Mathematical Tools for Physics

2—Infinite Series 47

Call the coordinate along the width of the slity, where 0 < y < a. I want to find the total light wave that
passes through the slit and that heads at the angleθaway from straight ahead. The light that passes through
between coordinatesyandy+dyis a wave

Adycos(kr−ωt)

Its amplitude is proportional to the amplitude of the incoming wave,A, and to the widthdythat I am considering.
The coordinate along the direction of the wave isr. The total wave that will head in this direction is the sum
(integral) over all these little pieces of the slit.
Letr 0 be the distance measured from the bottom of the slit to where the light is received far away. I can
find the value ofrby doing a little trigonometry to get

r=r 0 −ysinθ

The total wave to be received is now the integral

∫a

0

Adycos

(

k(r 0 −ysinθ)−ωt

)

=A

sin

(

k(r 0 −ysinθ)−ωt

)

−ksinθ

∣

∣

∣

∣

∣

a

0

Put in the limits to get
A
−ksinθ

[

sin

(

k(r 0 −asinθ)−ωt

)

−sin

(

kr 0 −ωt

)]

I need a trigonometric identity here, one that you can easily derive with the techniques of complex algebra in
chapter 3.

sinx−siny= 2 sin

(

x−y

2

)

cos

(

x+y

2

)

Use this and the light amplitude is

2 A

−ksinθ

sin

(

−

ka

2

sinθ

)

cos

(

k(r 0 −

a

2

sinθ)−ωt

)

(20)

Thewaveis the cosine factor. It is a cosine of(k.distance−ωt), and the distance in question is the
distance to the center of the slit. This is then a wave that appears to be coming from the middle of the slit, but
with an amplitude that varies strongly with angle. That variation comes from the other factors in Eq. ( 20 ).