PRACTICE FINAL 2
Solutions
1) Which one of the phasor diagrams shown below best represents a series LRC circuit driven at
resonance?
A) 1
B) 2
C) 3
D) 4
E) 5
Answer: C
2) In a series LRC circuit, the frequency at which the circuit is at resonance is f0. If you double the
resistance, the inductance, the capacitance, and the voltage amplitude of the ac source, what is the new
resonance frequency?
A) 4 f0
B) 2 f0
C) f0
D) f0/2
E) f0/4
Answer: D
3) An electromagnetic wave is propagating towards the west. At a certain moment the direction of the
magnetic field vector associated with this wave points vertically up. The direction of the electric field
vector of this wave is
A) horizontal and pointing south.
B) vertical and pointing down.
C) horizontal and pointing north.
D) vertical and pointing up.
E) horizontal and pointing east.
Answer: A
4) The energy per unit volume in an electromagnetic wave is
A) equally divided between the electric and magnetic fields.
B) mostly in the electric field.
C) mostly in the magnetic field.
D) all in the electric field.
E) all in the magnetic field.
Answer: A
5) A ray of light goes from one transparent material into another, as shown in the figure. What can you
conclude about the indices of refraction of these two materials?
A) n1 ≥ n2
B) n1 > n2
C) n1 = n2
D) n2 ≥ n1
E) n2 > n1
Answer: B
6) A light beam shines through a thin slit and illuminates a distant screen. The central bright fringe on
the screen is 1.00 cm wide, as measured between the dark fringes that border it on either side. Which of
the following actions would decrease the width of the central bright fringe? (There may be more than
one correct choice.)
A) increase the wavelength of the light
B) decrease the wavelength of the light
C) increase the width of the slit
D) decrease the width of the slit
E) put the apparatus all under water
Answer: B, C, E
7) A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of
this lens from a diffraction standpoint is
A) the same for all wavelengths.
B) in the near-ultraviolet.
C) in the visible.
D) in the near-infrared.
E) indeterminate.
Answer: B
8) Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at
an angle of 6.5° away from the central fringe. What is the separation between the double slits?
A) 5.0 µm
B) 10 µm
C) 15 µm
D) 20 µm
E) 25 µm
Answer: C
9) Which of the following statements are true concerning the reflection of light?
A) The reflection of light from a smooth surface is called specular reflection.
B) The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane
mirror.
C) For specular reflection, the angle of incidence is less than the angle of reflection.
D) The reflection of light from a rough surface is called diffuse reflection.
E) For diffuse reflection, the angle of incidence is greater than the angle of reflection.
Answer: A, D
10) A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam
experience a phase change, and if so, by how much is the phase of the beam changed?
A) The reflected beam does not experience a phase change.
B) The reflected beam experiences a 45° phase change.
C) The reflected beam experiences a 90° phase change.
D) The reflected beam experiences a 180° phase change.
Answer: D
2
Circuits
To the left is a diagram of a series RLC circuit. It is powered by an AC generator operating at a frequency of 2 Hz, with max output
voltage of 10 V. The resistor has R = 3 Ω, the
capacitor has C = 0.25 F, and the inductor has
L = 3H.
1. What is the total impedence for this circuit?
�
The equation for impdence Z is Z = R2 + (XL − XC )2 with XC = 1/ωC and XL = ωL.
The angular frequency ω = 2πf , where f is the generator frequency. Plugging in values we obtain
Z = 37.5 Ω.
2. What is the phase angle for this circuit?
�
�
The phase angle is given by φ = tan−1 (XL − XC )/R = 85.4◦ .
3. Draw the phasor diagram for the above circuit, where the dashed arrow represents I max .
4. What is the magnitude of Irms ?
Irms =
I√
max
2
√
so we need Imax , but we know that Vmax = Imax Z so that Irms = Vmax /(Z 2) = 189 mA.
2
A Fish’s View
There is a phenomenon called Snell’s Window which determines how fish (and divers) see
the world outside of their watery enclosure. Basically, if you are under water and you look out
into air you see the entire 180◦ world above you
compressed into a circle subtending an angle
which is less less than 100◦ . Everything outside
of this circle you see comes from under water (it
is bouncing off the surface via total internal reflection.) The index of refraction of water is 1.33.
1. What is the exact angle subtended by Snell’s window for someone in water looking into air?
The edge of Snell’s window is the point where
total internal reflection occurs, i.e. where the angle at which the light strikes the surface is at the
critical angle from the normal, so that φ = θc =
sin−1 (n2 /n1 ) where n1 is the index of refraction
of water and n2 is that of air. The total angle subtended is 2φ = 97.5◦
2. Would Snell’s window be larger or smaller if the substance above the water was not air, but
some gas with an index of refraction n = 1.2? What if (in addition to the gas) one was not in
water, but encased in glass (index of refraction n = 1.5)?
In the first case set n2 = 1.2, which increases the critical angle and hence makes the window larger.
In the second case also set n1 = 1.5 making the ratio n1 /n2 = .8 which again gives a larger window.
3. The index of refraction of water actually depends slightly on the frequency, with it having a
higher index for higher frequency light. Given this, describe how the colors one would see
would be different from normal when looking through Snell’s window.
The images will be blurred out, with each of the colors from any image appearing slightly displaced
relative to each other. Recalling Snell’s law, we know that higher index of refraction implies more
bending, so that higher frequency light will be closer to the center. This means that objects will be
bluer on their portion closest to the center of the window, and redder on their portion nearer the
outside.
4. If one were under water looking toward Snell’s window, at what angle (from the normal)
could all the light be blocked by wearing a pair of suitibly polarized sunglasses?
At Brewster’s angle reflected light is totally polarized and hence could be completely blocked by
polarized sunglasses. This angle is θB = tan−1 (n2 /n1 ) = 37◦ , where n2 is the index for the reflecting
material. Since this angle is less than θc not only reflected light comes from this angle (also refracted
light from above), and hence all light from this angle is not polarized and cannot be blocked, so there
is NO angle for which polarized sunglasses can block all the light.
Quiz 3: Solutions
PHYS 108 (Summer 2009)
July 22, 2009
Constants: c = 3.00 × 108 m/s; �0 = 8.85 × 10−12 C2 /(N·m2 ) ; µ0 = 4π × 10−7 T·m/A
1
Solar Sails
One method of propulsion that has been
proposed for interplanetary travel is solar sailing. The idea is to have a large sail that can
collect light from the sun or a laser on Earth,
and use the light pressure to move about the solar system. Assume that a spaceship called the
Dragonfly has a mass of 10,000 kg and a solar
sail with an area of 10 km2 . The energy density
of solar radiation at Earth’s orbit is 45.5 µJ/m3 .
1. What is the maximum acceleration the sun’s light can give to the Dragonfly while it is the
same distance from the sun as the Earth?
To get acceleration we need force (F = ma, with mass m) and to get force we need pressure (F = P A,
with area A). The pressure of EM radiation is given by P = I/c with intensity I. However intensity
is given by I = uc where u is the energy density. Plugging this all together we get
a = F/m = P A/m = IA/mc = ucA/mc = uA/m = .0455 m/s2
2. Assume that a laser with an output of 1.21 Gigawatts operating at a frequency of ν = 3×1018
Hz is used to propel the Dragonfly. What accleration could this laser impart to the craft?
1
)Δp/Δt),
Recall that force (ma) is change in momentum per unit time (ma = Δp/Δt, and so a = ( m
and also that an EM wave delivering energy U transfers momentum p = U/c, so that Δp = ΔU/c.
Also power is P = ΔU/Δt, and so Δp/Δt = ( 1c )ΔU/Δt = P/c. This gives a = P/mc =
0.403 mm/s2 .
3. Assume the Dragonfly is moving directly away from Earth at a velocity of 8000 km/s (about
100 times the speed of Voyager I). What frequency should the pilot turn his radio to in order
to listen to Q101 (which broadcasts at 101.1 MHz)?
An observer moving with velocity u away from a source emitting radiation with frequency ν observes
frequency
ν � = ν(1 − u/c) = 98.4 MHz
Quiz 5
Waves and Interference
Phys 106 — Fall 2011
SOLUTIONS
Constants: c = 3.0 × 108 m/s, (velocity of sound) vs = 343.0 m/s.
1
New Stealth Technology (5 pts)
Using the same principle as thin film reflection, attempts are being made to make airplanes invisible to radar. The idea is to coat
the plane with a material that will cause the
reflected waves to interfere destructively. You
may take the index of refraction of air to be 1.0,
and assume that n1 , the index of refraction of
the coating, satisfies n1 > 1.0, and also that the
coating does not have zero thickness. In the following, explain your reasoning to get full credit.
a) Is there a phase change when the incident light strikes the outer surface? (1 pt)
Yes. Whenever light moves from a substance to another substance with a higher index of refraction there
is a phase change.
b) Assuming that the designers have made the coating to be as thin as possible and still achieve
destructive interference, is there a phase change when the incident radar strikes the inner surface? (2 pts)
Yes. The condition for destructive interference with no reflection is given by the equation from your book
— 2n1 t/λ = m, where m = 0, 1, 2, . . . and t is the thickness of the coating. This is at a (nonzero)
minimum for m = 1 in which case d = λ/2n1 . If there is an additional phase change at the inner
surface, the effective path length increases by 1/2 (in units of wavelength), and so the condition for
destructive interference gets modified to 2n1 t/λ + 1/2 = m, again with m = 0, 1, 2, . . ., which is at a
(positive) minimum for m = 1. In this case we have d = λ/4n1 . Hence, the coating is as thin as possible
when there is a phase change at the inner surface.
c) Assuming that n1 = 2.0, n2 = 1.5, and that the radar waves have a wavelength of 3.0 m
and travel at the speed of light, what is the thickness of the coating (assuming it is as thin as
possible)? (2 pts)
This is the standard setup for thin films, and so we can simply solve the equation 2n1 t/λ = m for t so
that t = mλ/2n1 . This is at a (non-zero) minimum for m = 1 and so t = λ/2n1 = (3.0 m)/(2×2.0) =
.75 m.