Physics E1bx
April 14 – April 21, 2015
Physics E1bx: Assignment for April 14 – April 21
Homework #8: Waves and Interference
Due Tuesday, April 21, at 6:00PM
This assignment must be turned in by 6:00PM on Tuesday, April 21. Late homework will not be
accepted. Please write your answers to these questions on a separate sheet of paper with your name and
your section TF’s name written at the top. Turn in your homework to the mailbox marked with your
section TF’s name in the row of mailboxes outside of Sci Ctr 108.
You are encouraged to work with your classmates on these assignments, but please write the names of
all your study group members on your homework.
After completing this homework assignment, you should be able to…
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understand and be able to identify various parameters of a wave: wavelength, wavenumber,
frequency, angular frequency, period, amplitude, phase
determine the speed of a wave
calculate the intensity of a wave
combine waves using superposition and determine if the combination is constructive or
destructive interference
describe what happens and how the phase is affected when waves reflect from boundaries and
interfaces where the wave speed changes
determine the relative phase Δφ for two waves with a difference in path length.
identify what kinds of phase difference Δφ leads to constructive vs. destructive interference.
use the conditions on Δφ in double-slit interference to find where bright or dark bands will
appear on a screen
use the pattern of bands on a screen from a double-slit experiment to determine the wavelength
of light
identify the phase changes that occur when light is reflected from an interface between two
different media
use the phase changes, along with path length, to find constructive and destructive interference in
thin films (soap, water, oil, etc.)
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April 14 – April 21, 2015
Here are summaries of this module’s important concepts to help you complete this homework:
For transverse waves, the disturbance is perpendicular to the direction of propagation. Other waves are
longitudinal: the disturbance is parallel to the direction of propagation.
For each kind of wave, the wave speed is a characteristic of the wave and the medium. The most
important examples are:
Electromagnetic waves:
n = index of refraction
Sound waves in a fluid:
Waves carry energy as they travel. The rate at which energy travels is the power of the wave, and the
power per unit area is the intensity. All of these properties (energy, power, and intensity) are
proportional to the square of the amplitude of the wave. For a spherical wave traveling in three
dimensions, the intensity decreases in proportion to 1/r2, where r is the distance from the source.
A sinusoidal wave also requires a phase, φ, so we have
. In particular, if the
phase is equal to π, then the wave is “inverted” from its usual shape. If two waves are in phase, they
will combine favorably, with constructive interference. If they are exactly out of phase (relative phase
π), they will cancel each other out, with destructive interference.
When a wave reaches a boundary, the reflected wave may have a phase shift. Going from fast to slow,
the reflected wave will be inverted (phase shift of π). Going from slow to fast, the reflected wave will
have no phase shift.
All waves exhibit interference. The most important question is: what is Δ φ ?
If Δφ = 0, ±2π, ±4π, ±6π, … then you have constructive interference.
If Δφ = ±π, ±3π, ±5π, … then you have destructive interference.
If there is a difference in path length Δs, that contributes a phase difference
.
For two sources that are in phase (or two slits), with distances s1 and s2 from each source, a distance d
between the slits, and an angle θ from the normal:
Constructive interference occurs for
Destructive interference occurs for
When light from two slits travels a distance L and strikes a screen at a height y,
Constructive interference occurs for
Destructive interference occurs for
When light strikes a thin film, such as a soap bubble or oil slick, the reflected light will show thin-film
interference, and different colors will be reflected strongly (or not). Constructive or destructive
interference depends on the usual rules for Δφ. Notably, reflection off of an interface might result in an
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April 14 – April 21, 2015
additional phase change of Δφ = π, if the light is going from a region of low index n to an region of high
index.
1. Putting Everything Together (Exam-Type Question) (3 pts).
The diagram below represents a snapshot of two springs at an instant after a pulse has reached the
boundary between them.
a) On the diagram, clearly label which spring has the larger wave speed. Explain how you could tell
from the diagram.
Since the wave is wider on the left rope, it must have a larger wave speed.
b) Is the reflection at the boundary between the springs more like reflection from a fixed end or a free
end? What is the relative phase? Explain how you can tell from the diagram.
It is more like reflection from a free end because it reflects in the same direction it transmits (up).
c) On the diagram, label which pulse is the reflected pulse and which is the transmitted pulse. Explain
your reasoning.
To get a reflection that is up (just as the transmission) we must go from heavy rope (slow rope) to a
light rope (fast rope). In that case, the initial pulse isn’t inverted, which it isn’t here. So we must
have gone from the right, heavy rope, to the left, light rope.
2. A very wavy rope (3 pts). A person with a very nice hat shakes a long rope to make a transverse
traveling wave as shown. The graph on the right is a zoomed-in snapshot of the beginning of the rope at
time t = 0. The waveform travels down the rope to the right at a speed of v = 1.8 m/s. Even though the
wave is a little bit irregular, we will approximate it as a sinusoid.
a) At what frequency f is the person shaking the rope up and down?
The wavelength of the rope can be estimated from the graph. The peak-to-peak distance along x appears
to be about λ = 85 cm. From there, we can calculate f from the wave speed:
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Physics E1bx
a cont.)
April 14 – April 21, 2015
Calculate the wave number k and angular frequency ω of the transverse wave.
We can get the wave number k and angular frequency ω from λ and f, respectively:
b) Write an equation for the vertical displacement y of the rope as a function of position x (distance
from the shaker’s hand) and time t in the form y(x, t) = A sin(kx ± ωt – φ). Give the numerical values
(with correct units) of all of the constants in your equation.
The amplitude A is also easy to estimate from the graph: it appears to be about 13 cm. (Note that the
vertical scale on the graph is in centimeters, whereas the horizontal scale is in meters.)
The t = 0 snapshot of the string doesn’t start at y = 0 when x = 0, which means φ isn’t zero. We have to
go about 7 cm to the right before reaching zero, but the phase is an angle (expressed in radians) not a
distance. In order to get from a distance along the string (the x-coordinate) to an angle, we need to
multiply by k. (That’s what k is for, really.) The phase is about φ = (7.4 rad/m)(0.07 m) = 0.5 radians.
The sign in front of ωt should be negative because the wave moves to the right (towards +x). The sign in
front of φ should be negative also. To see why, try plugging in x = 0 and t = 0; you should be taking the
sine of a slightly negative angle, to reflect that the displacement of the rope is slightly below zero at that
specific time and place. Putting it all together gives us a final equation of:
.
3. Loudspeaker interference (2 pts). You are standing near two speakers wired to produce in-phase
sound signals at the same frequency, which can be varied. You are 6 m away from one and 4.5 m
away from the other. Assume that the speed of sound in air is v = 340 m/s.
a) You notice that certain frequencies are very quiet. What are the three lowest frequencies at which
you will observe destructive interference at your location?
The difference in path length between the sound waves from the two speakers at your position is
Δs = 1.3 m, so the phase difference is:
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Physics E1bx
April 14 – April 21, 2015
For destructive interference, this must be an odd multiple of π:
Since we want to know the frequency, we use
We can see that the lowest frequencies will come from the smallest values of m. Putting in the numbers,
for m = 0, 1, and 2 we get f = 113 Hz, 340 Hz, and 566 Hz.
b) Your friend flips a switch which causes the two speakers to become 180° out of phase. Now what are
the three lowest frequencies at which destructive interference occurs?
The path lengths are the same as in part a), so the only difference is that now there is also an intrinsic
phase difference of π (or 180°):
Applying the criterion for destructive interference gives
f must be positive, so m = 0 doesn’t represent a physical solution. So we use m = 1, 2, and 3, giving us
f = 226 Hz, 453 Hz, and 680 Hz.
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Physics E1bx
April 14 – April 21, 2015
4. Double-slit interference pattern (2 pts). When green light of wavelength 550 nm is incident upon a
double-slit apparatus, it produces a third-order maximum (i.e., m = 3) at a certain spot X on the
screen.
a) If point X is located 32° away from the central maximum, what is the spacing between the slits?
In a two-slit interference problem, the angle to the mth maximum is given by
where m = 0 (and θ = 0°) represents the central maximum, m = 1 is the first-order maximum, and m = 2
is the second-order maximum. The spacing d can be calculated from
b) If you want there to be a minimum at point X using the same slits and screen, what wavelength of
visible light should you use instead of blue light? What color is the light?
The locations of the minima are given by
The wavelength must satisfy
for any integer m. For m = 0, 1, 2, 3 and 4 we get λ = 3300 nm, 1100 nm, 660 nm, 471 nm, and 366 nm,
and for larger m we would get even shorter wavelengths. The two wavelengths that fall in the visible
spectrum are 660 nm and 471 nm, which correspond to blue and red light, respectively.
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Physics E1bx
April 14 – April 21, 2015
5. Beautiful plumage (4 pts). Thin-film interference is responsible for the fact that certain birds and
insects appear can different colors from different angles. Consider a very thin layer of oil (n = 1.30)
stretched between the wisps of a bird feather, so that it has air on either side. Assume that the oil
layer has uniform thickness t.
a) When you look at the feather with normal incidence, it strongly reflects green light of wavelength
550 nm. What is the minimum possible thickness of the oil layer?
Ray 1 reflects off the front surface of the oil, and ray 2 reflects off the
back surface. The path lengths for the two rays differ by 2t, and ray 1
has a 180° (or π radians) phase change upon reflection whereas ray 2
does not. The phase difference between the two rays is
,
where λn is the wavelength of the light inside the oil, which is λvacuum/n. In order for green light to be
strongly reflected, it must exhibit constructive interference,
,
for some integer m. Combining this with the equation for the phase difference gives
The smallest possible value of t corresponds to m = 0, giving t = λ/4n = 110 nm.
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