In a Single Slit Fraunhofer Diffraction pattern, let us have the wavelength lambda of the light comparable to the slit width a. Let the distance L between the screen and the slits. L >> a. We take the amplitude of the wave emerging from a slit, as being proportional (nearly) to slit width of the slit. Proof/reason is given below:
Let slit widths be a1 and a2 and the amplitudes from the two slits be A1 and A2.
Then: a1 / a2 = A1 / A2.
The intensity I in a light wave is proportional to amplitude square A².
Now in the double slit Young's Diffraction Experiment, the Intensity of the wave resulting from interference of two waves is given as:
I = I1 + I2 + 2 √I1 √I2 Cos Ф
A² = A1² + A2² + 2 A1 A2 Cos Ф
At Maximum intensity, Ф = 0. So A = A1 + A2
At minimum intensity, Ф = π. So A = A1 - A2, Assume A1 > A2.
We have : (A1 - A2)² / (A1 + A2)² = 9/25
=> (A1 - A2 ) / (A1 + A2 ) = 3 /5
=> 2 A1 = 8 A2
=> A1 : A2 = 4 : 1
Since the amplitudes are nearly proportional to slit widths,
w1 : w2 = 4 : 1
The Amplitude A of the wave that is emitted by the slit at an angle theta is the result of sum of all wave fronts from all points from y = 0 to a, along the slit width.
Sum using integration on [0, a] of A Sin[kL+ ky Sin θ - ωt] dy.
(L + y Sinθ) gives the path length of wave.
so y(t) = [ (a A Sin α) / α] Sin (K L + α - ω t) --- eq of wave emitted at y.
= a A Sinc (α) * Sin (K L + α - ω t)
where α = π a Sinθ / λ = 1/2 * k a Sinθ
Thus amplitude of a light wave emitted by a single slit of width a, is proportional to (sin Ca), where C is a constant. We will take it as proportional to a.