Physics - Geometrical Optics Question with Solution | TestHub

The angle of minimum deviation is given as $${\delta }_{\min }=i+e-A$$ for minimum deviation $$\begin{array}{rl}{\delta }_{\min }& =A\text{ then }\\ 2A& =i+e\end{array}$$ in case of ${\delta }_{\min }i=e$ $$\begin{array}{rl}2A& =2i{r}_1=r_2=\frac{A}{2}\\ i& =A=90^{\circ }\end{array}$$ from smell's law $$\begin{array}{l}1\sin i=n\sin r_1\\ \sin A=n\sin \frac{A}{2}\\ 2\sin \frac{A}{2}\cos \frac{A}{2}=n\sin \frac{A}{2}\\ 2\cos \frac{A}{2}=n\end{array}$$ $$\begin{array}{l}\text{ when }A=90^{\circ }=i_{\min }\\ \text{ then }{n}_{\min }=\sqrt{2}\\ i=A=0n_{\max }=2\end{array}$$