# Bragg's law

1. Aug 27, 2009

### manofphysics

We all know, Bragg's law , $$n\lambda=2dsin\theta$$
where $$2dsin\theta=path difference$$.
In the derivation of the path difference we take the two incident rays to be parallel which is perfect BUT we take reflected rays also to be parallel.How can this be since the reflected rays have to MEET for superposition to take place.
This is the same as in FRAUNHOFER diffraction, where we CAN take parallel rays due to the distance being effectively infinity or due to the use of LENS.
But we are not using a lens nor is the distance supposed to be very large(infinity) in the experimental methods of Xray diffraction ( rotating crystal, powder photograph etc.).
So, how is it that this formula works?

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2. Aug 27, 2009

### PhaseShifter

I'm missing something here.

You have parallel incident waves reflecting from parallel planes...why wouldn't the reflected rays also be parallel?

Or are you trying to treat the wavefront as a single point rather than a plane?

Last edited: Aug 27, 2009
3. Aug 27, 2009

### manofphysics

when an EM wave strikes an atom, a spherical wave front is emitted. ie waves are emitted in ALL directions . Now we are considering the emitted rays by two atoms which are parallel,
But the parallel rays will not meet for interference, we need to use a lens...
(I am using an analogy of diffraction in optics here. as both are almost similar, I think)

4. Aug 27, 2009

### Edgardo

The rays do meet because the rays are not exactly parallel. They are only approximately parallel. And the approxmation is good enough because the place where they meet is "far" away.

5. Aug 27, 2009

### manofphysics

So, I guess it is the same assumption that we take in Fraunhofer diffraction.