Is Bragg's Law Equation Modified for Different Incidence Angles?

[helloween0908]

problem: In Bragg's law equation, mormally, we measure the angle θ from the surface If instead the light strikes the grating at an angle of incident θ’ (measured from the normal), show that the condition for an intensity maximum is not 2dsin θ= mλ (m=1,2,3...)
but rather
d(sin θ + sin θ’ ) = mλ (m=0, ±1, ±2, ±3...)

No matter which way I tried, I finally ended up with 2dcosθ’ rather than d(sin θ + sin θ’ ) = mλ (m=0, ±1, ±2, ±3...).
Can anyone help me?

 

[nasu]

When you say "Bragg's law" usually you mean x-ray diffraction on a lattice.
Here it seems that you have something else: diffraction of light from a diffraction grating.
Or maybe a mix-up.
The formula with the sum of the two sines applies to the diffraction grating when the light hits it at an angle theta'.
There is a path difference between the incident rays hitting two different "holes" in the grating and this is given by d*sin(theta'). And then there is the path difference between the rays on the other side of the grating which is d*sin(theta).
I hope this helps.

 

[helloween0908]

Is this what you mean:

The Bragg's law becomes:
the length of the red+ blue lines = d(sin θ + sin θ’ ) = mλ

 

[nasu]

No. As I said, I was referring to an optical diffraction grating.
something like this:
http://en.wikipedia.org/wiki/Diffraction_grating
The math is similar though.

For x-ray diffraction from the crystal the maximum occurs when the two angles are equal.

 

[helloween0908]

oh, thanks.
The problem seems clear now :D