Week 13: Interference and Diffraction 443
is going on. In the end, of course, we will letN→ ∞so that wereallycover the slit
with a continuum of radiators^123 so no particular choice forNmuch matters.
We have to be able to “scale” the field result itself. After all, the light we shine on the
slit could be very intense or it could be weak. The slit could be large (letting a lot of light
through) or it could be very small (not letting a lot of light through).We need a single
parameter that indicates how strong theE-field is on the screen, or equivalently, how
intense. We choose to setE 0 to the value of theE-field that makes it through the slit to
the screen in thecenter of the principle maximumatθ= 0. With this interpretation, it
isexactlylike what we did for the interference ofN“narrow” slits above. Indeed, at the
end of this topic we can go back anda posterioriformally justify our narrow slit results,
and define precisely just what “narrow” means!
If we split the slit up intoNradiators, each with the same path length to the center of
the screen (in the Fraunhofer limit, recall), then from symmetry and superposition run
backwards each radiator must produce an individualE-field on the screen with strength
E 0 /N. That way, no matter whatNis, the superposition of the fields at the center will
remain equal toE 0 , the measured/known/observed/assumedE-field there. AsNgets
large, this field amplitude (per radiator) will get very small (but nonzero) but the larger
number of radiators will precisely compensate.
Next, let’s think about path differences and phase differences. Recall thatasin(theta)
is the total path difference to the pointP between the wave from the (radiator at the)
very top of the slit and the wave from the (radiator at the) very bottom of the slit. In
the figure above, the top and bottom radiators aren’t, of course, precisely “at” the top
and bottom of the slits, but as we increase the number of radiatorsthey will get closer
and closer, and any error we make in assuming that they are there already for a finiteN
will go away.
We therefore can splitasin(θ) up intoNpieces, and make the path difference between
adjacent radiatorsNasin(θ). A very astute student might observe that for the 7 slits
above, it really should bea 6 sin(θ) (or rather, that our general rule should beNa− 1 sin(θ)
because the top radiator is at “zero”) but in the limitN→ ∞we will make an error
of order 1/Nusing the first relation^124 so we’ll just ignore it and use the first (easier)
relation.
Let’s turn this path difference between waves from adjacent radiators into a phase dif-
ference between adjacent radiators (by multiplying it byk, as always). Recall that we
definedφ=kasin(θ), so the phase difference between adjacent slits is just ∆φ=φ/N.
This phase differenceaccumulatesas we count down the radiators from the top – the
first slit down has a phase difference ofφ/N, the second has a phase difference of 2φ/N,
the third 3φ/Nand so on.
The wave we have to sum – using our ever-so-useful phasors, of course – is then (for
N= 7):Etot =E 0
N sin(kr−ωt) +E 0
Nsin(kr−ωt+φ/N)
+E 0
N
sin(kr−ωt+ 2φ/N) +E 0
N
sin(kr−ωt+ 3φ/N)+E^0
Nsin(kr−ωt+ 4φ/N) +E^0
Nsin(kr−ωt+ 5φ/N)+E 0
Nsin(kr−ωt+ 6φ/N) (1069)(^123) ... or, if this were a course inopticsbeing given to majors or folks with mad math skills, we’d justwrite an
integralfor the field at an arbitraryPand not bother with all of this dividing up and summing...
(^124) As you can easily see by doing the binomial expansion ofa/(N−1) = (a/N)(1− 1 /N)− (^1) , right...?