Single slit intensity at given angle

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

Given: λ = 600 nm = 6×10⁻⁷ m, a = 0.1 mm = 1×10⁻⁴ m, θ = 0.2°.

1. Angle in radians: θ = 0.2 × π/180 = 3.491×10⁻³ rad, so sin θ ≈ 3.491×10⁻³
2. Compute α:

α = π a sin θ / λ
  = π × 1×10⁻⁴ × 3.491×10⁻³ / 6×10⁻⁷
  = 1.828 rad

3. sin α = sin(1.828 rad) = 0.9672, so sin α / α = 0.9672/1.828 = 0.5292
4. Intensity:

I = I₀ (0.5292)² = 0.280 I₀

Answer: I ≈ 0.28 I₀, i.e. about 28 % of the central maximum intensity.