NAME
BLAZER ID
Write your name also in the back of the last page.
PH202-U4 Test 2 (July. 19, 2012, 3:00PM-5:00PM)
• You may not open the textbook nor notebook.
• A letter size information sheet may be used.
• A calculator may be used. However, mathematics or physics formula programmed in a calculator may
not be used.
• Solve seven problems out of eight. Clearly cross out one problem you do not want to be graded. If no
problems is crossed out, the first seven problems that are not crossed out are graded even if the answer
is empty.
• Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary.
• Answers without reasonable explanation nor convincing mathematical derivation will receive no point
even if your answers coincide with the correct ones. Substantial partial credit may be given to correct
reasoning and mathematical procedures even when your final answers are wrong. However, if your
final answers are too obviously wrong, a partial credit may not be given.
Important Physical Constants:
elementary charge: e = 1.602176487 × 10−19 C
electron’s mass: me = 9.10938215 × 10−31 kg
proton’s mass: mp = 1.674927211 × 10−27 kg
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permittivity of free space: 0 = 4πk
= 8.854187817 × 10−12 C 2 /(N · m2 ) where
k ≈ 8.99 × 109 N · m2 /C 2
permeability of free space: µ0 = 4π × 10−7 T · m/A
acceleration due to gravity at the surface of the earth: g = 9.80 m/s2
speed of flight: c = 2.99792458 × 108 m/s
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Raw
Total
1. [10 pts.] In vacuum, an electromagnetic wave has speed c = 2.99792458 × 108 m/s and frequency
f = 6.00 × 1014 Hz. It enters a material whose index of refraction is n = 1.40.
(a) Find the speed of the wave in the material.
(b) Find the frequency of the wave in the material.
(c) Find the wavelength of the wave in the material.
(d) Human can see electromagnetic wave in the range of wavelengths from 390 nm to 750 nm in
vacuum. Is the electromagnetic wave of frequency 8.50 × 1014 Hz in vacuum visible light, infra red
or ultraviolet?
(a) c0 =
c
3.00 × 108
=
= 2.14 × 108 m/s .
n
1.40
(b) Frequency does not depend on the index of refraction. 6.00 × 1014 Hz .
(c) The wavelength in the material is λ =
c0
2.14 × 108
=
= 3.57 × 10−7 m = 357 nm .
f
6.00 × 101 4
3.00 × 108
c
=
= 3.53 × 10−7 = 353 nm. Hence, it is outside
f
8.50 × 1014
the visible range. Since it is shorter than the wave length of visible light, it is ultraviolet .
(d) The wavelength in vacuum is λ =
2
3.0
2.0
1.0
I(t) [A]
2. [10 pts.] When an alternating voltage is applied across a resister whose resistance is R = 40Ω, an alternating current flows
through the resister. Measurement shows that the instantaneous
current I(t) in the resistor oscillates as plotted in Figure. Find
the amount of energy dissipated in the resister during one hour of
operation.
0.0
-1.0
-2.0
-3.0
0.0
0.5
1.0
time [s]
1.5
The graph indicates that the amplitude of the current is I0 = 2.0 A. Hence its root mean square is
I0
Irms = √ . The average power is
2
I0 2
I 2R
(2.0)2 × 40
2
P̄ = Irms Vrms = Irms R = √
R= 0 =
= 80 W
2
2
2
Finally, we obtain the energy dissipated in the resister:
W = P t = 80 ∗ 3600 = 288000 J = 288 kJ
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2.0
3. [10 pts.] Two plane mirrors are separated by 120◦ and an object (arrow) is placed as Figure illustrates.
A person sees the image of the arrow through tye second mirror (M2). A ray of light emitted from the
arrow head and reflected to the eye of the observer is shown in Figure.
(a) In Figure, construct the image the person sees. Leave rays and other lines used to construct the image
in Figure so that the instructor knows how you find the image.
(b) If the ray strikes mirror M1 at an angle of incidence θ = 65◦ , at what angle φ does it leave mirror
M2?
(a) See the drawing.
(b) Consider angles α and β defined in Figure. Using the law of reflection, α = 90◦ − θ and
β = 90◦ − φ. Since the sum of inner angles in a triangle is 180◦ , α + β + 120◦ = 180◦ . Hence,
α + β = 60◦ . Eliminating α and β, we find 90◦ − θ + 90◦ − φ = 60◦ , which leads to
φ = 120◦ − θ = 102◦ − 65◦ = 55◦





M2

object
120o
M1
image1
image2
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4. [10 pts.] A boy is standing 2.00 m (x2 ) from the edge of a 1.20 m (d) deep pond and trying to see fish in
it as Figure illustrates. One fish (left one) is behind a rock and the other (right one) is 1.00 m (x1 ) from the
edge. Both are resting at the bottom. Assume that the index of refraction of water is n = 1.333.
(a) Assuming that the boy’s eye is 1.60 m (h) above the ground, does he see the right fish? Justify your
answer. You must show your mathematical work. (Points are given only for collect reasoning and
calculation.)
(b) The left fish cannot see the right fish directly due to the presence of the rock. However, the left fish is
able to see the image of the right fish above the water surface. At least how far are they separated?
Find the smallest possible distance (x3 ) between them.
(a) Applying Snell’s law to θ1 and θ2 defined in Figure,
x 2.00
2
θ2 = arctan
= arctan
= 0.896 rad.
h
1.60
sin θ1 =
sinθ2
sin(0.896)
=
= 0.586 → θ1 = arcsin(0.586) = 0.626 rad
n
1.333
x1 = tan θ1 d = tan(0.626) × 1.2 = 0.867 m
Since this is smaller than 1 m, the fish is visible to the boy .
(b) Total internal reflection happens only when the incidence angle is larger than a critical angle given by
θc = arcsin(1/n) = arcsin(1/1.333) = 0.848 rad
The shortest distance between the fish is
x2 = 2d tan θc = 2 × 1.2 × tan(0.848) = 2.72 m .
h
image

c c
d

x3
x1
5
x2
5. [10 pts.] An observer sees an image of a toy car through a spherical mirror. The car is moving along the
axis of the mirror. Initially the image is inverted. When the car is 30 cm from the mirror, the image
disappears. After that the image becomes upright.
(a) Is the mirror concave or convex? Justify your answer briefly with a few sentences.
(b) Is the car moving away from the mirror or toward the mirror. Justify your answer briefly with a few
sentences.
(c) When the car is 20 cm in front of the mirror, where is the image formed? Find the image distance
including its sign.
(d) When the car is 20 cm in front of the mirror, how big is the car compared to the original size? Find
the magnification.
(a) Convex mirror forms only a upright virtual image. Since the image changes from inverted to upright,
the mirror must be a concave mirror.
(b) When an object is between the focal point and the mirror, the image is virtual and upright. On the
other hand, when the object is outside the focal point, the image is real and inverted. Hence, the car
moves toward the mirror .
(c) When the object is located at the focal point, the image disappears. Since the image of the car
disappears at 30 cm from the mirror, the focal length is f = 30 cm. Using the thin lens equation,
1
1
1
1
1
1
= −
=
−
=−
di
f
do
30 20
60
(d)
m=−
→
di
−60
=−
=3
do
20
Hence, the image is 3 times bigger than the car.
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di = −60 cm
6. [10 pts.] An object (an arrow in Figure) is placed do = 15.0 cm from a lens. An upright image is formed
at di as shown in Figure. The image is twice as big as the object.
(a) Is the image real or virtual? Justify your answer briefly with a few sentences.
(b) Find the image distance di . An appropriate sign, + or −, must be included in the distance.
(c) Find the focal length of the lens.
(d) Determine the focal points by drawing rays in Figure. Note that there are two focal points. Leave
rays and other lines in the figure so that the instructor knows how you find the focal points.
(a) Virtual . Drawing shows that the actual light is not coming from the image. Hence, it is virtual.
(b) Since the image is upright and twice bigger than the object, its mangification is m = 2.
di = −mdo = −2 × 15 = −30 cm
(c)
1
1
1
1
1
1
=
+
=
−
=
f
do di
15 30
30
Hence, f = 30 cm .
(d) See drawing. Note that one of the focal point coincides with the location of the image since
|di | = f = 30 cm.
image
object
F
F
do
di
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7. [10 pts.] A soap film is surrounded on both sides by
air as shown in Figure. The soap film has an index of
refraction n = 1.333. When sunlight strikes the film
nearly perpendicularly, a color corresponding to wavelength 665 nm in vacuum look bright in the reflected
light due to constructive interference.
(a) Find the smallest possible thickness of the soap
film.
(b) Assuming that the soap film has thickness obtained in part (a), no other visible light appears
bright. However, it is possible that the intensity of other electromagnetic wave is enhanced
by constructive interfered. Find which of ultraviolet waves has significantly stronger intensity,
λ1 = 444 nm or λ2 = 222 nm (λ1 and λ2 are
wave lengths in vacuum.) Show your mathematical work to justify your answer.
(a)
2t +
λfilm
= mλfilm
2
m = 1, 2, 3, · · ·
Hence,
λfilm
t=
2
1
m−
2
For the smallest thickness, m = 1. Substituting λfilm =
t=
λair
,
n
λair
665
=
= 125 nm
4n
4 × 1.333
(b) The next possible constructive interference occurs at m = 2.
λfilm =
2t
m−
1
2
=
2 × 125
= 166.7 nm
3/2
The corresponding wavelength in air is
λair = nλfilm = 1.333 × 166.7 = 222 nm
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8. [10 pts.] In a Young’s double-slit experiment shown
in Figure, one of the slits is covered by a transparent
glass whose index of refraction is 1.46. When the double
slit is illuminated by monochromatic light whose wave
length in vacuum is 586 nm, the center of the screen
remains bright. What is the minimum thickness of the
glass?
d
d
d
nd
and φglass =
.
=
λvacuum
λglass
λvacuum
= m where m is a positive integer. For
The number of wavelength fits in the length d is given by φvacuum =
Since the interference patterns remain unchanged, φglass − φvacuum
minimum possible thickness, m = 1.
nd
d
−
=1
λair λair
d=
→
d
(n − 1) = 1
λair
λair
586
=
= 1270 nm
n−1
1.46 − 1
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NAME
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