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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?
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?