Question

An em wave of frequency $\nu$ = 3.0 MHz passes from vacuum into dielectric medium with $\epsilon_r$ = 4.0. Then
A. wavelength $\lambda$ is halved and frequency $\nu$ remains unchanged
B. $\lambda$ doubles and $\nu$ becomes half
C. $\lambda$ is doubled and $\nu$ remains same
D. $\lambda$ and $\nu$ both remain unchanged.

Answer 1

Hint: As the wave travels in a medium with different refractive index, refraction occurs and speed of the light in that medium changes. The refractive index is the ratio of velocity of light in vacuum to velocity of light in medium.

Formula used:
Refractive index can be written as:
$\eta = \dfrac{c}{v} = \dfrac{1}{\sqrt{\mu_r \epsilon_r}} $
The velocity of light in a medium can be written as product of wavelength and frequency i.e.,
$v = \lambda \nu$

Complete step by step answer:
We know that relative permittivity of a medium is defined as the ratio of its permittivity in medium to permittivity of the vacuum:
$\epsilon_r = \dfrac{\epsilon}{\epsilon_0}$.
This relation can be transformed as:
$\epsilon = \epsilon_r \epsilon_0$
Same will be the case for permeability.
The relation between speed of light in a medium with the permittivity and permeability can be written as:
$v = \dfrac{1}{\sqrt{\mu \epsilon}}$
or we may write:
$v = \dfrac{1}{\sqrt{\mu_0 \epsilon_0 \mu_r \epsilon_r}}$ .
For the case of vacuum velocity is c,
$\epsilon_r$ = 1 and $\mu_r$ = 1, so we have:
$c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0 }}$.
So, we can write the formula for v as:
$v = \dfrac{c}{\sqrt{\mu_r \epsilon_r}}$
We are given $\epsilon_r$ = 4 and we keep $\mu_r$ = 1, we get:
$v = \dfrac{c}{\sqrt{4}} = \dfrac{c}{2}$.

Now, we consider the relationship between velocity wavelength and frequency:
For vacuum:
$c = \lambda \nu$
For medium, we suppose:
$v = \lambda ' \nu$
as the frequency will not change in going from vacuum to the media.

Making substitutions in the velocity formula we derived previously, we get:
$\lambda ' \nu = \dfrac{\lambda \nu}{2}$
$\lambda ' = \dfrac{\lambda}{2}$;
which means that the wavelength in the medium has been halved and the frequency remains the same.

So, the correct answer is “Option A”.

Note:
Our assumption that relative permeability is equal to 1 helped us in getting the right answer. It happens so that most substances have permeability same as that of permeability of free space.