Single slit diffraction

Intensity distribution

The slit of width "a" is divided into an infinite elementary slits "dy". Each "dy" correspond to the elementary electric field dE. We will add all these dE over the range [-a/2, +a/2] to find the resultant field E at a point on the screen; then we square its amplitude to find its intensity at this point.

If E_c, the amplitude of the electric field of the wave emerging from the center of the slit, is uniformly distributed along "y", we can write at y = y on the screen:
dE = E_c (dy/a) sin (ωt + φ). φ is the phase difference between the wave from y = 0 and from y.
We have:
2π/φ = λ/y sin θ, thus: φ = (2 π/λ) y sin θ Then: dy = dφ (λ/2 π sin θ), and
dE = E_c [ (λ/2 π sin θ)/a] sin (ωt + φ) dφ

Let's write:
β = π a sin θ/λ, thus:
dE = (E_c /2 β) sin (ωt + φ) dφ

If "y" varies from: -a/2 to +a/2, then φ varies from: - β to + β
We have then:
E = ∫ dE = ∫ (E_c/a) sin (ωt + φ) dy from -a/2 to +a/2
= ∫ (E_c /2 β) sin (ωt + φ) dφ from: - β to + β

We have the following trigonometric identity:
sin (ωt + φ) = sin ωt cos φ + cos ωt sin φ
Thus:
E = (E_c /2 β) [sin ωt ∫ cos φ dφ + cos ωt ∫ sin φ dφ] from: - β to + β
= (E_c /2 β) [sin ωt (sin φ) + cos ωt (- cos φ)] from: - β to + β
= (E_c /2 β) [sin ωt (2 sin β) + cos ωt (0)] = (E_c / β) sin β sin ωt
Thus:
I = (1/2)cε_o E² = (1/2)cε_o(E_c²) sin² β / β² = I_c sin² β / β²
With I_c = (1/2)cε_o(E_c²)

Finally:
I = I_c sin² β / β²
I_c = (1/2)cε_o(E_c²),
β = (π a/λ) sin θ