Question

a ray of light enter into benzene from air. if the refractive index of benzene is 1.50 by what percent is the speed of light reduce on entering the Benzene

Answer 1

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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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 % .