W9_parallel_resonance.eps

Week 13: Interference and Diffraction 433

or these angles withany integer multiple of 2 πadded (or subtracted). If we multiply this

out and turn it into a rule, it becomes:

δmin=kdsin(θ) =

2 π

3

,

4 π

3

,

8 π

3

,

10 π

3

,

14 π

3

, ...

2 π

λ

dsin(θ) =

2 π

3

,

4 π

3

,

8 π

3

,

10 π

3

,

14 π

3

, ...

dsin(θ) = mλ

3

m=⊗, 1 , 2 ,⊗, 4 , 5 ,⊗, 7 , 8 ... (1040)

Note that this isalmostthe integer multiples of 2π/3 (where 3, recall, is thenumber

of slits – hmmm, one wonders if this rulegeneralizes...). However, we have toskip

the multiples of 2π/3 that are also multiples of 2πbecause we already know that the

multiples of 2πare principlemaxima. I remind you of this by putting⊗’d outholesin

them-sequence in the final result. We’ll continue this practice in the nextsection.

Finally, consider the last phasor diagram, which coorresponds to asecondary maximum.

If we set:

δsecondary max=π, 3 π, 5 π... (1041)

then this phasor diagram results. Although at the moment there isn’t any compelling

reason to see why (there will be shortly) let’s write this as:

δsecondary max = π, 3 π, 5 π...

=^2 πm

2

m=⊗, 1 ,⊗, 3 ,⊗, 5 ... (1042)

which looks like itmightbe a rule involving 2πm/(N−1) with the usual skip-all-m-that-

lead-to-a-multiple-of-2πconstraint.

The expressions forθminm andθsecondary maxm are now pretty obvious, and I’ll leave you to

find them on your own. A typical problem for multiple slits would have you build a table

of angles (or sines of angles) for the principle maxima, the minima, andthe secondary

maxima, and then to draw a “generic” graph of the intensity using this information.

Unfortunately, just looking at two and three slits isn’tquiteenough to infer a trustworthy

rule, especially for the secondary maxima. We’ll therefore (in the next section) skip 4

slits and jump right on up to 5 slits, and from there to an arbitrary finite numberNof

slits. We won’t quite prove that the rules we infer (which all work forN= 2, 3 ,5 in our

examples) work for any number of slits, but wealmostprove it and cleaning up what we

do and turning it into a formal proof isn’t difficult, just a bit beyond the scope of this

course. Unless, perhaps, you are a physics major and need the practice...

13.4: Interference from 4, 5, N Narrow Slits

Now let’s look at one more particular case (just to be sure) and thengeneralize the

above results. We will not worry (inthiscourse) about actually finding the explicit total

electric field and squaring it and factoring it to find the intensity for more than two

slits, even though I derived the intensity for three just to show you one way that it can

be done. Instead we will (continue to) focus on just finding the angles of the principle

maxima, the minima, andapproximatelyfinding the angles of the secondary maxima. In

all cases we will be graphically “evaluating”:

Etot = E 0 sin(kr−ωt) +E 0 sin(kr−ωt+δ) +E 0 sin(kr−ωt+ 2δ) +...

+E 0 sin(kr−ωt+ (N−2)δ) +E 0 sin(kr−ωt+ (N−1)δ) (1043)