# Dodgy step in the Far field approximation

by Loro
Tags: approximation, dodgy, field, step
 P: 61 The Fresnel diffraction integral is: $A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2]}$ When we want to obtain the Fraunhofer diffraction integral from here, we need to somehow convert it to: $A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{+ik}{z} [x x_0 + y y_0]}$ So I thought we should do it as follows: $\frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2] = \frac{-ik}{2z} [x^2 + x_0^2 + y^2 + y_0^2 - 2x x_0 - 2y y_0 ]$ And then it seems that we should neglect: $x^2 + x_0^2 + y^2 + y_0^2$ since they're all much smaller than z. Then we get the correct solution. But I don't see why we could do that, and leave out the $- 2x x_0 - 2y y_0$. After all they are of the same order... Please help!
 P: 67 There might be an assumption that the aperture is small compared to the image space (x0,y0). Considering this is a far-field approximation, that tends to make sense.
 P: 61 Thanks, It does, but then we couldn't neglect $x_0^2 + y_0^2$
Mentor
P: 10,499

## Dodgy step in the Far field approximation

Those terms do not depend on the integration variables, it is possible to pull them out of the integral. They give a prefactor, which might be irrelevant, or accounted for in some other way.
 P: 61 They're just a part of a phase! Got it. Thanks :)
 P: 61 Hold on, but wouldn't that mean that Fraunhofer approximation works best away from the optical axis - where we're allowed to say: $x_0 , y_0 >> x , y$ ? (I don't think that's the case)

 Related Discussions Atomic, Solid State, Comp. Physics 2 Advanced Physics Homework 5 Computing & Technology 2 Special & General Relativity 3 Precalculus Mathematics Homework 12