Simple Diffraction: Deriving Equations with Phasors

[Manchot]

Hello, I've recently been studying simple diffraction for an upcoming proficiency test, and in the lecture notes for the class, a simple equation regarding single-slit Fraunhoffer diffraction was derived. The derivation was a little weird, so as an exercise, I just decided to go ahead and use phasors to derive the formula myself. Anyway, when I was finished, I nearly had the exact same result as the notes, except for one thing: a factor of a, the width of the slit. Because I couldn't find my mistake anywhere, I decided to look up the result on Wikipedia, which I suspected would do it the same way that I did. Anyway, in the step where the factor of a in the denominator was introduced, there seems to be an integration error. Am I just missing something here?

= C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^\frac{ikxx^\prime}{z} \,dx^\prime<br /> =C \frac{\left(e^\frac{ikax}{2z} - e^\frac{-ikax}{2z}\right)}{\frac{2ikax}{2z}}

 

[marlon]

where do you think there is a mistake ?

marlon

 

[Doc Al]

Looks to me like they snuck in that factor of a in the denominator where it doesn't belong.

 

[marlon]

you are right Doc Al

i did not even see that...i must be getting delirious again

marlon

 

[Claude Bile]

In my notes, the LHS should be;

C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^{ikxsin\theta} dx

Which is just a Fourier integral (u = xsin\theta is the conjugate variable), whereby the correct result is the sinc function given.

It would appear that you should be integrating with respect to x rather than x-prime.

Claude.