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A light ray enters into a medium whose refractive index varies along x-axis as $y(x) = \mu_{circ} \sqrt {1+ 0.25 x}.$ Where $\mu_0=1$.The medium is bounded by the planes $x=0,x=1\;cm$ and $y=0$ . If the ray enters at an angle $\theta=30^{\circ}$ with horizontal . Then the trajectory of light ray is


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

1 Answer

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.
answered Feb 20, 2014 by meena.p

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