First subsidiary maxima intensity percentage

In single-slit Fraunhofer diffraction, I = I₀ (sin α / α)², where α = π a sin θ/λ.

Positions of maxima

1. For a maximum, dI/dα = 0: dI/dα = I₀ · 2 (sin α/α) · (α cos α − sin α)/α² = 0
2. Either sin α = 0 (minima) or α cos α − sin α = 0, i.e. tan α = α.
3. The roots (graphically from the intersections of y = tan α and y = α) are close to α ≈ ±3π/2, ±5π/2 ... midway between the minima at ±π, ±2π, ... Using the textbook approximation, the mth secondary maximum is at α ≈ (2m+1)π/2.

 y
 |   y = α   /  /   /
 |          /  /   /     tan α curves
 |-----π---/--3π/2--2π--5π/2----> α

Intensity of the first secondary maximum

Put α = 3π/2: sin α = −1, so

I₁ = I₀ (1/α)² = I₀ / (3π/2)² = 4 I₀ / (9π²)
   = 4 / (9 × 9.8696) I₀ = 0.045 I₀

So I₁/I₀ ≈ 4.5 %. (Solving tan α = α exactly gives α = 4.493 and 4.7 %; the commonly quoted textbook value is about 4.5 %.)

Hence the first subsidiary maximum has only about 4.5 % of the intensity of the principal (central) maximum.