- #1

- 4

- 0

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?

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- Thread starter helloween0908
- Start date

- #1

- 4

- 0

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?

- #2

nasu

Gold Member

- 3,776

- 432

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.

- #3

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The Bragg's law becomes:

the length of the red+ blue lines = d(sin θ + sin θ’ ) = mλ

- #4

nasu

Gold Member

- 3,776

- 432

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.

- #5

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oh, thanks.

The problem seems clear now :D

The problem seems clear now :D

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