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# A ray of light falls on a transparent glass slab with refractive index (relative to air) of $1.62$ . The angle of incident for which the reflected and refractive rays are mutually perpendicular is :

$(a)\;\tan ^{-1} (1.62) \\ (b)\;\sin ^{-1} (1.62) \\ (c)\;\cos^{-1} (1.62) \\ (d)\;None\;of \;these$

We know that $\mu= \large\frac{\sin r}{\sin r}$ and $i+r=90^{\circ}$
or $r=90^{\circ}-i$
$\mu= \large\frac{\sin i}{\sin (90-i)}$
$\qquad=\tan i$
or $i= \tan ^{-1} (\mu)$
$\qquad= \tan ^{-1}(1.62)$
Hence a is the correct answer.