
#1
Dec1612, 03:36 PM

P: 61

The Fresnel diffraction integral is:
[itex] 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]} [/itex] When we want to obtain the Fraunhofer diffraction integral from here, we need to somehow convert it to: [itex] 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]} [/itex] So I thought we should do it as follows: [itex] \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 ] [/itex] And then it seems that we should neglect: [itex] x^2 + x_0^2 + y^2 + y_0^2 [/itex] 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 [itex]  2x x_0  2y y_0 [/itex]. After all they are of the same order... Please help! 



#2
Dec1612, 10:43 PM

P: 67

There might be an assumption that the aperture is small compared to the image space (x0,y0). Considering this is a farfield approximation, that tends to make sense.




#3
Dec1712, 06:10 AM

P: 61

Thanks,
It does, but then we couldn't neglect [itex] x_0^2 + y_0^2 [/itex] 



#4
Dec1712, 09:27 AM

Mentor
P: 10,840

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.




#5
Dec1712, 11:57 AM

P: 61

They're just a part of a phase! Got it. Thanks :)




#6
Dec1712, 05:06 PM

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: [itex] x_0 , y_0 >> x , y [/itex] ? (I don't think that's the case)



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