a ray of light enters into benzene from air.If the refractive index of benzene is 1.5,by what percent does the speed of light reduces on entering the benzene?
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Answered by
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We know speed of light in vacuum is 3 x 10^(8) m/s ,
and we know refractive index of benzene = 1.5
So using the absolute refractive index formula :-
n = c/v ( c is speed of light in vacuum and v is speed of light in medium)
=> 1.5 = 3 x 10^(8)/ v
Thus, v = 2 x 10^(8) m/s
Therefore speed of light in benzene reduces by a factor of 2 x 10^(8) m/s
sinhaaaryan27:
Oops! almost forgot, if u need to find the % of decrease all u need to do is Change/ Actual speed x 100, So it is {2 x 10^(8) / 3 x 10^(8) } x 100 , which is 200/3 = 66.667%
Oops! my mistake.... its not 10^(8) but 10^(6)...... just change 10^(8) to 10^(6)..... wont make a difference though.....but anyways the % looks correct
Answered by
0
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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