JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 4)
A prism of refractive index $$\mu$$ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index
$$2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$
$${\sin ^{ - 1}}\left( {{\mu \over 2}} \right)$$
$${\sin ^{ - 1}}\left( {\sqrt {{{\mu - 1} \over 2}} } \right)$$
$${\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$
Explanation
$$\mu = {{\sin \left( {{{A + {\delta _{\min }}} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$$
$$\mu = {{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$$
$$\mu = {{\sin A} \over {\sin {A \over 2}}} = 2\cos {A \over 2}$$
$$A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$
$$\mu = {{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$$
$$\mu = {{\sin A} \over {\sin {A \over 2}}} = 2\cos {A \over 2}$$
$$A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$$
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