W9_parallel_resonance.eps

448 Week 13: Interference and Diffraction

1.0

I 0

−1.0

I 0

−π/2 π/2

sinθ

θ

Figure 190: Typical graphs of the diffraction intensity from a single slit of widtha= 4λ. Note the
distortion of the horizontal scale by the inverse sine in the lower graph – the top graph is much
easier to draw and requires no calculator.

Nothing to it. Note well that as always,δ=kdsin(θ) andφ=kasin(θ), so this is an

indirect function ofθlinked by inverse sines.

Example 13.9.1: Two Slits of Separationd= 8λand widtha= 4λ

We proceed exactly the same way we did for the previous example, except now we add

two more tables: The angles of theinterference maximaand theinterference minima.

We find these (as usual) from:

sin(θm) =

mλ

d

=

m

8

(1085)

for maxima and

sin(θm) =(2m+ 1)λ

2 d

=^2 m+ 1

16

(1086)

for minima. The result is displayed in table 7. Using these numbers we can easily

enough construct a combined interference/diffraction pattern,displayed in figure 191.

For simplicity I only present the graph for sin(theta) – you can easily visualize or fill in

a graph as a function ofθusing the previous example as a guide to the distortion (or a

piece of paper with an accurate graph scale on it). Note well the “squashed” interference

that occur where there are diffractionminima. This illustrates a simple rule – when one

of the two functions in the product above inItotare zero, zero wins!

Problems like this are graded on the basis of whether or not they contain the essential

features illustrated herein. The various min’s and max’s should be correctly tablulated

and located approximately correctly on the graph. The diffraction envelope should be

qualitatively as shown, and the interference pattern should be drawn “under” it. If max’s

and min’s occur at the same angle, the minimum wins. The maximum central intensity

should be 4I 0 , whereI 0 is the central intensity produced by a single slit.

Nothing to it!