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
A. 1
B. 1.5
C. 3
D. Sq. Root(1.5)

Answer 1

Answer:

The correct option is (B) 1.5.

Step-by-step explanation:

Given:

• Angle with which the plane wave incidents on the dielectric material with respect to its normal, $\theta_i= 60^\circ.$.

• Dielectric constant of the given dielectric = $\epsilon_r$.
• Angle with which the transmitted wave propagates in the dielectric with respect to its normal, $\theta_t=45^\circ.$

According to the Snell's law,

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

where,

$n_1,\ n_2$ are the refractive indices of the media from where light incidents and through which the light propagates after transmitting respectively.

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

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

Therefore,

$n_2 = \sqrt \epsilon_r$.

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

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

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

Thus, option B is correct.