$(a)\;-h \\ (b)\;\frac{h}{\mu} \\ (c)\;\frac{h}{\sqrt {\mu-1}} \\ (d)\;\frac{h}{\sqrt{\mu^2-1}} $

Rays coming out of source and incident at an angle greater than $\theta_c$ will be reflected back into liquid therefore, the corresponding region on surface will appear dark.

As is obvious from the figure the radius of bright spot is

$R= h \tan \theta_c=\large\frac{h \sin \theta_c}{\cos \theta_c}$

or $R= \large\frac{h \sin \theta}{\sqrt {1- \sin ^2 \theta _c}}$

Since $\sin \theta _c =\large\frac{1}{\mu}$

$\therefore R= \large=\frac{h}{\sqrt {\mu^2-1}}$

Hence d is the correct answer.

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