The refractive index of the medium is:

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(D) From the given figure,the light ray is incident at the interface of the medium and air at an angle $	heta$ and grazes the surface,which means $

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$\sqrt{3}$

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(D) From the given figure,the light ray is incident at the interface of the medium and air at an angle $	heta$ and grazes the surface,which means $	heta$ is the critical angle $(C)$.In the right-angled triangle formed by the light ray,the vertical depth,and the horizontal distance,the hypotenuse is $\sqrt{3} \ m$ and the base is $1 \ m$.Using trigonometry,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{1}{\sqrt{3}}$.The refractive index $\mu$ of the medium with respect to air is given by $\mu = \frac{1}{\sin C} = \frac{1}{\sin 	heta}$.Substituting the value of $\sin 	heta$,we get $\mu = \frac{1}{1/\sqrt{3}} = \sqrt{3}$.

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