Download 1 Practice Exam B

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1.1
Practice Exam B
Polarizers
A clear sheet of polaroid is placed on top of a similar sheet so that their polarizing
axes make an angle of 60 degrees with each other. The ratio of the intensity of
emerging light to the incident unpolarized light is:
Solution: Remember Malus’ law:
I = I0 cos2 θ.
(1)
We have two polarizers in this problem, so we need to apply Malus’s law first. Since
the incident light is unpolarized, it is reduced by 50% as it goes through the first
polarizer. Remember that this is because we integrate over θ in Malus’s law since
unpolarized light is basically light polarized at all possible angles. Now, the light
from this polarizer goes through a second polarizer at 60 degrees. Since cos 60 = 12 ,
this then reduces the intensity down 25% more since we have to square this one-half.
So, putting all these pieces together, we have light that is reduced by .5 × .25 = 1/8.
1.2
Acceleration of a ball due to light beam
In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8.0 mW/m2 . The sphere is totally
absorbing and has a radius of 1.0 microns and a uniform density of 4500.0 kg/m3 .
What is the magnitude of the sphere’s acceleration (in m/s2 ) due to the light?
Solution: This problem deals with the pressure that light exerts on objects. This
is too small of a pressure for us to feel normally, as we’ll see in this problem, but it
is there!
In your book you should have some formulas for light intensity pressure to use here we’ll just figure this out based on units. Remember Newton’s second law:
F = ma,
(2)
so to find a we need the force that the light exerts. We can get the mass of the sphere
by multiplying its volume by its density. What about the force? Well, think about
what might affect the force. The brightness of the beam (intensity) definitely will.
What else? It would make sense that the cross-sectional area of the thing being hit
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by intensity should matter, so probably the cross-sectional area of this square will
factor in. Notice that if we multiply intensity by area, that gives us units of power.
Power divided by speed gives us force. What speed constant comes up when we talk
about light? The speed of light! So it makes sense that we have:
a=
1.3
I · πr2 /c
3 I
F
= 4 3 =
= 4.45 × 10−9 .
m
4c
rρ
πr
ρ
3
(3)
Displacement Current
A sinusoidally-varying voltage V (t) = V0 sin(2πf t) with amplitude V0 = 10 V and
frequency f = 100 Hz is impressed across the plates of a circular-shaped parallel
plate air-gap capacitor of radius a = 1.0 cm and plate separation d = 0.01 mm. The
amplitude of Maxwell’s displacement current ID flowing across the gap between the
plates of this capacitor is:
Solution: As we discussed in recitation, the displacement current can be found
in two ways. One way is to use the flux formula we derived:
ID = 0
dΦE
.
dt
(4)
Let’s try it this way first. The electric flux is
~ ·A
~ = EA = V (t) πa2 ,
ΦE = E
d
(5)
where I’m using the approximation that the electric field inside of a parallel plate
capacitor points in straight lines of constant field from one plate to the other. Using
this form, we can take a derivative to quickly obtain:
Id = 0
2π 2 a2 0 V0 f
V0 2 d
πa
sin(2πf t) =
cos(2πf t).
d
dt
d
(6)
Plugging in numbers gives 1.75 × 10−6 .
The other approach would be to go back to the definition of capacitance, Q = CV ,
and calculate a “current” that way:
ID =
dQ
dV
0 A d
=C
=
V0 sin(2πf t).
dt
dt
d dt
Doing this will get you the right answer as well. Whatever you like best!
(7)
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1.4
Reflection of light from thin film
White light, with uniform intensity across the visible wavelength range of 400 nm 690 nm, is perpendicularly incident on a water film, of index of refraction 1.33 and
thickness 300 nm. At what wavelength (in nanometers) is the light reflected by the
film brightest to an observer?
Solution: Let’s use the thin film equation. For light passing from air to water,
we have the constructive interference condition:
1
λ.
(8)
2t = m −
2
Here I’m assuming that, since we know nothing else about the setup, whatever is
underneath the water has a lower index of refraction than water. This means that
there is one phase shift for the initially reflected ray, but none for the second ray,
hence why we use m − 12 instead of m for a bright spot.
Don’t forget that λ here is the wavelength in the water. If λ0 is its wavelength
in air, then we have
1
λ0 .
(9)
2nt = m −
2
Now just start trying values of m and solving for λ0 to see what the wavelength must
be in order for this condition to be satisfied. I get:
λ0 =
1.5
2nt
798
= 532, m = 2.
1 =
m− 2
m − 12
(10)
Electromagnetic Wave
The next three questions pertain to the following situation: An electromagnetic
plane wave is propagating in empty space. The electric field at time t=0 over two
wavelength is sketched in the figure below. The E-field is given by
~
E(x,
y, z, t) = îE0 sin(kz + ωt).
(11)
where, î is the unit vector in the +x direction, ĵ is the unit vector in the +y direction,
and k̂ is the unit vector in the +z direction.
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(a) Which of the following expressions describes the magnetic field of this electromagnetic wave?
Solution: We know that the light wave is traveling in the negative z direction
since the argument is kz + ωt, so the position displacement is in z and, since
it’s +ωt rather than −ωt, it is moving in the negative z direction. This tells
us that the Poynting vector is also pointing in the negative z direction. Since
~∝E
~ × B,
~ and E
~ is in the +x direction, we get that B
~ must point in the −y
S
direction via the right-hand rule. And so,
~
B(x,
y, z, t) = −ĵB0 sin(kz + ωt).
(12)
(b) If the average intensity of the wave is 1 Watt/m2 , what is the average energy
per unit volume of the wave?
Solution: The average intensity of an EM wave is given by
I = cn0 E 2 .
(13)
The average energy per unit volume (energy density) is
u = 0 E 2
(14)
By comparing these two equations, we see that u is a part of I:
I = cnu =⇒ u =
I
.
cn
(15)
And so, since n = 1 in this case, the average energy density is just I/c =
3.33 × 10−9 .
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(c) If the average intensity of the wave is 1 Watt/m2 , what is the peak value of
the magnetic field, B0 , of the wave?
Solution: Let’s use the equation from part (b):
I = cn0 E 2 = cn0 c2 B 2
(16)
since E = cB in a light wave. And so,
r
B=
I
0 c3
(17)
Plugging in values gives B = 6.47 × 10−8 - not an answer! What went wrong?
Well, remember that the formula we just used is dealing with the RMS value
√
of B. We want the peak value here! That’s just the value we got times 2,
which gives B0 = 9.15 × 10−8 , which works.
1.6
Concave Mirror
How far (in cm) must a man’s face be located in front of a concave spherical make-up
mirror of radius 40 cm for him to see an erect image of his face five times its real size?
Solution: Just use the equations. We know that we need m = 5, which means
m=5=−
S0
=⇒ S 0 = −5S.
S
(18)
Now we can use the lens equation to solve for S:
1
1
1
+ 0 =
S S
f
1
1
2
=⇒
−
=
S 5S
40
=⇒ S = 16.
Draw a good ray tracing diagram to check your work!!!!
1.7
Equivalent Resistance
A diffraction grating 1.00 cm wide has 10,000 parallel slits. Monochromatic light
that is incident normally is diffracted through 30degrees in the first order. What is
the wavelength of the light?
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Solution: Ignore the title. Anyways, we’re told that the first dark spot happens at
θ = 30 (yes, the question wording is confusing. Ask if you’re confused on the exam!!)
We know that, for single-slit diffraction, we get a dark spot whenever
a sin θm = mλ.
(19)
Since the grating is 1 cm wide and has 10,000 slits, each one must be 1/10000 cm
wide. Thus,
1
1
sin 30 = λ =⇒ λ =
sin 30 = 0.00005cm.
10000
10000
(20)
This is 500 nm.
1.8
RLC circuit
A series RLC circuit with L = 25 mH, C = 0.8 µF , and R= 7Ω is driven by a
generator with a maximum emf of 12 V and a variable angular frequency ω.
a) At ω = 8000 rad/sec the impedence of the circuit (in ohms), is...?
Solution: Remember your impedance!
p
Z = R2 + (XL − XC )2 ,
where
XL = ωL,
Now it’s just plug in chug. I get 44.3 Ω.
b) Skip this one. We didn’t talk about it.
XC =
1
.
ωC
(21)
(22)
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1.9
Laser beam through water
A laser sends a beam of light from water toward a plastic slab at the surface of water.
Above the plastic slab is air.
a) What is the maximum value of the angle, θ, that the laser beam can make
with the vertical and still have the beam of light emerge into the air above the
plastic?
Solution: You should recognize this as a total internal reflection problem!
Remember that the threshold final angle for total internal reflection to happen
is 90degrees. So, let’s work backwards from the air/plastic boundary. Applying
snell’s law there gives:
nplastic sin θp = nair sin θa =⇒ sin θp =
1
2
sin(90) = .
1.5
3
(23)
Check out this picture to see what we’re talking about. θa = theta plastic.
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Now we apply Snell’s law at the plastic/water boundary:
nwater sin θ = nplastic sin θp =⇒ sin θ =
1.5 2
= 0.76.
1.33 3
(24)
And so, solving for θ, I get:
θ = sin−1 (0.76) = 49.1.
(25)
I don’t know why I don’t exactly the right answer, but it works.
b) Assume the light beam is shone so that θ = 38degrees. If the plastic slab is
10 cm thick in the vertical direction, how long did it take the beam to travel
through the plastic?
Solution: Let’s apply Snell’s law at the boundary of plastic and water:
1.33
−1
1.33 sin(38) = 1.5 sin θ =⇒ θ = sin
sin(38) = 33.1
(26)
1.5
Now let’s do some trigonometry. We want to find time, which is:
t=
length
.
speed
(27)
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c
, we just need the
Since we know the speed of light in the plastic is v = 1.5
distance, which is the hypotenuse of the triangle formed with angle θ. Thus.
t=
1.10
10cm/ cos θ
0.1m · 1.5
=
= 5.97 × 10−10 s.
c/1.5
c cos(33.1)
(28)
diverging lens
An object (bold arrow) is located at a distance of (3/2) times the focal length in
front of a diverging lens as shown in the Figure, where the focal points are labeled
as f.
a) What is the location, s’, of the image?
Solution: We can just use the lens equation. Don’t forget that, following the
sign conventions, f must be negative for this diverging lens!
1
1
1
+ 0 =
s s
f
1
1
1
=⇒
+ 0 =−
(3/2)f
s
f
1
5
=⇒ 0 = −
s
3f
3
=⇒ s0 = − f.
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Again, ray trace and check your answer!!!!!!!
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b) Choose the correct description of the image in the situation above.
0
Solution: Since we got a negative s0 , it must be virtual. Since m = − ss > 0,
it must be upright.
c) Suppose the object is now placed at the focal point. What is the magnification,
M?
Solution: We can solve like part a:
1
1
1
+ 0 =
s s
f
1
1
1
+ 0 =−
=⇒
f
s
f
1
2
=⇒ 0 = −
s
f
1
=⇒ s0 = − f.
2
Now we can calculate M :
M =−
−1f
s0
1
=− 2 = .
s
f
2
(29)