Thank you for reporting, we will resolve it shortly
Q.
A plane of spacing $'d'$ shows first order Bragg diffraction at angle $\theta$. A plane of spacing $2d$
The Solid State
Solution:
From Bragg's equation, we have, $2d \,sin\,\theta= n \lambda$
$\lambda=2d \,sin \theta$ (for first order reflection)
$d=\frac{\lambda}{2\,sin\,\theta}$
Then for spacing $2d$, we have
$2 (2d) \,sin\theta=\lambda$
$sin \,\theta=\frac{\lambda}{4d}=\frac{\lambda}{4\left(\lambda 2 sin \,\theta\right)}=\frac{sin\,\theta}{2}$
$\theta =sin^{-1}\left(\frac{sin\,\theta}{2}\right)$