W9_parallel_resonance.eps

442 Week 13: Interference and Diffraction

fourth quarter cancel. Again, no field amplitude arrives atP– this is a minimum with

zero intensity. Multiplying out we get a second condition for a minimum:

asin(θ) = 2λ (1067)

If we consider dividing the strip up into sixths, the conditiona 6 sin(θ) =λ/2 and the exact

same argument shows thatasin(θ) = 3λis a minimum. If we divide it into eights we get

asin(θ) = 4λ. Clearly we can continue indefinitely; the general rule for a minimum is:

asin(θ) =mλ m=⊗, 1 , 2 , 3 , ... (1068)

where I’ve used⊗again to indicate thatm= 0 is the principlemaximumat the center,

not aminimumand so must be skipped.

Finally, we know that diffraction will be symmetric, so that we have minima at all of

the negative anglesasin(θ) =−mλbut as before we’ll manage this by hand to keep the

equation simple.

Alas, no such simple argument can be made in order to find the angles of the diffraction

maxima(except for the central principle maximum, already considered). We know there

mustbemaxima in between each of the minima above but we expect from our discussion

ofN-slit interference that they won’t occur at any “simple” values of the phase angleφ

any more than they did at simple values ofδ. We therefore abandon heuristics at this

point and proceed to solve for theexactdiffraction intensity as a function of phase angle

φ(and henceθ, via the usual kind of inverse sines).

13.8: Exact Solution to Diffraction by a Single Slit

θ

to P

to center of screen

N = 7 radiators

Path difference a/N sin

at screen is E 0 /N

E field strength

per radiator

θper radiator

Figure 186: If we split the slit up intoNradiators, the field amplitude at the maximum in the center
of the screen fromeachradiator isE 0 /N, whereE 0 is the maximum amplitude from the entire slit
there. When we consider the waves emerging at an angleθdirected towards pointP, each radiator
travels an additional distance of ∆r =Nasin(θ) compared to the radiator immediately above it.
Both of these relations scale withN, and hence will be useful when we try to letN→∞and fill in
the entire slit with radiators.

In figure 186 you can see a single slit withNradiators neatly drawn out. I choseN= 7

because it is enough to “cover” the slit without being so many that you can’t see what