2—Infinite Series 49Remember,ka/ 2 is big! This means that it makes sense to keep only one term of the sine expansion forsinθ
itself, but you’d better keep an extra term in the expansion of thesin^2 (ka...).
I≈
sin^2(
(ka/2)θ)
θ^2=
1
θ^2[(
ka
2θ)
−
1
6
(
ka
2θ) 3
+···
] 2
=
1
θ^2(
ka
2θ) 2 [
1 −
1
6
(
ka
2θ) 2
+···
] 2
=
(
ka
2) 2 [
1 −
1
3
(
ka
2θ) 2
+···
]
When you use the binomial expansion, put the binomial in the standard form,(1 +x)as I did in the second line
of these equations. What is the shape of this function? Forget all the constants, and it looks like 1 −θ^2. That’s
a parabola.
The dots are the points where the intensity goes to zero,nλ/a. Between these directions it reaches a
maximum. How big is it there? These maxima are about halfway between the points where(kasinθ)/2 =nπ.
This is
ka
2
sinθ= (n+^1 / 2 )π, n=± 1 ,± 2 , ...At these angles the value ofIis, from Eq. ( 21 ),
I=(
ka
2) 2 (
1
(2n+ 1)π/ 2) 2
The intensity atθ= 0is by Eq. ( 22 ),(ka/2)^2 , so the maxima off to the side have intensities that are smaller
than this by factors of
1
9 π^2 / 4
= 0. 045 ,
1
25 π^2 / 4