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Second maxima intensity 1.62 percent

Short answer

In single-slit diffraction, I = I₀(sinα/α)². The second secondary maximum occurs at α ≈ 5π/2 where sinα=1, giving I₂/I₀ = 4/(25π²) ≈ 0.0162 or 1.62%.

Show that the intensity of second primary maxima is 1.62% of central maxima in Fraunhofer's single slit diffraction.

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

  1. Maxima: dI/dα = 0 gives α cos α − sin α = 0, i.e. tan α = α.
  2. The roots lie close to the odd multiples of π/2 (midway between the minima α = mπ): α ≈ ±3π/2 for the first, ±5π/2 for the second secondary maximum.
 I/I0
 1 |        /\
   |       /  \
.045|  _   /    \   _
.016|_/ \_/      \_/ \_
   +--------------------> α
   -5π/2 -3π/2  0  3π/2 5π/2
  1. At α = 5π/2, sin α = 1, so
I₂ = I₀ (1/α)² = I₀ / (5π/2)² = 4 I₀ / (25 π²)
   = 4 / (25 × 9.8696) I₀
   = 0.0162 I₀
  1. As a percentage: I₂/I₀ ≈ 1.62 %.

(The first secondary maximum, at 3π/2, similarly gives 4/(9π²) ≈ 4.5 %.)

Answer: the second secondary maximum has about 1.62 % of the central maximum intensity.

✓ Answer checked by Claude Opus 5.5 (high effort)

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