Mathematical Tools for Physics

2—Infinite Series 48

It’s the variation with angle that I want to look at. The intensity of the wave, the power per area, is
proportional to the square of the wave’s amplitude. I’m going to ignore all the constant factors, so I won’t worry
about the constant of proportionality. The intensity is then (up to a factor)

I=

sin^2

(

(ka/2) sinθ

)

sin^2 θ

(21)

For light, the wavelength is about 400 to 700 nm, and the slit may be a millimeter or a tenth of a millimeter.
The size ofka/ 2 is then about

ka/2 =πa/λ≈ 3. 0. 1 mm/ 500 nm≈ 1000

When you plot this intensity versus angle, the numerator vanishes when the argument ofsin^2 ()isnπ, withnan
integer,+,−, or 0. This says that the intensity vanishes in these directionsexceptforθ= 0. In that case the
denominator vanishes too, so you have to look closer. For the simpler case thatθ 6 = 0, these angles are

nπ=

ka

2

sinθ≈

ka

2

θ n=± 1 , ± 2 ,...

Becausekais big, you have many values ofnbefore the approximation thatsinθ=θbecomes invalid. You can
rewrite this in terms of the wavelength becausek= 2π/λ.

nπ=

2 πa

2 λ

θ, or θ=nλ/a

What happens at zero? Use power series expansions to evaluate this indeterminate form. The first term in
the series expansion of the sine isθitself, so

I=

sin^2

(

(ka/2) sinθ

)

sin^2 θ

−→

(

(ka/2)θ

) 2

θ^2

=

(

ka

2

) 2

(22)

What is the behavior of the intensitynear θ= 0? Again, use power series expansions, but keep another
term

sinθ=θ−

1

6

θ^3 +···, and (1 +x)α= 1 +αx+···