Fraunhofer Diffraction for two unequal slit widths

[devaultc]

Homework Statement

Consider a Fraunhofer diffraction pattern due to two unequal slits. Let a and b be the unequal slit widths and c the distance between their centers. Derive an expression for the intensity of the pattern for and diffraction angle $$\theta$$, assuming the arrangement to be illuminated by perpendicular light of wavelength $$\lambda$$.

Homework Equations

Fresnel-Kirchoff formula: U_p = C$$\int$$ E^(ikr)dA
E(k_z) = FT{A(z)} where A(z) is the aperture function and FT is Fourier Trnasform

The Attempt at a Solution

My attempt has been to express the double slit as an aperture function A(z) as the sum of the rectangle function rect|(z+c)/a| + rect|(z-c)/b| and take the FT of A(z). I know the general FT of a rectangle function is sinc() but I am not sure how to show that and if I do, how do I interpret my results? Thanks

 

[mark726]

for any help!
Thank you for your interesting question. The Fraunhofer diffraction pattern due to two unequal slits is a classic problem in optics and can be solved using the Fresnel-Kirchoff formula and Fourier transforms.

To start, we can define the aperture function A(z) as the sum of the two rectangle functions representing the two slits:

A(z) = rect[(z+c)/a] + rect[(z-c)/b]

Taking the Fourier transform of A(z), we get:

E(kz) = FT{A(z)} = FT{rect[(z+c)/a]} + FT{rect[(z-c)/b]}

Using the property that the Fourier transform of a rectangle function is a sinc function, we can write the above equation as:

E(kz) = sinc[(ka/2)sin\theta] + sinc[(kb/2)sin\theta]

where \theta is the diffraction angle and k is the wavenumber given by k = 2\pi/\lambda.

Now, using the Fresnel-Kirchoff formula, we can write the intensity of the diffraction pattern at a point P on the screen as:

I_P = C\int E^(ikr)dA

where C is a constant and r is the distance between the point P and the aperture A(z).

Substituting the expression for E(kz) in the above equation, we get:

I_P = C\int [sinc[(ka/2)sin\theta] + sinc[(kb/2)sin\theta]] e^(ikr) dA

Integrating over the aperture A(z), we get:

I_P = C\{sinc[(ka/2)sin\theta] FT{e^(ikr)} + sinc[(kb/2)sin\theta] FT{e^(ikr)}\}

Using the Fourier transform property that FT{e^(ikr)} = \delta(k), where \delta(k) is the Dirac delta function, we can simplify the above equation to:

I_P = C\{sinc[(ka/2)sin\theta] + sinc[(kb/2)sin\theta]\}

Finally, we can substitute the values of a, b, and c in terms of the diffraction angle \theta to get the final expression for the intensity of the diffraction pattern:

I_P = C\{