Question

A ray of light enters into benzene from air .if the refractive inden of benzene is 1.50 by what prevent does the speed of light reduce on entering the benezero?

Answer 1

The answer to the question is $=\ 33.3%$

CALCULATION:

As per the question the refractive index of the benzene is given as $\mu=\ 1.50$

Initially the light was travelling in air. After travelling in air, the light enters benzene.

Let the velocity of the light in benzene is v .

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

The refractive index of the benzene is calculated as -

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

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

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

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

Hence, the change in velocity of light  = c - v

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

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

Hence, the percentage of reduction of speed  will be-

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

                                                  $=\ 33.3%$   [ans]

Answer 2

Answer:

33.3 % .

Explanation:

Given :

Refractive index of benzene = 1.50

We know :

Speed of light = 3 × 10⁸ m / sec

We know formula for n :

n = speed of light in vacuum / speed of light in benzene

n = c / v

1.5 = 3 × 10⁸ / v

v = 2 × 10⁸ m / sec

Now ,

Percentage ( % ) decrease = 1  × 10⁸ / 3  × 10⁸  × 100 %

% decrease = 33.3 % .