Question

a ray of light enters benzene from air , if the refractive index of benzene is 1.50 , by what percentage does the speed of light reduce on entering the benzene????

Answer 1

Let c be speed of light in vaccum and v be speed of light in benzene.

Refractive index of benzene=1.5=speed of light in vaccum/speed of light in benzene.

c/v=1.5

v=2/3c

Percentage reduce in speed of light

=(c-v/c)*100

=(1/3)*100=33.33%

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 % .