- #1
aoner
- 11
- 0
Hi there,
So in bragg's law [itex]2d\sin \theta =n\lambda[/itex], 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
So in bragg's law [itex]2d\sin \theta =n\lambda[/itex], 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