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| from wikipedia's diffraction article |
The double slit is covered in detail on the AP Physics B exam. Students should be able to explain why an interference pattern is formed, cite an equation for the positions of the maxima, tell a bit about where that equation comes from, and describe qualitatively how the brightness of the pattern varies along a screen.
The single slit is not covered in quite so much detail. Good thing, too. I spent some time last week learning the details about single slit diffraction. You see, one of the problems for the USIYPT next year is about a pinhole camera. To understand the image formed by a circular slit, one first has to understand how a one-dimensional single slit works.
At the Physics B level, we have to know that the positions of single slit minima are identical to the positions of double-slit maxima. That is, the relevant equation x=mλL/d for double slits gives bright-spot positions for whole number m's; for single slits, the same equation gives dark-spot positions for whole number m's. The central maximum for single slit diffraction is much wider than for double slits.
Why? I don't go into that in AP Physics B. I tell the convenient mistruth that diffraction occurs simply because light diffracts around the edges of the slit, and interferes similarly to the double slit situation. The true reasoning is very, very complicated.
At a deeper-than-introductory level, the single slit can be considered as an infinite set of point sources. Constructive interference occurs at the central maximum. Diffraction minima can be found by trigonometric analysis of when each of the infinite point sources can be paired with another whose path difference to the location on the screen is λ/2. The diffraction maxima are not halfway between the minima; rather, their positions are determined by the solution to a complicated differential equation whose solutions are Bessel functions.
(Confused? Me too. I spent a couple of class periods and finally understood the derivation of the minima, but the maxima derivation is still flummoxing.)
This year's operational (i.e. regular form, not "form B") AP Physics B exam asked about a single slit. Students were to design an experiment to measure the width of the slit. The fact that it was a single, not double, slit was nearly irrelevant in this case. Using x=mλL/d still works, as long as x and m are described appropriately. Measuring the distance between dark spots with m = 1 works just fine whether this is a single or double slit.
The Form B exam, though, really hit at the qualitative difference between the single and double slit. It asked to graph the intensity as a function of location on a screen for a double slit; then for a single slit under the same conditions. Answer: the central maximum remains, double-slit-maxima become single-slit-minima.
And this is as deep as a Physics B question will get about single slits.
GCJ
Based on your experience (I'm assuming you've been a reader at one point), how many points, out of 15, do you think a student would earn if their solution was perfectly correct for a DOUBLE slit?
ReplyDeleteWell... I don't want to speak for the table leader on that question. And I'll certainly know more in a couple of weeks, when the reading starts.
ReplyDeleteThat said, my instincts are a totally correct double-slit solution on the operational exam might be worth all but one or two points.
As a comparison, consider 1999 B6, which was essentially this experiment with a double slit. An answer using the small-angle approximation (x=m(lambda)L/d) earned all but one point; that final point was only earned if the small-angle assumption was explicitly stated. I myself would use a similar approach here.
Caveat, as always -- the specific rubric is a product of the table leader's whims, coupled with intense discussion with other table leaders. There's no way to predict the rubric. And, it DOESN'T MATTER in the end -- as long as the rubric is applied consistently (which it is, trust me) then scores are assigned in a fair and appropriate manner.
GCJ