Question

A cubical block of glass of refractive index $n_1$ is in contact with the surface of water of refractive index $n_2.$ A beam of light is incident on vertical face of the block (see figure). After refraction, a total internal reflection at the base and refraction at the opposite vertical face, the ray emerges out at an angle $\theta$. The value of $\theta$ is given by

• A. $sin\theta <\sqrt{n_1^2-n_2^2}$
• B. $tan\theta <\sqrt{n_1^2-n_2^2}$
• C. $sin\theta <\frac{1}{\sqrt{n_1^2-n_2^2}}$
• D. $tan\theta <\frac{1}{\sqrt{n_1^2-n_2^2}}$

Answer 1

The correct option is A $sin\theta <\sqrt{n_1^2-n_2^2}$

Let us consider the angle of incidence at glass air medium be $\alpha$
applying Snell's law
$n_{1}sin\alpha =1\times sin\theta$

$sin\theta =n_{1}sin\alpha$ $\to 1$

Now angle of incidence at water glass medium will be $90-\alpha$

Here the ray is refracted back , total internal reflection has took place
so let the $90-\alpha$ be the critical angle

Applying snell's law

$n_{1}sin(90-\alpha )=n_{2}sin90$

$cos\alpha =\frac{n_2}{n_1}$

Substitute this in $e{q}_1$

$sin\theta =n_1\frac{\sqrt{n_2^2-n_1^2}}{n_1}=\sqrt{n_2^2-n_1^2}$

This tis the maximum value of $sin\theta$ as we have assumed critical angle

Hence $sin\theta <\sqrt{n_2^2-n_1^2}$