Question

The $X-Y$ plane be taken as the boundary between two transparent media $M_1$ and $M_2$. $M_1$ in $Z\ge 0$ has a refractive index of $\sqrt{2}$ and $M_2$ with $Z<0$ has a refractive index of $\sqrt{3}$. A ray of light travelling in $M_1$ along the direction given by the vector $\vec{A}=4\sqrt{3}\hat{i}-3\sqrt{3}\hat{j}-5\hat{k}$, 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.

Answer 1

Answer: 15

As incident vector A makes $i$ angle with normal z-axis \& refracted vector $R$ makes $r$ angle with normal $z$ - axis with help of direction cosine
$i={\cos }^{-1}\left(\frac{A_z}{A}\right)={\cos }^{-1}\left(\frac{5}{\sqrt{(4\sqrt{3}{)}^2+(3\sqrt{3}{)}^2+5^2}}\right)$
$={\cos }^{-1}\left(\frac{5}{10}\right)\Rightarrow i=60^{\circ }$
$\sqrt{2}\sin 60=\sqrt{3}\times \sin r$
$r=45^{\circ }$
Difference between$i\&r=60-45=15$