Physics - Geometrical Optics Question with Solution | TestHub

As we know, angle of minimum deviation is given by, $\Rightarrow \mu =\frac{\sin {\displaystyle \frac{\left({\delta }_m+A\right)}{2}}}{\sin {\displaystyle \frac{A}{2}}}$

$\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin {\displaystyle \frac{\left({\delta }_m+A\right)}{2}}}{\sin {\displaystyle \frac{A}{2}}}$

$\Rightarrow \frac{\cos \left({\displaystyle \frac{A}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)}=\frac{\sin {\displaystyle \frac{\left({\delta }_m+A\right)}{2}}}{\sin {\displaystyle \frac{A}{2}}}$

$\Rightarrow \sin \left(90°-\frac{A}{2}\right)=\sin \left(\frac{{\delta }_m+A}{2}\right)$

Therefore, $90°-\frac{A}{2}=\frac{{\delta }_m+A}{2}$  $\Rightarrow {\delta }_m=180°-2\mathrm{A}$