Question

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 $Z <0$ 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

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$