Second order maxima in single slit

Intensity in single-slit diffraction

For a slit of width a, the resultant amplitude at angle θ is R = R₀ (sin α)/α, and the intensity is

I = I₀ (sin α / α)², where α = (π a sin θ) / λ.

The central maximum is at α = 0, where sin α / α = 1, so I = I₀.

Position of the secondary maxima

Secondary maxima occur where dI/dα = 0, i.e. tan α = α. Solving graphically gives α ≈ ±1.43π, ±2.46π, ... A very good approximation, which is the usual textbook one, is that the maxima lie midway between the minima:

α ≈ (2n + 1)π / 2, n = 1, 2, 3, ...

n = 1 is the first secondary (first order) maximum with α = 3π/2; n = 2 is the second-order maximum with α = 5π/2.

Second-order maximum

sin α  = sin (5π/2) = 1
sin α / α = 1 / (5π/2) = 2 / (5π)       (amplitude ratio)

So the amplitude of the second-order maximum is 2/(5π) times the central amplitude (≈ 0.127).

The intensity is the square of this:

I₂ / I₀ = (2 / 5π)² = 4 / (25π²) ≈ 0.0162 (about 1/62).

Remark: the figure 2/(5π) is the ratio R₂/R₀ of the amplitudes (the ratio sin α / α). The ratio of the intensities is its square, 4/(25π²). The first-order maximum similarly gives 2/(3π) for amplitude and 4/(9π²) ≈ 0.045 for intensity.

 I/I0
 1.0 |        *
     |       * *
     |      *   *
 0.045|  *  *     *  *   <- 1st secondary
 0.016|*                  <- 2nd secondary
     +-------------------> sin(theta)