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

The speed of light in air is c = $3\times 10^{8}\ m/s$

The refractive index of the benzene $[\mu]\ =\ 1.50$

Let the velocity of light in benzene is v.

The refractive index of the benzene is defined as -

                                  $\mu\ =\ \frac{velocity\ of\ light\ in\ air}{velocity\ of\ light\ in\ benzene}$

                                  $i.e\ \mu\ =\ \frac{c}{v}$

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

                                        $=\ \frac{3\times 10^{8}} {1.5}\ m/s$

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

Hence, the decrease in value of speed dv = c - v

                                                                      = $(3\times 10^8\ -\ 2\times 10^8)\ m/s$

                                                                      = $1\times 10^8\ m/s$

The percentage loss of speed = $\frac{dv}{c}\times 100$

                                                   = $\frac{1\times 10^8}{3\times 10^8}\times 100$

                                                  = $0.333\times 100$

                                                  = 33.3%

Hence, the speed of light will be reduced by 33.3 % when it will enter the benzene.

Answer 2

Given:

Refractive index = 1.50

Light enters benzene.

To find:

The reduction in the speed of the light.

Solution:

By formula,

Refractive index = c / v

Where,

c - speed of light in air

c = 3 * 10^8 m / s

v -  speed of light in benzene.

v = c / Refractive index

Substituting,

We get,

v = 3 * 10^8 / 1.50

Hence, the speed of the light in benzene = 2 * 10^8 m / s

Difference in speeds =  3 * 10^8 - 2 * 10^8 / 3 * 10^8

Hence, the speed reduces by 33%  

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