Question

The speed of light in diamond is found to be $124,000\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. What is the refractive index of diamond?

Answer 1

Step 1: Given data

Speed of the light in the diamond,

$$\begin{gathered} \nu =124,000\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ \nu =124,000\times 1000\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ \nu =1.24\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Consider, speed of the light in the vacuum, $c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2: Formula used

Refractive index, $n=\frac{Speedoflightinvacuum\left(c\right)}{Speedoflightinanymedium\left(\nu \right)}$

Step 3: Refractive index of diamond

$$\begin{gathered} n=\frac{Speedoflightinvacuum\left(c\right)}{Speedoflightindiamond\left(\nu \right)} \\ \Rightarrow n=\frac{3\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{1.24\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}} \\ \Rightarrow n=2.5 \end{gathered}$$

Hence, the refractive index of the diamond is $n=2.5$.