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5. Two-Slit Interference

  • Page ID
    • Wendell Potter and David Webb et al.
    • UC Davis
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    Single-Slit Introduction

    When a wave passes through a wide slit, the pieces of the wave that hit the boundary stop, but the rest of the wave goes through more or less unaffected. This effect is why shadows exist – objects abruptly stop light, and leave a dark region. When the slit is smaller than or roughly the same size as the wavelength, something interesting happens: instead of waves passing through the gap in the boundary and continuing in the same direction, the waves travel outward in all available directions. Practically, this means the slit in the boundary acts as if it is a source of the wave in the region on the other side of the boundary. Two slits exhibiting this feature are shown in the picture below. To reiterate, this effect is only seen when the wavelength of the wave is small or around the same size as the slit.

    Path Length Difference

    If we have two slits in a screen the waves coming from one slit can interfere with the waves coming from the other slit. To figure out the sort of interference we get, we will use the phase charts that we have already introduced in Superposition of Harmonic Waves. The most common situation we will discuss is one in which a single wave hits two slits at different points. We can treat this situation like two sources with the same separation \(d\), as shown below.

    Screen Shot 2016-04-21 at 12.03.58 PM.png

    This special setup gives us two extra advantages:

    1. The incident wave is the same for both slits; we know that the frequencies are the same.
    2. As we've described the situation, incoming wave peaks hit the slits at the same time. Hence we can treat the two “sources” as having the same phase constant \(\phi\). (This would not necessarily be true if the wave hit the screen at an angle).

    In this case, our phase chart is given by

    \(2 \pi \frac{t}{T}\) \(\pm 2 \pi \frac{x}{\lambda}\)


    Wave 1 \(2 \pi \frac{t}{T}\) \(-2 \pi \frac{x_1}{\lambda}\) \(\phi\)
    Wave 2 \(2 \pi \frac{t}{T}\) \(-2 \pi \frac{x_2}{\lambda}\) \(\phi\)
    Change 0 \(-2 \pi \frac{x_1 - x_2}{\lambda}\) 0 \(\Delta \Phi\)

    \(\implies \Delta \Phi = -2 \pi (x_1-x_2)/ \lambda\)

    Because \(x_1-x_2\) is the difference in the distance each wave has to travel to the detector, we call this quantity the path length difference \(\Delta x \equiv x_1-x_2\)

    Approximating the Path Length Difference

    Finding the path length difference exactly can be quite lengthy. Fortunately if we are interested in the type of interference a far distance from the slits we can use an approximation that makes it considerably easier. Consider the following set up.

    Screen Shot 2016-04-21 at 12.23.34 PM.png

    Here we are interested at the type of interference we get at the location of the white circle, located a distance \(y\) from the center on a screen a distance \(L\) from the slits. The approximation that we are going to use will involve the angle \(\theta\) between the dotted lines. If we wish to relate this back to \(y\) and \(L\) we may use trigonometry: \( \tan \theta = y/L\).

    We start by drawing part of a circle with its origin on the hollow dot and that goes through the closest slit as shown below on the left. Because this is a circle we know that all points on the circle are an equal distance \(x_2\)from the hollow point. The path length difference \(\Delta x\) is precisely that small distance that lies between the edge of the dotted arc and the furthest slit.Screen Shot 2016-04-21 at 12.23.39 PM.png

    Now we make an appoximation: if the circle was instead a straight line, the shaded region would be a right-angled triangle. As long as the white circle is a far distance from the slits, ignoring the curvature of the circle is not a bad approximation. Once we make that approximation we are lead to the simpler picture above on the right. Looking at this right-angled triangle we have \[ \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \approx \dfrac {\Delta x}{d}\] Rearranging we find that \(\Delta x \approx d \sin \theta\). We must remember this relationship is only approximate as the triangle is not precisely a right-angled triangle.


    Prove that the angle \(\theta\) shown in the previous diagram on the right is the same angle \(\theta\) in the picture above it.

    Example 1

    A speaker in the distance is playing a single note. The peaks of the waves hit the wall together, and pass through the two holes separated by 1 m. At different locations behind the wall the note heard is either loud or quiet. What can we tell about the frequency of the note being played if we know that the location of the hollow circle (shown below) is quiet?


    Hint: you will need to know that the speed of sound is roughly 340 m/s, and that the range of human hearing is 20 – 20,000 Hz.


    We know that the two waves will have the same frequency, and as the peaks are arriving together they must also have the same constant phase . Plugging this into the phase chart like before we find

    \[\Delta \Phi = -2 \pi \dfrac{\Delta x}{\lambda}\]

    Because we know this is a quiet location we know that

    \[\Delta \Phi = - (odd) \pi.\]

    From the geometry of the problem we can calculate \(\Delta x\). The distance from the slit on the left to the hollow circle is 8 m. The distance to the right slit
    can be found by

    \[\Delta x = (8.062 - 8) \text{ m} = 0.062 \text{ m}\]

    Using that \(\Delta \Phi = - (odd) \pi\) and \(\Delta x = 0.062 \text{ m}\) we have

    \[(odd) \pi = 2 \pi \dfrac{0.062 \text{ m}}{\lambda} \implies \lambda = \dfrac{2 \Delta x}{odd} = \dfrac{0.124\text{ m}}{odd}.\]

    This gives us information on the wavelength, to get to the frequency we use the fact that the speed of sound is \(v_{sound} = 340 \text{ m/s} = f \lambda\). The frequencies the speaker could be playing are

    \[f = \dfrac{v_{sound}}{\lambda}= 2740 \times (odd) \text{ Hz}\]

    We cannot pick a unique answer; we can only find possible answers. If we know that humans cannot hear beyond 20,000 Hz, we can list every possible answer (because the speaker wouldn't play a note we couldn't here). If we couldn't limit possible answers and list all of them, an acceptable solution would be to stop here.
    Plugging in values we have (to three sig. fig.)

    (\(odd)\) frequency \(f\)
    1 2740 Hz
    3 8230 Hz
    5 13700 Hz
    7 19200 Hz
    9 24700 Hz

    So the possible solutions are 2740, 8230, 13700, and 19200 Hz.

    This page titled 5. Two-Slit Interference is shared under a not declared license and was authored, remixed, and/or curated by Wendell Potter and David Webb et al..

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