Chapter 23, Problem 38P

To determine

The refractive index of the liquid.

Answer to Problem 38P

Solution:

The refractive index of the liquid $n_{liquid}=1.36$

Explanation of Solution

Given Info:

Let $n_{liquid}$ be the refractive index of liquid

Refractive index of air $n_{air}=1.0$

Let the angle of incident (critical angle)be $\angle i=\angle c={47.2}^0$

Angle of refraction $\angle r={90}^0$

Formula Used:

Apply Snell’s Law

$\begin{array}{l}{n}_{liquid}\sin \angle i=n_{air}\sin \angle r\\ n_{liquid}\sin \angle c=n_{air}\sin {90}^0\end{array}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law for liquid-air surface

$\begin{array}{l}{n}_{liquid}\sin \angle i=n_{air}\sin \angle r\\ n_{liquid}\sin \angle c=n_{air}\sin {90}^0\\ or,n_{liquid}=\frac{n_{air}\sin {90}^0}{\sin \angle c}\\ or,n_{liquid}=\frac{1.0\sin {90}^0}{\sin {47.2}^0}\\ or,n_{liquid}=1.36\end{array}$

Hence, the refractive index of the liquid $n_{liquid}=1.36$