Question

A uniform plane wave in air (refractive index = 1) is incident at an angle of 60∘ with respect to normal on a lossless dielectric material with dielectric constant ϵr. The transmitted wave propagates at an angle 45∘ with respect to normal. The value of ϵr is

Answer 1

Answer:

$\epsilon_r = 1.5$.

Explanation:

Given:

• Incidence angle of the plane wave on the surface of dielectric material with respect to its normal, $\theta_i= 60^\circ.$

• Dielectric constant of the given dielectric = $\epsilon_r.$

• Transmitted angle of the wave with which it propagates in the dielectric with respect to its normal, $\theta_t=45^\circ.$

The Snell's law states that,

$n_1\sin\theta_i=n_2\sin\theta_t.$

where,

• $n_1\ \text{and}\ n_2$ = the refractive indices of the air and the dielectric respectively.

The refractive index of air, $n_1 = 1$.

The dielectric constant of a material ( medium ) is related with its refractive index as

$\text{Refractive index}=\sqrt {\text{Dielectric constant}}$

Therefore,

$n_2 = \sqrt \epsilon_r.$

Putting these values in the expression of the Snell's law, we get,

$1\cdot \sin(60^\circ)=\sqrt \epsilon_r \sin(45^\circ) \\\sqrt \epsilon_r = \dfrac{\sin(60^\circ)}{\sin(45^\circ)}\\=\dfrac{\dfrac{\sqrt 3}2}{\dfrac 1{\sqrt 2}}\\=\dfrac{\sqrt 3}{2}\times \dfrac{\sqrt 2}{1}\\=\sqrt {\dfrac{ 3}{2}}\\\text{On squaring both the sides, we get, }\\\epsilon_r = \dfrac 32\\ = 1.5$