Question

A ray of light enters into benzene from air if the refractive index of benzene is 1.50 then by what percent did the ray of light reduce after entering into benzene

Answer 1

Answer:

Refractive Index = Speed of light in Air / Speed of light in medium

Speed of light in medium = Speed of light in air / Refractive index

=> Speed of light in medium = 3 × 10⁸ / 1.5

=> Speed of light in medium = 2 × 10⁸ m/s

So, Percentage Change = ( Change in Speed / Speed of light in air ) * 100

=> Change in Speed = 3 × 10⁸ - 2 × 10⁸

=> Change in speed = 10⁸ ( 3 - 2 ) = 10⁸

=> Percentage change = ( 10⁸ / 3 × 10⁸ ) * 100

=> Percentage change = ( 1 / 3 ) * 100

=> Percentage change = 33.34 %

This is the required answer !!

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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