Question

A prism of refractive index m and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of m is –

• A. sin^–1$\left( \frac { \mu } { 2 } \right)$
• B. sin^–1$\left( \sqrt { \frac { \mu - 1 } { 2 } } \right)$
• C. 2cos^–1$\left( \frac { \mu } { 2 } \right)$
• D. cos^–1$\left( \frac { \mu } { 2 } \right)$

Answer 1

Answer: (C)

2cos^–1$\left( \frac { \mu } { 2 } \right)$

Answer 2

m = sin $\frac { \sin \left( \frac { A + \delta _ { m } } { 2 } \right) } { \sin \frac { A } { 2 } }$ , d_m = A

\ m = $\frac { \sin \mathrm { A } } { \sin \mathrm { A } / 2 }$ = 2cos A/2

\ A = 2cos^–1(m/2)