Chapter 23, Problem 28P

To determine

The refractive index of substance.

Answer to Problem 28P

Solution:

The refractive index of substance $n_s=1.62$

Explanation of Solution

Given Info:

Speed of light in substance $v_s=80\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$0$}\right.of{v}_w$

Let the speed of light in water be $v_w$

The refractive index of water $n_w=1.33$

Let the refractive index of substance be $n_s$

Formula Used:

The refractive index of any medium $n=\frac{c}{v}$

Here, all alphabets are in their usual meanings.

Calculation:

The refractive index of water $n_w=\frac{c}{v_w}$

$or,v_w=\frac{c}{n_w}-----(i)$

The refractive index of substance is given as

$\begin{array}{l}{n}_s=\frac{c}{v_s}\\ n_s=\frac{c}{82\%v_w}\\ n_s=\frac{c}{82\%\times \left(\frac{c}{n_w}\right)}-----\text{Using equation (i)}\end{array}$

$\begin{array}{l}{n}_s=\frac{100}{82}\times n_w\\ n_s=\frac{100}{82}\times 1.33\\ n_s=1.62\end{array}$