W9_parallel_resonance.eps

Week 13: Interference and Diffraction 445

Etot

E 0

r

r

φ

φ

φ/2

= 2

rsinφ/2

rsinφ/2

Figure 188: The phasor diagram forN→ ∞Huygens radiators distributed acrossa. The “phasor
snake” bends smoothly around into a circular arc of lengthE 0 , where we need to determine the
length of the secant that cuts across,Etot.

accumulated between the radii. From this we see that the arc lengthE 0 can be related

torby:

E 0 =rφ (1071)

If we drop a perpendicular bisector (dashed line) from the center of the circular arc to

the total field phasorEtot, we make two simple right triangles with vertex angleφ/2.

The opposite side of each of them has lengthrsin(φ/2) so that:

Etot= 2rsin(φ/2) (1072)

We substituter=E 0 /φinto this (eliminatingrin favor ofE 0 ) to get:

Etot=

2 E 0 sin(φ/2)

φ

=E 0

(

sin(φ/2)

φ/ 2

)

(1073)

Finally, we go through the usual ritual to convert the field amplitudes to intensities:

I 0 =

1

2 μ 0 cE

2

0 (1074)

so that:

Itot=

1

2 μ 0 c

Etot^2 =

1

2 μ 0 c

E^20

(

sin(φ/2)

φ/ 2

) 2

(1075)

or

Itot(θ) =I 0

(

sin(φ/2)

φ/ 2

) 2

. (1076)

This is what we have been trying to get – an exact formula for the intensity of the

diffraction pattern as a function ofθ(yes, it is actually given as a function ofφbut recall

thatφ=kasin(θ) so we also know it as a function ofθ, at the expense of a little extra

(and tedious, admittedly) arithmetic. But arithmetic isn’t tedious tohumans any more

as long as an equation can be programmed into a computer, and this one is easy to code.

At a glance, this equation has all of the right features. Atθ= 0 (and henceφ= 0) we

get an intensity ofI 0127. At all the other places where sin(φ/2) = 0, we get a minimum.

(^127) We avoid the problem of “division by zero” calculus-fashionby taking thelimit
xlim→ 0 sinx(x)=x−x
(^3) /3! +x (^5) /5!−...
x = 1−x
(^2) /3! +x (^4) /5!−...= 1