Chapter 25, Problem 1OQ

To determine

The critical angle for total internal reflection of a light ray when incident on a liquid to glass surface.

Answer to Problem 1OQ

Option (b).

Explanation of Solution

Write the expression for critical angle.

    ${\theta }_{\text{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$        (I)

Here, ${\theta }_{\text{c}}$ is the critical angle, $n_2$ is the refractive index of lighter medium 2, and $n_1$ is the refractive index of the denser medium 1.

Given that the refractive index of carbon disulfide is $n_1=1.63$, and the refractive index of crown glass is $n_2=1.52$.

Conclusion:

Substitute $1.63$ for $n_1$, and $1.52$ for $n_2$ in equation (I), to find ${\theta }_{\text{c}}$.

    $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1.52}{1.63}\right)\\ =68.8°\end{array}$

Since the critical angle of a light ray when incident on a liquid to glass surface is $68.8°$. option (b) is correct.

The critical angle of a light ray when incident on a liquid to glass surface is $68.8°$. Thus, option (a) is incorrect.

The critical angle of a light ray when incident on a liquid to glass surface is $68.8°$. Thus, option (c) is incorrect.

The critical angle of a light ray when incident on a liquid to glass surface is $68.8°$. Thus, option (d) is incorrect.

The critical angle of a light ray when incident on a liquid to glass surface is $68.8°$. Thus, option (e) is incorrect.