Question

determine the coordinates ($x_1,y_1$) of the point P, where the ray intersects the upper surface of the slab-air boundary.

Answer 1

Given: $\mu =[k{y}^{\frac{3}{2}}+1{]}^{\frac{1}{2}}$

$k={1.0}^{\frac{-3}{2}}$

Now lets find x-coordinate;

From Snells Law at initial point;

${\mu }_{air}\sin {90}^0=\mu \sin \theta$

$\sin \theta =\frac{1}{\mu }$

$\cos \theta$$=\sqrt{1-\frac{1^2}{k{y}^{\frac{3}{2}}+1}}$

$=\frac{\sqrt{k{y}^{\frac{3}{2}}}}{\mu }$ ...................(1)

$\cot \theta =\frac{\cos \theta }{\sin \theta }=k^{\frac{1}{2}}.y^{\frac{3}{4}}$ $=\frac{dy}{dx}$
$\frac{dy}{y^{\frac{3}{4}}}=k^{\frac{1}{2}}dx$

Integrating

${\int }_0^y\frac{dy}{y^{\frac{3}{4}}}={\int }_0^x{k}^{\frac{1}{2}}dx$

$y=\frac{k^{{}^2}{x}^4}{256}$

$y=\frac{x^4}{256}$

$x=4m$