Question

arey enters into beneze from air if the refractive index of lens is 1.50 by what percent does the speed of light reduce or entering the beneze

Answer 1

refractive index of lens of benzene =1.5
c=v/u
c = ,3.0×10^8m/s ÷ 1.5 =2.0×10^8m let's calculate percentage
2.0 ×10^8m/s ×100 ÷ 3.0×10^8m/s
=66.6%
if is reduced by;
(100-66.6)%
= 33.3%
light rays speed on passing through benzene from air will reduced by 33.3%
I hope it will help you

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 % .