$(a)\;y=\sqrt x+3 \\ (b)\;y= 2[ \sqrt {x+3} -\sqrt 3] \\ (c)\;y=2 \sqrt {x+3} \\ (d)\;y=2\sqrt 3 $

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Assume a vertical strip of thickness dx at a distance x as shown.

The slope of trajectory of ray at P is

$\large\frac{dy}{dx}=$$\tan \theta=\large\frac{\sin \theta}{\sqrt {1- \sin ^2 \theta}}$

According to Snell's law,

$\mu \sin \theta =(\mu_0) \sin \theta _0 =constant$

or $ \sin \theta =\large\frac{(1) \sin 30^{\circ}}{\mu}$

$\qquad= \large\frac{1}{2 \mu}$

$\large\frac{dy}{dx} =\large\frac{1}{\sqrt {4y^2-1}}$

$\therefore \mu =\mu_0 \sqrt {1+0.25 x}$

$\therefore \large\frac{dy}{dx}=\frac{1}{\sqrt {4(1+0.25 x)-1}}$

$\qquad= \large\frac{1}{\sqrt {3+x}}$

or $dy =\large\frac{dx}{\sqrt {3+x}}$

Integrating $y= 2 \sqrt {3+x}+c$

At $ x=0,y=0$

$c=-2 \sqrt 3$

$ y= 2[ \sqrt {x+3} -\sqrt 3]$

Hence b is the correct answer.

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