W9_parallel_resonance.eps

444 Week 13: Interference and Diffraction

This is looking really tedious, and we’re only atN= 7. However, if wedraw the phasor

diagram for this sum, it isn’t so bad:

E 0 /N

Etot

∆φ

Figure 187: The phasor diagram forN= 7 Huygens radiators distributed acrossa. The amplitude
of each radiator isE 0 /N, and the phase ∆φ=φ/Naccumulates.

The diagram in figure 187 (which we might have drawn for a 7-slit interference pattern!)

shows us that as long as ∆φissmall, the phasors gently arc up into what looks almost

like a smooth curve even for onlyN= 7. In a sevenslitproblem however, as we increase

θthenδbetween two slits gets bigger and soon isn’t small at all – we expect toget things

like seven-pointed stars and so on that don’t at all look like a smooth curve.

In this case of asingleslit, however, as we makeφlarge, we can make ∆φas small as

we likeby increasingN! In fact, we can make itinfinitesimallysmall, accumulatingdφ

as we go around asmoothcurve. We won’t actually do the following sums algebraically

(so don’t be intimidated by the notation) but we can in fact write the total field at the

pointPat the angleθin the Frauhofer approximation as^125 :

Etot= limN→∞

∑N

i=0

E 0

N

sin(kr−ωt+iφ/N) (1070)

This sort of sum, accumulating infinitesimal chunks ofEat infinitesimally different phase

angles, is begging to be turned into an integral^126 , but we will stop here and turn back

to our user-friendly phasors. In this limit, the line ofE 0 /N-length phasors will form a

smooth arcwith a fixed length ofE 0. The total angle accumulated between the beginning

of the arc and the end will beφ, the total phase difference between the top and bottom

of the slits. Our “discrete” phasor diagram for 7 slits above will become the continuous

phasor diagram illustrated in figure 188.

Almost all of our work has been done for us in this diagram! Let’s go over its features

and results so that you understand them as we derive our final result. Note that the

length of the arc isE 0 (we are just “bending it around”, but all the superposition of all

of theamplitudesof the infinitesimal phasor chunks still has to add up toE 0 ). The total

phase difference between (a tangent to) the beginning of the arc and (a tangent to) the

end of the arc is justφ, as illustrated with the lowerφangle. Thissameangleφis the

angle subtended by the circular arc as illustrated at the top – you can “see” by noting

that the tworradii are perpendicular to the arc at both ends, so as we swing out the

secondrthe angle accumulated by the tangent at the bottom has to match the angle

(^125) Note that we are still ignoring that extraO(N) term on the end as there areN+ 1 terms in the sum.
(^126) Ideally a complex exponential integral. Who actuallylikesto integrate sines and cosines and remember all of
those silly sign change?∫eudu=eu, all we ever really need to know...