Physics - Geometrical Optics Question with Solution | TestHub

As we know, $\mu =\frac{\sin \left({\displaystyle \frac{A+{\delta }_m}{2}}\right)}{\sin {\displaystyle \frac{A}{2}}}$

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

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

$\Rightarrow \frac{\pi }{2}-\frac{A}{2}=\frac{A}{2}+\frac{{\delta }_m}{2}$

$\Rightarrow {\delta }_m=\pi -2A=180°-2A$