Question

If a distance of ${x_1}$ in a rarer medium and a distance ${x_2}$ in a denser medium contain the same number of waves of a given monochromatic light, then the refractive index of the denser medium relative to the rarer medium is:
A) $\dfrac{{{x_2}}}{{{x_1} - {x_2}}}$
B) $1 - \dfrac{{{x_2}}}{{{x_1}}}$
C) $\dfrac{{{x_2}}}{{{x_1}}}$
D) $\dfrac{{{x_1}}}{{{x_2}}}$

Answer 1

Hint: Before we understand the question, it is important to understand the phenomenon of refraction. Refraction is defined as the phenomenon in which there is a change in the direction of propagation of a light wave when light enters another medium of different optical density. It is important to understand the mathematical expression of refraction to solve this problem.

Complete step by step answer:
The phenomenon of refraction is governed by the Snell’s law of refraction, which states that–
The ratio of sine of angle of incidence to the sine of angle of refraction is always constant and is equal to the refractive index of the pair of media.
If $i$ is the angle of incidence and $r$ is the angle of refraction of a light wave travelling from a rarer medium 1 to a denser medium 2, as per Snell’s law, we have –
$\dfrac{{\sin i}}{{\sin r}} = {n_{21}}$
where ${n_{21}} = \dfrac{{{n_2}}}{{{n_1}}}$; ${n_2}$ and ${n_1}$ are the absolute refractive indices of the media with respect to air.
The velocity of a wave is defined as the product of frequency and its wavelength.
$v = f\lambda $
During refraction, there is a change in the direction of the light wave. Hence, we can say that there is a change in the velocity. However, the change in the velocity is due to the change in the wavelength and not the frequency since the frequency of the wave depends on the source and never changes in the journey of the wave.
The refractive index can also, be written as –
$\Rightarrow {n_{21}} = \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$
where ${\lambda _1}$ and ${\lambda _2}$ are the wavelengths of light wave in the medium 1 and 2 respectively.
The number of waves per unit length is given by the quantity called wave number, represented by $x$ where –
$\Rightarrow x = \dfrac{1}{\lambda }$
$ \Rightarrow \dfrac{{{x_1}}}{{{x_2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$
Thus, substituting the value of wave number in the equation of refractive index, we have –
$\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}} = \dfrac{{{x_1}}}{{{x_2}}}$
Therefore, the refractive index of the denser medium relative to the rarer medium,
$\Rightarrow {n_{21}} = \dfrac{{{n_2}}}{{{n_1}}} = \dfrac{{{x_1}}}{{{x_2}}}$

Hence, the correct option is Option D.

Note: The material with the lowest refractive index is air, whose value is equal to 1. The material with the highest value of refractive index is diamond, whose value is about 2.4. Thus, diamond is regarded as the material with the highest optical density, which contributes greatly to its characteristic lustre.