Question

A ray of light enters into benzene from air. If the refractive index of benzene is $1.50$. By what percent does the speed of light reduce on entering the benzene?

Answer 1

Step 1 : Concept :

1. The refractive index is defined as the ratio of the speed of light in a vacuum to the speed of the light in the medium.
2. Refractive index is denoted by $\mu$.
3. The mathematical expression for refractive index is $\mathrm{\mu }=\frac{\mathrm{c}}{\mathrm{v}}$.
4. The refractive index determines how much the light changes its path when it travels from one medium to another medium.

Step 2 : Calculation :

We have been given that the refractive index of benzene is $1.50$.

The formula to calculate the speed of light in benzene is :

$\mathrm{v}=\frac{\mathrm{c}}{\mathrm{\mu }}$

Where $\mathrm{v}$ = speed of light in medium, $\mathrm{c}$ = speed of light in vacuum = $3\times {10}^8\mathrm{m}/\mathrm{s}$, $\mathrm{\mu }$ = refractive index of medium.

Substituting the values in the formula :

$\mathrm{v}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{1.50}$

$\mathrm{v}=2\times {10}^8\mathrm{m}/\mathrm{s}$

Thus, the change in velocity of light is = $\mathrm{c}-\mathrm{v}$

The change in velocity of light is = $3\times {10}^8\mathrm{m}/\mathrm{s}-2\times {10}^8\mathrm{m}/\mathrm{s}$

The change in velocity of light = $1\times {10}^8\mathrm{m}/\mathrm{s}$

Hence, the percentage in speed reduction will be :

$\frac{1\times {10}^8\mathrm{m}/\mathrm{s}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\times 100=33.3\%$

The speed of light reduce by $33.3\%$ on entering the benzene.