Question

Light enters from air to kerosene having a refractive index of 1.47. What is the speed of light in kerosene, if the speed of light in air is $3 \times {10^8}$ m/s?

Answer 1

Hint: The refractive index of a medium can be defined as the ratio of the velocity of light in vacuum to the velocity of light in the given medium. Two of these values are given to us and third can be calculated easily.

Complete step-by-step solution -
The phenomenon of bending of light when it travels from one medium to another is called refraction.
Refraction is governed by two laws of refraction given below:
1. The incident and refracted light ray and the normal drawn perpendicular at the point of incidence, all lie in the same vertical plane.
2. If we take ratio of the sine of the angle of incidence to the sine of the angle of refraction, we get a quantity called the refractive index which decides the amount of refraction of light and is given as
$\mu = \dfrac{{\sin i}}{{\sin r}} = \dfrac{{{\text{Velocity of light in air or vacuum}}}}{{{\text{Velocity of light in a medium}}}}{\text{ }}...{\text{(i)}}$
where $\mu $ is called the refractive index, i is the angle of incidence and r is the angle of refraction.
We are given that the refractive index of kerosene has the following value
$\mu = 1.47$
We are also given that the velocity of light in air or vacuum is
$c = 3 \times {10^8}m/s$
Now, we need to calculate the velocity of light in kerosene which can be calculated using given information from equation (i) as follows:
$
  \mu = \dfrac{c}{{\text{v}}} \\
  \therefore {\text{v}} = \dfrac{c}{\mu } \\
$
Substituting the given values, we get
${\text{v}} = \dfrac{{3 \times {{10}^8}}}{{1.47}} = 2.04 \times {10^8}m/s$
This is the required answer.

Note: The student should note that the refractive index of any medium can never be less than 1 because in equation (i), we can see that no velocity can be faster than the velocity of light which will always give us $\mu \geqslant 1$.