Physics - Geometrical Optics Question with Solution | TestHub

The refractive index of a prism is given by

$\mu =\frac{\sin \left(\frac{A+5_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Where, A = angle of prism

${\delta }_m=$angle of minimum deviation

Given,$\mu =\cot \left(\frac{A}{2}\right)=\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

So, from equation (i)

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

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

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

$\Rightarrow {\delta }_m=180°-2A$