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Physics 102-1 Spring 2009 Final Exam
Solution
Grading note: There are 11 problems on 11 pages. Point values are given with
each problem. They add up to 100 points. In multi-part problems, points are not
necessarily evenly divided between the parts.
It was possible to get 3 “extra points” on this exam, beyond the nominal 100 points.
These extra points will not be factored into the initial calculation of your overall course
grade. They will be considered (along with lab performance and class participation) in
setting your final grade. See the solution for problem 6 part c for further explanation.
10
cm
10
cm
laser
5m
1. [5 points] Light from a green laser pointer, λ = 532 nm, shines onto a pair of
slits, producing a pattern of bright spots on a wall 5 m away. A bright spot
is seem immediately across from the pair of slits, and the next nearest bright
spots are 10 cm away from this central point. What is the distance between the
two slits?
Note: y = 10 cm = 0.10 m.
λ
y=n L
d
(1) 532 × 10−9 m (5 m)
nλL
d=
=
= 2.7 × 10−5 m = 27 µm
y
0.10 m
1
2. [5 points] For an MRI imaging apparatus, it is desired to create a magnetic field
of 1 T within a solenoid of length 2.00 m and radius 0.20 m, with 1000 coils per
centimeter over the length of the solenoid. What current should run through
the solenoid?
With a length of
2.0 m=200 cm, the 5total number of coils is
N = 1000 coils
cm (200 cm) = 2.0 × 10 .
B=
µ0 IN
L
⇒
I=
BL
(1)(2.0)
=
= 8.0 A
µ0 N
(4π × 10−7 ) (2.0 × 105 )
3. [5 points] Some airport immigration stations use thermal imaging cameras to
detect incoming passengers who have fevers. The idea is to test and isolate
anyone who carrying a highly contagious disease. In practice, such detection
may be of limited use in containing an epidemic, such as the present-day H1N1
scare, because carriers of the disease are highly contagious well before they show
any symptoms of the disease.
A typical healthy face is 34◦ C=307 K. Thermal imaging systems are capable of
distinguishing feverish people whose faces are 3 degrees warmer than this.
What is the ratio of the power emitted by a feverish face to the power emitted
by a healthy face?
The Kelvin temperatures are 310 K (feverish) and 307 K (healthy). The
areas and efficiencies are the same for both cases. Thus
e1 A1 σT14
P1
=
=
P2
e2 A2 σT24
T1
T2
4
=
2
310 K
307 K
4
= 1.040
4. [10 points]
(a) The Earth’s distance from the Sun is 1.5 × 1011 m.
Imagine that the Earth is suddenly removed from our own solar system
and put in orbit around a star which has twice the power output of our
own Sun (but with the same color spectrum). At what distance from that
star should the Earth be placed in order to be at the same temperature
that it has in its present orbit around the Sun?
The temperature of the Earth is determined by the balance between power
falling onto the Earth (sunlight) and power emitted by the Earth (thermal
radiation). If the Earth’s temperature is to be the same, then its output
power must be the same, which means the input power must be the same,
which in turn means the intensity of the incoming starlight must be the
2
same (since Pin = (I)(πr⊕
), where r⊕ , the radius of the Earth, does not
change).
r
Pstar
Pstar
I=
⇒
r=
2
4πr
4πI
√
If Pstar is doubled and I is the same, then distance r increases by 2 from
its previous value, going from 1.5 × 1011 m to 2.1 × 1011 m.
(b) If that star has twice the power output of our own Sun but the same color
spectrum, what would have to be different about that star (compared to
our Sun), and what would have to be the same?
The color spectrum of a thermal emitter (such as a star) depends on the
temperature of the object. Thus if the color spectrum is the same, the
temperature is the same.
Pstar
eσT 4
Since e, T , and σ do not change, doubling Pstar must
double the surface area. The surface area is A = 4πr∗2 , where r∗ is the
p
√
radius of the star, so r∗ = A/4π, and the radius must increase by 2.
Pstar = eAσT 4
Either answer is acceptable.
3
⇒
A=
5. [10 points] A 1 µF capacitor was charged up using a 5 V power source. The
power source was then disconnected. Later on, at time t = 0, a 10 kΩ resistor
was connected to the capacitor, and the capacitor began to discharge through
the resistor. What was the capacitor voltage at time t = 5 ms? What was the
current through the resistor at this time? What was the charge on the capacitor
at this time?
The time constant is
τ = RC = (10 000) 1 × 10−6 = 0.01 s = 10 ms.
The capacitor voltage, capacitor charge, and current all decay away with
this time constant.
The initial voltage is V0 = 5 V.
The initial charge is Q0 = CV0 = 1 × 10−6 (5) = 5 × 10−6 C = 5 µC.
The initial current is I0 =
V0
R
=
5
10000
= 5 × 10−4 A = 0.50mA.
V = V0 e−t/τ = (5 V)e−5/10 = (5 V)e−0.5 = (5 V)(0.606) = 3.0 V
I = I0 e−t/τ = (0.50 mA)e−5/10 = (0.50 mA)e−0.5 = (0.50 mA)(0.606) = 0.30 mA
Q = Q0 e−t/τ = (5 µC)e−5/10 = (5 µC)e−0.5 = (5 µC)(0.606) = 3.0 µC
4
6. [10 points] Rear-view mirrors on the passenger sides of cars are convex
(diverging) mirrors. They have the words “objects in mirror are closer than
they appear” written on them.
You are riding in a car. You are 1.0 m from the passenger-side rear-view mirror.
The mirror has focal length 0.50 m.
(a) There is a truck 10 m behind the car. (More specifically, the distance from
the truck to the mirror is 10 m.) The truck is 3 m high. You see an image
of the truck in the mirror. Is the object (the truck) actually closer than
the image of the truck, as the words on the mirror seem to imply, or is the
image closer than the object?
Since it is a diverging mirror, the focal length is negative, f = −0.50 m.
The object distance is s = 10 m.
1
1
1
+ 0 =
s s
f1
⇒
1
1
1
+ 0 =
10 s
−0.5
⇒
s0 = −0.476 m
Since s0 < s, the image is closer than the object.
Note: because this problem did not ask for a quantitative justification of
the answer, papers which correctly stated that the image was closer than
the object were given full credit whether or not the value of s0 was properly
calculated. (Of course, an incorrect s0 usually lead to an incorrect answer
for the next part of the problem.)
(b) What is the height of the image?
s0
h0
=−
h
s
⇒
s0
−0.476
h0 = − h = −
(3) = 0.143 m
s
10
(c) Consider what you see as you view the image of the truck, and as you
view the truck itself. Is the angular size of the image of the truck smaller,
larger, or the same size as the angular size of the truck itself? (Another
way to ask this: is the relative angular magnification less than one, greater
than one, or equal to one?)
Note: as you might conclude from this problem, it is angular size, not image
distance, which is responsible for the warning message on the mirror.
The answer to this problem is on the following page.
5
Answer to problem 6 part c.
Image
Mirror
Viewer
Object
θo
1m
3m
10m−1m=9m
0.146m
θi
0.476 m
1m
1+0.476=1.476m
The above figures are not to scale. They show the angles subtended by the object and
the image as viewed by a rider in the car 1 m from the mirror. (The car itself is not
shown.)
There is some latitude in interpreting the distance from viewer to the object (truck). If
the viewer, object, and mirror do not all lie on a line, the viewer-object distance could
be more than 9 m. In contrast, the viewer, image, and mirror always lie on a line, so
the viewer-image distance must be 1.476 m.
The angular size of the truck is θo = tan−1 39 = 18.4◦ .
0.146
The angular size of the image of the truck is θi = tan−1 1.476
= 5.6◦ .
The angular size of the image is smaller than the angular size of the truck itself.
This is the real reason the truck “appears smaller.”
Note: because this problem did not ask for a quantitative justification of the answer,
papers which correctly stated that the angular size of the image was smaller than the
object were given the correct answer whether or not the relevant angles were calculated.
However, I felt that some credit should be given for correct calculation of the angles.
Therefore papers were awarded 3 “extra points” if the angles were calculated correctly.
These “extra points” have been recorded separately from the exam grade and will only
be considered after the rest of the course grading is complete.
6
7. [10 points] A light bulb blew out in the dining room of my house a while back.
Normally, a bulb just goes dead, but this time something funny happened.
About half of the light bulb filament broke off and fell to the bottom of the
bulb. One end of the remaining half of the filament remained attached to one
of the leads through which the bulb gets power. The other end of the remaining
half fell, by chance, on the other lead. The light bulb continued to make a
complete circuit, but the filament was, in effect, only half its usual length.
Assume the filament is the only resistive part of the light bulb. The bulb was
plugged in to a standard 120 V socket.
Before:
After:
(a) How does the resistance of the bulb with the half-length filament compare
to the resistance of the bulb when it had a full-length filament? (Be
quantitative, e.g., an answer might be “its resistance was 3 times higher”.)
R=ρ
L
A
Since L is half its original value, R is half its original value.
(b) How does the brightness of the bulb with the half-length filament compare
to the brightness of the bulb when it had a full-length filament? (Again,
be quantitative)
V2
R
The voltage is the same, 120 V, but R has half its original value. Thus P
is doubled, and the light is twice as bright.
P = IV =
7
8. [15 points] An infinitely long, straight wire carries an upward current, I1 , which
steadily increases, going from 0 to 6 A in 0.2 seconds. A coil is 0.1 m away from
the wire. The coil consists of 1000 turns with radius 0.01 m. It is oriented as
shown, with the coils parallel to the plane of the paper. The ends of the coil
are attached to a resistor of resistance R = 1Ω.
I1
R
0.1 m
(a) A current flows through the coil. What direction does it flow, clockwise or
counterclockwise?
By the right hand rule, the current I1 produces a magnetic field into the
page at the location of the coil. Since I1 is increasing, the inward flux
through the coil is increasing. By Lenz’s law, the current induced in the
coil flows so that the field it generates within the coil opposes the externally
imposed change. Thus the coil current must flow counterclockwise in order
to produce a magnetic field pointing out of the page, in opposition to the
increasing inward field.
(b) How much current flows through the coil?
Note: Because the size of the coil is much smaller than its distance from
the wire, you may assume that, at any moment, the magnetic field is the
same across the entire area of the coil, and that it is equal to the magnetic
field at the center.
Let d be the distance from the wire to the coil and r be the radius of the
coil. The area of the coil is A = πr2 . We will ignore signs.
B=
µ0 I1
2πd
µ0 I1 πr2
µ0 I1 r2
=
2πd
2d
µ0 I1 r 2
∆ 2d
∆Φ
N µ0 r2 ∆I1
E =N
=N
=
∆t
∆t
2d∆t
E
N µ0 r2 ∆I1
(1000)(4π × 10−7 )(0.01)2 (6)
I=
=
=
= 1.9 × 10−5 A = 19 µA
R
2dR∆t
2(0.1)(1)(0.2)
ΦB = BA =
8
9. [10 points] For each of the three resistors in the circuit shown below, give the
voltage difference across the resistor and the current through the resistor.
100Ω
300Ω
45V
150Ω
100Ω
300Ω
45V
75Ω
45V
45V
150Ω
225Ω
150Ω
The 100 Ω and 300 Ω resistors parallel to one another have equivalent
1
1
resistance 100
+ 300
= R1e ⇒ Re = 75 Ω.
The 75 Ω equivalent resistance is in series with the 150 Ω resistor for a total
resistance of 225 Ω.
A 225 Ω resistor powered by a 45 V power source draws current I = V /R =
45/225 = 0.20 A.
All 0.20 A of current flows through the 150 Ω resistor, so it has voltage
V = IR = (0.2)(150) = 30 V.
Since there is 30 V across the 150 Ω resistor, the remaining 45-30=15 V must
be across the parallel combination of 100 Ω and 300 Ω. Since there is 15 V
across each of these resistors, they draw currents I = V /R = 15/100 =
0.15 A and I = V /R = 15/300 = 0.05 A, respectively.
Summary:
Resistor
100 Ω
300 Ω
150 Ω
Voltage
15 V
15 V
30 V
9
Current
0.15 A
0.05 A
0.20 A
10. [10 points] A cathode ray tube (CRT) is an old-style television set. A beam
of electrons is emitted in the back of the television set and it hits the front of
the set. The electrons are absorbed by phosphor compounds which glow when
struck by electrons; this light forms the television picture.
You have a CRT in a laboratory, oriented horizontally. The controls of the
CRT have been adjusted so that, if it were working correctly, the electron beam
should make a single spot of light exactly in the center of the screen (the dot
in the diagram below). You observe, however, that the spot is deflected to the
left when viewed from the front of the screen (the × int the diagram below).
You are told that there are three possible reasons for this:
(a) The CRT is broken (e.g., the beam of electrons might be mis-aimed).
(b) There is a uniform electric field throughout the laboratory.
(c) There is a uniform magnetic field throughout the laboratory.
Assuming that you do not have compasses, magnets, charged objects, etc., how
can use the CRT itself to determine which of these three possibilities is the
cause of the beam being off position? You can move the CRT around (changing
its position or orientation), but you cannot remove it from the room. Describe
how you would figure out which cause is the correct one.
You should ignore gravity in this problem.
beam is supposed follow the dashed path
and hit the front of the CRT here...
...but it is observed to hit here instead
Answer this question on the following page (which is otherwise blank).
10
This space is for answering question 10.
z
y
x
Define coordinates as in the figure. The electron beam is supposed to travel straight in the +x direction
but it has been deflected by a small amount in the −y direction.
The orientations of x, y, and z remain fixed relative to the room even if the CRT is moved or reoriented.
Electrons are negatively charged. If they are deflected in the −y direction due to an electric field, the
field must be in the +y direction. (There could also be a component of the electric field in the +x or
−x direction; a ±x component would have essentially no effect on where the spot landed on the screen.)
If the deflection is due to a magnetic field, we can find the direction using the right hand rule. The
deflection is in the −y direction, which means that ~v × B must be in the +y direction (since the charge
is negative). With ~v in the +x direction, a little playing around with your right hand will convince you
~ must be in the −z direction in order for ~v × B
~ to be in the +y direction. (Actually,
that the field B
~
to be completely accurate, B could also have a component in the ±x direction; such a component
would impose no force on electrons traveling in the +x direction. A ±x component would have to be
in addition to a component in the −z direction.)
So: we need to check for the possibilities of (a) no field at all, (b) an electric field in the +y direction,
and (c) a magnetic field in the −z direction.
Here are experiments we could do. (There are many variations on this answer which are also correct.)
Experiment 1. Turn the CRT around so that the beam goes in the −x direction instead of the +x
direction.
ˆ If the deflection is due to an electric field, it will still be deflected in the −y direction.
ˆ If the deflection is due to a magnetic field or a broken CRT, it will be deflected in the +y
direction.
We have found a way to determine whether it is due to an electric field (“explanation b”). If we
determine that it is not due to an electric field, we must do another experiment.
Note: if there is an electric or magnetic field component in the ±x direction, it would have no effect on
the results of this experiment.
Experiment 2. Turn the CRT vertically, so that its beam shoots upward, in the +z direction.
ˆ If the deflection is due to a magnetic field, it electron motion is now parallel to the field, so
~ = 0 and there is no deflection.
~v × B
ˆ If the deflection is due to a broken CRT, the beam will still be deflected.
We have found a way to distinguish between a magnetic field (“explanation c”) and a broken CRT
(“explanation a”).
Note: if the magnetic field had a component in the ±x direction, we could detect it by rotating the
CRT into different orientations while it is oriented vertically. Deflection of the beam due to a magnetic
field in the ±x direction would retain the same orientation relative to the room as the CRT was rotated;
deflection of the beam due to a broken CRT would change orientation relative to the room as the CRT
was rotated.
11
11. [10 points] Two glass microscope slides sit atop each other as shown in the
diagram. On the left side, the slides are touching each other. The top slide
is at a slight angle from the horizontal, θ = 0.001◦ (see figure). The index of
refraction of the glass is 1.50. If the slides are illuminated form above with light
of wavelength 500 nm, a series of dark and bright bands are seen (when viewed
from above).
Let x be the distance from the left end. How far from the left end is the first
bright band, that is, what is the smallest x for which there is a bright reflection?
Note: Assume that only the inner two surfaces are reflective, i.e., you can ignore
the top surface of the upper slide and the bottom surface of the lower slide.
500 nm light
θ=0.001o
x
The answer to this problem is on the following page.
12
Answer to problem 11.
Path 2
Path 1
d
θ=0.001o
x
First, notice that the angle θ = 0.001 is very small, so the light going into and out of
the glass slides will hardly be refracted at all, and we can neglect refractive effects.
Now consider the figure above, in which angle θ is greatly exaggerated. For a bright
reflection, there must be constructive interference between the two paths shown. The
usual constructive interference condition (which we will have to modify) is that the
difference in path lengths, measured in wavelengths, must be an integer n:
∆L1
∆L2
−
= n.
λ1
λ2
For this problem, path 1 is longer than path 2 by distance 2d. We set ∆L1 = 2d and
∆L2 =0, and hence we drop the ∆L2 term from the equation. The extra length of path
1 is entirely in the air, so use λ = 500 nm. Path 1 reflects off a medium of higher
index of refraction, while path 2 does not, so path 2 has the equivalent of half an extra
wavelength, so we add a “ 21 ” term to the equation, giving us
2d 1
+ = n.
λ
2
(We could equally well have put the
gives
1
2
on the other side of the equation.) Solving for d
d=
n−
1
2
λ
2
Every integer n which gives a positive value of d corresponds to a bright band. The
leftmost band
has the smallest value of d. Find it by plugging in n = 1 and calculating
d = 1 − 21 500
2 = 125 nm.
Finally, do a little trig to find x. It is equally valid to use tangent or sine in the following
calculation; for small θ they give the same answer.
d
125 × 10−9 m
d
= tan θ ⇒ x =
=
= 0.0072 m = 7.2 mm
x
tan 0.001◦
1.745 × 10−5
13