Question

A water film is formed on a glass-block. A light ray is incident on water film from air at an angle of ${60}^{\circ }$ with the normal. The angle of incidence on glass slab is
(${\mu }_g=1.5,{\mu }_w=\frac{4}{3}$)

• A. ${sin}^{-1}\left(\frac{3\sqrt{3}}{8}\right)$
• B. ${sin}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. ${sin}^{-1}\left(\frac{4\sqrt{3}}{9}\right)$
• D. ${sin}^{-1}\left(\frac{9\sqrt{3}}{16}\right)$

Answer 1

The correct option is A ${sin}^{-1}\left(\frac{3\sqrt{3}}{8}\right)$

The angle on incidence on glass slab will be equal to the angle of refraction on the air water interface.

Applying Snell's Law on air water interface, we get

${\mu }_{a}sini={\mu }_{w}sinr$

$1\times sin{60}^0=\frac{4}{3}sinr$

$sinr=\frac{3\sqrt{3}}{8}$

$r={sin}^{-1}\frac{3\sqrt{3}}{8}$