Question

The speed of light in a transparent medium is $2.44\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. Find the refractive index of the medium?

Answer 1

Step 1: Given data

Speed of the light in the transparent medium, $\nu =2.44\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

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 transparent medium

$$\begin{gathered} n=\frac{Speedoflightinvacuum\left(c\right)}{Speedoflightinanymedium\left(\nu \right)} \\ \Rightarrow n=\frac{3\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{2.44\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}} \\ \Rightarrow n=1.23 \end{gathered}$$

Hence, the refractive index of the transparent medium is $n=1.23$.