Physics - Geometrical Optics Question with Solution | TestHub

Given, the refractive index$=\sqrt{2}$

Prism angle,$A=60°$

For a prism of the position minimum deviation,

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

$\sqrt{2}=\frac{\sin \left(\frac{60^o+{\delta }_m}{2}\right)}{\sin \left(\frac{60^o}{2}\right)}$

$=\frac{\sin \left(\frac{60^o+{\delta }_m}{2}\right)}{\sin {30}^o}$

$\sin \left(\frac{60^o+{\delta }_m}{2}\right)=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{60+{\delta }_m}{2}=45°\Rightarrow {\delta }_m=30°$