Question 97 of 295

The $X-Y$ plane be taken as the boundary between two transparent media ${M}_{1}$ and ${M}_{2}$. ${M}_{1}$ in $Z\geq 0$ has a refractive index of $\sqrt{2}$ and ${M}_{2}$ with $Z\lt 0$ has a refractive index of $\sqrt{3}$. A ray of light travelling in ${M}_{1}$ along the direction given by the vector $\vec{A}=4\sqrt{3i}-3\sqrt{3}j-5\hat{k}$, is incident on the plane of separation. The value of difference between the angle of incident in ${M}_{1}$ and the angle of refraction in ${M}_{2}$ will be _____ degree.

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Answer: 15

Given: $\vec{A}=4\sqrt{3i}-3\sqrt{3j}-5\hat{k}$

As incident vector $A$ makes $i$ angle with normal $Z$-axis and refracted vector $R$ makes $r$ angle with normal $Z$- axis.
The angle made by the incident ray with the normal is given as
$i={\cos }^{-1}\left(\dfrac{{A}_{z}}{A}\right)={\cos }^{-1}\left(\dfrac{5}{\sqrt{(4\sqrt{3}{)}^{2}+(3\sqrt{3}{)}^{2}+{5}^{2}}}\right)$
$={\cos }^{-1}\left(\dfrac{5}{10}\right)\Rightarrow i=60^{\circ}$
Applying Snell's law of refraction at the interface of two media
${\mu }_{1}\sin i={\mu }_{2}\sin r$
$\Rightarrow \sqrt{2}\sin 60^{\circ}=\sqrt{3}\times \sin r$
$\Rightarrow r=45^{\circ}$
Difference between $i$ and $r$ $=60^{\circ}-45^{\circ}=15^{\circ}$