a ray of light entered into beneze from air. if the reflective index of beneze is 1.50 by what percent does the speed of light reduce on entering the beneze.
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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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