Physics, asked by srikanth2783, 1 year ago

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 benzrne​

Answers

Answered by manavjaison
1

Refractive index, n = v_{air} / v_{medium}

Now,

Refractive index of benzene = 1.50

and,

Speed of light in air / vacuum = 3 × 10^{8} m/s

So,

v_{medium} = \frac{3 * 10^{8} }{1.5}  m/s

                        = 2 *  10^{8} m/s  

Now,

The percentage decrease in the speed of light =

= 3 * 10^{8} - 2 * 10^{8} / 3 * 10^{8}   * 100 %

= 33.3 %

Hence,

The percentage decrease when light enters benzene from air is 33.3 %

Answered by BendingReality
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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