Question

A green light is incident from the water to the air – water interface at the critical angle . Select the correct statement
(A) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
(B) The entire spectrum of visible light will come out of the water at various angles to the normal.
(C) The entire spectrum of visible light will come out of the water at an angle of .
(D) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

Answer 1

Hint : To answer this question, we need to use the basic formula of the refractive index to calculate its expression in terms of the wavelength. Then using it we can determine the expression for the critical angle. Then comparing the wavelengths of the various components of the visible light we can compare the critical angles for each of them, and hence conclude our answer..

Formula used: The formulae used for solving this question are given by
$\mu = \dfrac{c}{v}$, here $\mu $ is the refractive index of a medium in which the velocity of light is equal to $v$, and $c$ is the velocity of light in vacuum.
${i_c} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$, here ${i_c}$ is the critical angle of incidence when it travels from a medium of refractive index $\mu $ to the air.
$v = \lambda f$, here $v$ is the velocity of a light having the wavelength of $\lambda $, and frequency of $f$.

Complete step by step answer
We know that the refractive index of a medium is the ratio of the velocity of light in vacuum to the velocity of light in that medium, that is,
$\mu = \dfrac{c}{v}$.......................(1)
Now, we know that the velocity of light is given by
$v = \lambda f$.......................(2)
Substituting (2) in (1) we get
$\mu = \dfrac{c}{{\lambda f}}$.......................(3)
Now, according to the question, the light is travelling from the water into the air. We know that the frequency of light remains constant and does not change with medium. Also the speed of light is a constant. So from (3) we can write
$\mu = \dfrac{k}{\lambda }$, where $k$ is a constant.......................(4)
Now, we know that the critical angle is given by
${i_c} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)$
From (4)
${i_c} = {\sin ^{ - 1}}\left( {\dfrac{\lambda }{k}} \right)$
From the above relation, the critical angle of incidence increases with the increase in the wavelength. So the spectrum of visible light whose wavelength is more than the green light will have a critical angle more than that of the green light. As the wavelength is inversely proportional to the frequency, so we can say that the spectrum of visible light whose frequency is less than that of green light will have a critical angle more than that of the green light.
According to the question, the green light is incident at its critical angle of incidence at the air water interface. So the spectrum of the visible light whose frequency is less than that of green light will come out of the interface.
Hence, the correct answer is option D.

Note
The phenomenon discussed in this question is called the total internal reflection. It occurs only when light travels from an optically denser to an optically rarer medium. An important application of this phenomenon is the optical fibre cables used for the communication purpose.