A Ray of light enters into 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
Answers
Refractive index of a medium 1 wrt medium 2 is always equal to the ratio of speed of light in medium 1 to that of speed of light in medium 2
n(C6H6) = c(vacuum)/c(C6H6)
Since,
Medium 1 is vacuum and Medium 2 is C6H6
Given
n of benzene = 1.50
c( speed of light in vacuum) = 3 x 10^8 m/s
c(C6H6) = c(air)/n(C6H6)
= 3 x 10^8/1.5 = 2 x 10^8
Reduction in speed of the light entering to the medium = Actual speed of light in air - Speed of light entering benzene
= 3 x 10^8 m/s - 2 x 10^8 m/s = ( 3 - 2 ) x 10^8 m/s = 10^8 m/s
Percentage of decrease in speed of light = Decrease in speed of light/ actual speed pf light x 100 = 10^8/3 x 10^8 x 100
=100/3 = 33.3%
Therefore, the speed of light is reduced by 33.3%.
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 % .