Let the $x - z$ plane be the boundary between two transparent media. Medium 1 in $z\ \geq \ 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z\

Question

Let the $x - z$ plane be the boundary between two transparent media. Medium 1 in $z\ \geq \ 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z\ < \ 0$ has a refractive index of$\sqrt{3}$. A ray of light in medium 1 given by the vector$\mathbf{A}\ = \ 6\sqrt{3}\mathbf{î} + \ 8\sqrt{3}\mathbf{ĵ}\ –\ 10\ \widehat{\mathbf{k}}$ is incident on the plane of separation. The angle of refraction in medium 2 is

Dr. Abubacker · Accepted Answer

(a) As refractive index for $z > 0$ and $z \leq \ 0$ is different $X - Y$ plane should be boundary between two media
Angle of incidence,
$\cos i = \left| \frac{A_{z}}{\sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}} ight| = \frac{1}{2}$
$	herefore i = 60{^\circ}$
From Snell’s law
$\frac{\sin i}{\sin r} = \frac{\sqrt{3}}{2}$
$\Rightarrow r = 45{^\circ}$

Explanation

Reviewed by Dr. Abubacker

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