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A Simple Introduction to Interference
Learning Goal: To understand the basic principles underlying interference.
One of the most important properties of waves is the principle of superposition. The principle of superposition for waves states that when two waves occupy the same point, their effect on the medium adds algebraically. So, if two waves would individually have the effect "+1" on a specific point in the medium, then when they are both at that point the effect on the medium is "+2." If a third wave with effect "-2" happens also to be at that point, then the total effect on the medium is zero. This idea of waves adding their effects, or canceling each other's effects, is the source of interference.

First, consider two wave pulses on a string, approaching each other. Assume that each moves with speed 1 meter per second. The figure shows the string at time t=0. The effect of each wave pulse on the string (which is the medium for these wave pulses) is to displace it up or down. The pulses have the same shape, except for their orientation. Assume that each pulse displaces the string a maximum of 0.5 meters, and that the scale on the x axis is in meters.

Part A
At time t=1\, \rm s, what will be the displacement Deltay at point x=2.5\, \rm m?
Express your answer in meters, to two significant figures.
  Deltay =  0   \rm m
Part B
Choose the picture that most closely represents what the rope will actually look like at time t=0.75\,\rm s.
The same process of superposition is at work when we talk about continuous waves instead of wave pulses. Consider a sinusoidal wave as in the figure.
Part C
How far Deltax to the left would the original sinusoidal wave have to be shifted to give a wave that would completely cancel the original? The variable lambda in the picture denotes the wavelength of the wave.
Express your answer in terms of lambda.
  Deltax =  \frac{{\lambda}}{2} 
Part D
In talking about interference, particularly with light, you will most likely speak in terms of phase differences, as well as wavelength differences. In the mathematical description of a sine wave, the phase corresponds to the argument of the sine function. For example, in the function y=A \sin (k x), the value of k x at a particular point is the phase of the wave at that point. Recall that in radians a full cycle (or a full circle) corresponds to 2 pi radians. How many radians would the shift of half a wavelength from the previous part correspond to?
Express your answer in terms of pi.
  phase difference =  {\pi}   radians
Part E
The phase difference of pi radians that you found in the previous part provides a criterion for destructive interference. What phase difference corresponds to completely constructive interference (i.e., the original wave and the shifted wave coincide at all points)?
Express your answer as a number in the interval [-\pi , \pi].
  phase difference =  0   radians
Part F

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus a cycle. For example, if two waves have a phase difference of \frac{\pi}{4}, the interference effects would be the same as if the two waves had a phase difference of \frac{\pi}{4} + 2\pi. The complete criterion for constructive interference between two waves is therefore written as follows:

{\rm phase\ difference} = 0 + 2 \pi n \qquad {\rm for\ any\ integer}\ n

Write the full criterion for destructive interference between two waves.

Express your answer in terms of pi and n.
  phase difference =  {\pi}+2{\pi}n   {\rm for\ any\ integer}\ n
The phase for a plane wave is a somewhat complicated expression that depends on both position and time. For most interference problems, you will work at a specific time and with coherent light sources, so that only geometric considerations are relevant. Consider two light rays propagating from point A to point B in the figure, which are \frac{3 \lambda}{4} apart. One ray follows a straight path, and the other travels at a 60^\circ angle to that path and then reflects off a plane surface to point B. Both rays have wavelength lambda.
Part G
Find the phase difference between these two rays at point B.
Part G.1
Find the difference in distance
Find the difference in length between the direct path and the reflected path. You can use the fact that triangle ABC is an equilateral triangle.
Express your answer in terms of lambda.
  path length difference =  \frac{3{\cdot}{\lambda}}{4} 
Now that you have the difference in path length, convert that to radians. Recall that every cycle of 2 pi radians is equivalent to one wavelength.
Express your answer in terms of pi.
  phase difference =  \frac{3{\pi}}{2}   radians
Part H
Suppose that the reflected ray receives an extra half-cycle phase shift when it reflects. What is the new phase shift at point B?
Hint H.1  

Hint not displayed

Express your answer in terms of pi.
  phase difference =  \frac{5{\pi}}{2}   radians
Whenever light reflects from a transparent interface, moving from lower index of refraction to higher index of refraction, it gets an extra half cycle phase difference. Being able to accurately find the phase differences between waves at various points will be useful in both interference and diffraction problems.
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