question_answer

A ray of light falls on a transparent glass slab of refractive index 1.52. If the reflected ray and refracted ray are mutually perpendicular, then the angle of incidence is

A) \[58{}^\circ \text{ }15'\]

B)        \[57{}^\circ \text{ }19'\]

C) \[58{}^\circ \text{ }19'\]

D)        \[57{}^\circ \text{ }12'\]

Correct Answer: C

Solution :

Given\[\angle BOC={{90}^{o}}\] \[\therefore \]  \[\angle i+{{90}^{o}}+\angle i={{180}^{o}}\] \[\Rightarrow \]               \[\angle (i+r)={{90}^{o}}\] \[\Rightarrow \]               \[r=(90-i)\] Now      \[\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin (90-i)}\]                 \[=\frac{\sin i}{\cos i}=\tan i\] \[\therefore \]Angle of incidence,\[i={{\tan }^{-1}}(\mu )\]                 \[={{\tan }^{-1}}(1.52)\]                 \[\mathbf{=5}{{\mathbf{8}}^{\mathbf{o}}}\mathbf{19'}\]