Question

Find the equation for $\theta$ for which intensity is maximum for electron waves of wavelength $\lambda$

• A. $d\sin \theta =n\lambda$
• B. $2d\sin \theta =n\lambda$
• C. $2d\sin \theta =(2n+1)\lambda /2$
• D. $\sin \theta =(2n+1)\lambda /2$

Answer 1

The correct option is B $2d\sin \theta =n\lambda$

Difference between the rays after reflection from the first surface and the one that passed in to crystal to reflect and then come out of the surface=$2\times d\sin \theta$.

For the intensity to be maximum, they should have a constructive interference. Thus $2d\sin \theta =n\lambda$