Why Does Bragg's Law Require Integers?

[aoner]

Hi there,

So in bragg's law 2d\sin \theta =n\lambda, n needs to be an integer. Can anyone explain why? I mean, what if the extra path 2d that the 'second beam' has is not dividable by a wavelength?
Not sure if this is asked before but could not find it!

Cheers,
Adnan

 

[Redbelly98]

Welcome to Physics Forums.

If the extra path 2dsinθ is not an integer multiple of the wavelength, then you don't get constructive interference, hence no wave for that angle θ.

 

[aoner]

Ah, yes! Sorry just also came up with that haha. Thanks!