Question

A ray of light enters into benzene from air. If the refractive index if benzene is 1.50,by what percent does the speed of light reduce on entering the benzene?

Answer 1

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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

refractive index is the ratio of speed in air by speed in medium
as you all know light is fastest in vacuum and only slows down in other medium
this explains why refractive index can't be less than 1
now let speed in benzene be v
c/v = refractive index = 1 .5
where c is speed of light in vacuum = 3*10^8 m/s
so v = c*2/3
v= 2 * 10^8
now reduction is final - initial = c-v= 10^8
percentage reduction = (10^8/3*10^8 )*100
= 1/3*100
= 33.33%