Given –

$\lambda =1.549 \times 10^{-10}$

$d=4.255 \times 10^{-10} $

To find – $\theta$, highest order

Solution –

For smallest glancing angle,

For $n=1$,

According to Bragg’s $n\lambda=2d\sin\theta$

$\sin\theta =\dfrac{n\lambda }{2d}$

$=\dfrac {(1)\times 1.549 \times 10^{-10}}{2\times 4.255 \times 10^{-10}}$

$\theta =10.49^{\circ}$

For highest order, $\sin\theta =1$

According to Bragg’s law,

$\therefore n\lambda=2d \sin\theta$

$n=\dfrac {2d(1)}{\lambda}$

$=\dfrac {2\times 4.255 \times 10^{-10}}{1.549 \times 10^{-10}}=5.49$

  $n\approx 5$

∴ highest order of reflection that can be observed is 5.