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
Solution
$$\mu = {{\sin \left( {{{A + {\delta _{\min }}} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$$<br><br>$$\mu = {{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$$<br><br>$\mu = {{\sin A} \over {\sin {A \over 2}}} = 2\cos {A \over 2}$<br><br>$A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$
About this question
Subject: Physics · Chapter: Optics · Topic: Refraction and Snell's Law
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