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Unformatted text preview: tum mechanics. But the
theory is not intuitive, and its description of matter is very diﬀerent from our
To understand what seems to be a paradox, we look to Young’s double-slit
experiment. Here’s the set up: a source of light is shone at a screen with two
very thin, identical slits cut into it. Some light passes through the two slits
and lands upon a subsequent screen. Take a look at Figure 1.1 for a diagram
of the experiment setup.
First, think about what would happen to a stream of bullets going through
this double slit experiment. The source, which we think of as a machine gun, is
unsteady and sprays the bullets in the general direction of the two slits. Some
bullets pass through one slit, some pass through the other slit, and others
don’t make it through the slits. The bullets that do go through the slits then
land on the observing screen behind them. Now suppose we closed slit 2. Then
the bullets can only go through slit 1 and land in a small spread behind slit 1.
If we graphed the number of times a bullet that went through slit 1 landed at 1.1. THE DOUBLE SLIT EXPERIMENT 3 Figure 1.1: Double- and single-slit diﬀraction. Notice that in the double-slit
experiment the two paths interfere with one another. This experiment gives
evidence that light propagates as a wave. the position y on the observation screen, we would see a normal distribution
centered directly behind slit 1. That is, most land directly behind the slit,
but some stray oﬀ a little due to the small amount randomness inherent in
the gun, and because of they ricochet oﬀ the edges of the slit. If we now close
slit 1 and open slit 2, we would see a normal distribution centered directly
behind slit 2.
Now let’s repeat the experiment with both slits open. If we graph the
number of times a bullet that went through either slit landed at the position
y , we should see the sum of the graph we made for slit 1 and a the graph for
Another way we can think of the graphs we made is as graphs of the
probability that a bullet will land at a particular spot y on the screen. Let
P1 (y ) denote the probability that the bullet lands at point y when only slit 1 is
open, and similarly for P2 (y ). And let P12 (y ) denote the probability that the
bullet lands at point y when both slits are open. Then P12 (y ) = P1 (y )+ P2 (y ).
Next, we consider the situation for waves, for example water waves. A
water wave doesn’t go through either slit 1 or slit 2, it goes through both.
You should imagine the crest of 1 water wave as it approaches the slits. As it
hits the slits, the wave is blocked at all places but the two slits, and waves on
the other side are generated at each slit as depicted in Figure 1.1.
When the new waves generated at each slit run into each other, interference
occurs. We can see this by plotting the intensity (that is, the amount of energy 4 CHAPTER 1. INTRODUCTION carried by the waves) at each point y along the viewing screen. What we
see is the fa...
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- Fall '13