.67K. Then I1 = 10/2.67K = 3.75 mA; this current then divides in the ratio
1:2 with the top branch getting 1 part and the middle branch getting 2 parts – therefore
the top gets I2 = (1/3)I1 = 1.25 mA and the middle gets I3 = (2/3)I1 = 2.50 mA.
b. Suppose that in the circuit of part (a) the battery and central branch are removed
so that there is a continuous single outer loop with only 3 resistors. If the square
loop has dimensions of 10 cm x 10 cm and is in a region of space where there is a
uniform B field directed into the paper and increasing at a rate of 0.1 T/s, find the
current (magnitude and direction) that flows through the 4 K resistor.
 1 ( BA)
1 B
1
I 

(0.1x0.1 m2 ) 
 0.1 0.01  1.7 x107 A
R R t
6K  t
6000
6.
i. a. A vertically polarized beam of light of 350 nm wavelength passes through a
polarizer with its transmission axis rotated at 30o to the vertical in the clockwise
direction, and then through a second polarizer with its transmission axis also
oriented at 30o to the vertical but in the counter-clockwise direction. If the
transmitted intensity of light is 1.0 W, find the intensity incident on the first
polarizer.
With Io = incident intensity, after the first polarizer the transmitted intensity is
I1 = Io[cos30]2 and after the second polarizer, making a 60o angle with
the first polarizer, the transmitted intensity is
I2 = Io[cos30]2 [cos60]2 = 1 W (as given, so we can solve for Io to find
Io = 1/{[cos30]2[cos60]2} = 5.3 W
ii. In a double slit experiment with a beam of electrons, the slit widths are 1 m and
the slit spacing is 10 m. The electrons are accelerated through a potential
difference of 104 V starting from rest before arriving at the slits.
a. What is the wavelength of the electrons?
h
h
= 1.22 x 10-11 m
 
p
2meV
b. What is the separation between the interference fringes formed on a screen 10 m
away from the slits?
D
 y
gives spacing = 1.22x10-5 m
d sin   m  d    m or y 
d
D
 
iii. An electron traveling north in a horizontal plane at a speed of 10 7 m/s enters a
region where there is a uniform 0.05 T magnetic field pointing vertically down.
a. Find the diameter of the orbit that the electron travels in and indicate its direction
of travel as viewed from above (specify either clockwise or counterclockwise).
The force on the electron is given by evB and is directed to the East
(remember q<0)
So F = ma = mv2/r gives us that
r = mv/eB = 0.0011 m or diameter = 0.0022 m and the electron goes in a CW
circle
b. What is the maximum accuracy with which the position of the electron can be
determined?
Since xp ≥h, we have that x≥h/p = h/mv. The max accuracy occurs when
the equality sign is used (minimum x) and when the maximum v is used. If we
take the max v as v itself, then we find x minimum = 7.3 x 10-11 m = 0.73 x
10-10 m = 0.73 Angstroms or just about the size of an atom
Part II Multiple-Choice – Circle the best answer to the following questions. Each
question is worth 2 points (20 points total).
1 Suppose that a charged particle of charge q and mass m moves in a magnetic field of
strength B with a velocity v that is perpendicular to the magnetic field. If particles with
twice the mass (with the same charge q traveling in the same magnetic field with the
same velocity) are used, what will happen to the trajectory of the more massive particles?
a) The two trajectories will be the same.
b) The more massive particles will follow an orbit with a radius twice as large as
the smaller mass particles.
c) The more massive particles will follow an orbit with a radius half as large as
the smaller mass particles.
d) There will be no way to determine the orbital radii of the particles.
2 Suppose that two wires are carrying currents I in opposite directions. The magnitude
of the magnetic field created at the center of the two wires separated by a distance d is
 I
a) 2 0 .
d
 I
b) 0 .
2d
 I
c) 0 .
4d
d) zero.
3 If it takes 1 second for blue light ( = 450 nm) to travel straight through a block of
glass (nglass = 1.5) and it takes 0.87 seconds for the same blue light to travel through a
block of ice of the same thickness as the glass, what is the index of refraction of the ice?
a) 1.64
b) 1.53
c) 1.42
d) 1.31
4. Suppose that polarized light with intensity I0, has its axis of polarization oriented at
45o with respect to the vertical. If this light is incident on a polarizer whose transmission
axis is oriented at 65o with respect to the vertical (and to the same side of the vertical as
the axis of polarization), what percent of the incident light is transmitted?
a) 0%
b) 12%
c) 18%
d) 88%
5. If a particle with charge 3e and mass m is accelerated from rest through a region of
space where there is a difference in potential of 100V, what will the particle’s
approximate speed be when it leaves the accelerating region?
e
a) 25 .
m
e
b) 25
.
m
m
c) 0.04 .
e
e2
d) 625 2 .
m
6. The total number of different quantum mechanical states available in the n = 3 shell of
an atom is
a. 18
b. 16
c. 10
d. 24
e. none of these
7. The number of photons in a beam of light is proportional to
a. E, the electric field
b. , the wavelength
c. I, the intensity
d. f, the frequency
e. p, the momentum
8. The Rayleigh criterion refers to
a. scattering of light
b. polarization of light
c. diffraction of light
d. refraction of light
e. aberration of light
9. Contrast in a microscope refers to
a. forming a clear image of two nearby points
b. forming a very large image
c. distinguishing objects of different colors
d. seeing brightness variations at edges
e. forming a very bright image
10. An electron in an excited state of an atom has a lifetime of 1 ns. The uncertainty
in the energy of the emitted photon when it makes a transition out of this state (in eV) is
f. 6.6 x 10-7
g. 4.1 x 10-6
h. 1.1 x 10-25
i. 6.6 x 10-25
j. none of these
11. Which of the following is a false statement about conductors: (a) just outside, the
electric field is perpendicular to the surface, (b) at equilibrium, the net charge inside is
zero, (c) a charge located within a hole inside the conductor feels no force from charges
outside the conductor, (d) the electric field inside is always zero.