Question

(b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits

Answer 1

If the width of the slits are A₁ and A₂, the amplitude will also be A1 and A2.

The maximum amplitude, A' will happen when constructive interference happens
A' = (A₁ + A₂)
The maximum amplitude, A will happen when destructive interference happens
A = (A₁ - A₂)

We know that intensity is proportional to the square of amplitude.
$\frac{I_1}{I_2} = \frac{A'^2}{A^2} = \frac{(A_1+A_2)^2}{(A_1-A_2)^2} \\ \\ \frac{9}{25} =(\frac{A_1+A_2}{A_1-A_2} )^2\\ \\ \sqrt{ \frac{9}{25} }=\frac{A_1+A_2}{A_1-A_2} \\ \\\frac{A_1+A_2}{A_1-A_2} = \frac{3}{5} \\ \\ \frac{A_1}{A_2}= \frac{5+3}{5-3} = \frac{8}{2} \\ \\ \boxed{A_1:A_2= 4:1}$

Answer 2

      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

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proof:

    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.

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