The X - Y plane be taken as the boundary between two transparent...

Given: \overset{\to}{A} = 4\sqrt{3i} - 3\sqrt{3j} - 5\overset{^}{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

$i = \cos^{- 1}\left(\dfrac{{A}_{z}}{A}\right) = \cos^{- 1}\left(\dfrac{5}{\sqrt{( 4\sqrt{3}{)}^{2} + ( 3\sqrt{3}{)}^{2} + {5}^{2}}}\right)$

$= \cos^{- 1}\left(\dfrac{5}{10}\right) ⇒ i = 60 °$

Applying Snell's law of refraction at the interface of two media

${\mu}_{1}sini = {\mu}_{2}sinr$

$⇒ \sqrt{2}\sin 60 ° = \sqrt{3} \times sinr$

$⇒ r = 45 °$

Difference between $i$ and $r$ $= 60 ° - 45 ° = 15 °$