Question

In Fig. $36-48,$ let a beam of x rays of wavelength $0.125nm$ be incident on an $NaCl$ crystal at angle $\theta ={45.0}^{\circ }$ to the top face of the crystal and a family of reflecting planes.
Let the reflecting planes have separation $d=0.252nm.$
The crystal is turned through angle $\varphi$ around an axis perpendicular to the plane of the page until these reflecting planes give diffraction maxima.
What are the
(a) smaller and
(b) larger value of $\varphi$ if the crystal is turned clockwise and the
(c) smaller and
(d) larger value of $\varphi$ if it is turned counterclockwise?

Answer 1

We want the reflections to obey the Bragg condition
$2d\sin \theta =m\lambda$, where $\theta$ is the angle between the incoming rays and the reflecting planes, $\lambda$ is the wavelength, and $m$ is an integer.
We solve for $\theta$
$\theta ={\sin }^{-1}\left(\frac{m\lambda }{2d}\right)={\sin }^{-1}\left(\frac{(0.125\times {10}^{-9}m)m}{2(0.252\times {10}^{-9}m)}\right)=0.2480m$
(a) For $m=2$ the above equation gives $\theta ={29.7}^{\circ }.$
The crystal should be turned
$\varphi ={45}^{\circ }-{29.7}^{\circ }={15.3}^{\circ }\text{clockwise}$
(b) For $m=1$ the above equation gives $\theta ={14.4}^{\circ }.$
The crystal should be turned
$\varphi ={45}^{\circ }-{14.4}^{\circ }={30.6}^{\circ }\text{clockwise}$
(c) For $m=3$ the above equation gives $\theta ={48.1}^{\circ }.$
The crystal should be turned
$\varphi ={48.1}^{\circ }-{45}^{\circ }={3.1}^{\circ }\text{counterclockwise.}$
(d) For $m=4$ the above equation gives $\theta ={82.8}^{\circ }.$
The crystal should be turned
$\varphi ={82.8}^{\circ }-{45}^{\circ }={37.8}^{\circ }\text{counterclockwise}$
Note that there are no intensity maxima for $m>4,$ as one can verify by noting that $m\lambda /2d$ is greater than 1 for $m$ greater than 4