- #1

- 468

- 4

[tex]= C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^\frac{ikxx^\prime}{z} \,dx^\prime

=C \frac{\left(e^\frac{ikax}{2z} - e^\frac{-ikax}{2z}\right)}{\frac{2ikax}{2z}}[/tex]

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- Thread starter Manchot
- Start date

- #1

- 468

- 4

[tex]= C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^\frac{ikxx^\prime}{z} \,dx^\prime

=C \frac{\left(e^\frac{ikax}{2z} - e^\frac{-ikax}{2z}\right)}{\frac{2ikax}{2z}}[/tex]

- #2

- 3,763

- 9

where do you think there is a mistake ?

marlon

marlon

- #3

Doc Al

Mentor

- 45,065

- 1,376

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

- #4

- 3,763

- 9

you are right Doc Al

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

marlon

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

marlon

- #5

Claude Bile

Science Advisor

- 1,471

- 19

In my notes, the LHS should be;

[tex]C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^{ikxsin\theta} dx[/tex]

Which is just a Fourier integral ([itex] u = xsin\theta [/itex] 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.

[tex]C \int_{-\frac{a}{2}}^{\frac{a}{2}}e^{ikxsin\theta} dx[/tex]

Which is just a Fourier integral ([itex] u = xsin\theta [/itex] 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.

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