2—Infinite Series 362.8 Diffraction
When light passes through a very small opening it will be diffracted so that it will spread out in a
characteristic pattern of higher and lower intensity. The analysis of the result uses many of the tools
that you’ve looked at in the first two chapters, so it’s worth showing the derivation first.
The light that is coming from the left side of the figure has a wavelengthλand wave number
k= 2π/λ. The light passes through a narrow slit of width=a. The Huygens construction for the
light that comes through the slit says that you can effectively treat each little part of the slit as if it is
a source of part of the wave that comes through to the right. (As a historical note, the mathematical
justification for this procedure didn’t come until about 150 years after Huygens proposed it, so if you
think it isn’t obvious why it works, you’re right.)
y
dy
ysinθ
θ
r
r 0
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. Find the value ofrby doing a little trigonometry, getting
r=r 0 −ysinθ
The total wave to be received is now the integral
∫a0Adycos
(
k(r 0 −ysinθ)−ωt
)
=A
sin(
k(r 0 −ysinθ)−ωt
)
−ksinθ
∣
∣∣
∣
∣
a0Put 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