A Ray of light enters into benzene from air if the refractive index of benzene is 1.50 by what percent does the speed of light reduce on entering the Benzene
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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%
shiziii3:
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Answered by
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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 % .
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