# Problem with angled diffraction

1. Jan 30, 2010

### ibc

2 plane waves do 1 slit diffraction, one comes perpendicularly to the slit, the other with an angle theta.
Rayleigh criterion says that in order to distinguish between them one's primary maximum must be further than the other wave's first minimum, therefore (after calculating and getting that the intensity of the light from the angled wave goes as:
sin[(kd/2)(sin(phi) - sin(theta)]/(kd/2)(n*sin(phi) - sin(theta))
) we get that the criterion is:
sin(theta)>lambda/d
where lambda is the wavelength and d is the width of the slit.
Now, instead of air, I put after the slit another material with a different refractive index, now what would be the angle at which one's maximum lands on the other one's minimum?
So one way of doing so is saying, when the light goes into the new material, by snell's law we get: sin(theta)=n*sin(phi)
so the angle inside the material is phi, and by the previous observation we know that sin(phi)=lambda/d
so we get that the initial angle should be: sin(theta)=n*lambda/d.
However, another way is to directly calculate the diffraction (of the waves that comes with an angle), so I calculate the phase differences and integrate, the part of the phase difference that comes from the angle (the extra length the light has to go outside) is just the same, but the part of the phase difference that comes from the length differences the light has to do inside the new material must be multiplied with the refractive index, therefore I get that the power of the light as function of phi - the angle of the point at which I'm looking at (angle from the slit's center) and phi(i) - the initial angle outside the material (theta):
sin[(kd/2)(n*sin(phi) - sin(phi(i))]/(kd/2)(n*sin(phi) - sin(phi(i)))
and of course, for the wave that comes perpendicularly, the primary maximum will still be at phi=0
therefore, if we want Rayleigh criterion, we need to demand that the first minimum of the other wave will fall on the point phi=0, therefore:
sin[(kd/2)(n*sin(0) - sin(phi(i))]=0
(kd/2)*sin(phi(i))=pi
therefore: sin(phi(i))=lambda/d
just like before, without the factor of n.

So my problem is, which of the above is correct? I have a greater belief in the second one, because the first one seems to ignore other effects. The only thing that troubles me with the second one, which seems to be thorough and I can't see anything wrong with it, is that the answer turns out to be independent of the refractive index, n, and it doesn't seem too reasonable.

anyone got an idea?

Thanks

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