Question

A ray of light travelling in glass medium $({\mu }_g=\frac{3}{2})$ is incident on a horizontal glass-air interface at the critical angle ${\theta }_c$. If a thin layer of water $({\mu }_w=\frac{4}{3})$ is now poured on the glass-air interface , then the angle at which the ray of light emerges into water at glass-water surface will be

• A. ${\sin }^{-1}(\frac{3}{4})$
• B. ${\sin }^{-1}(\frac{3}{5})$
• C. ${\cos }^{-1}(\frac{3}{4})$
• D. ${\cos }^{-1}(\frac{3}{5})$

Answer 1

The correct option is A ${\sin }^{-1}(\frac{3}{4})$

For glass -air interface,

$\sin {\theta }_c=\frac{{\mu }_a}{{\mu }_g}$

Substituting the given data we get,

$\sin {\theta }_c=\frac{1}{3/2}\Rightarrow \sin {\theta }_c=\frac{2}{3}.......\left(1\right)$

For glass-water interface, applying Snell's law we get,

${\mu }_g\sin i={\mu }_w\sin r$

$\Rightarrow \frac{3}{2}\times \sin {\theta }_c=\frac{4}{3}\times \sin r$

$\Rightarrow \frac{3}{2}\times \frac{2}{3}=\frac{4}{3}\times \sin r\left[\text{from}\right(1\left)\right]$

$\Rightarrow \sin r=\frac{3}{4}$

$\Rightarrow r={\sin }^{-1}\left(\frac{3}{4}\right)$

Hence, option (a) is the correct answer.

Why this question ? This question helps in building concept required to solve similar JEE mains level problems.