Question

If a material possesses relative permittivity $3$ and relative permeability $\frac{4}{3}$ for a given wavelength, the critical angle for that wavelength is.

• A. $15°$
• B. $30°$
• C. $45°$
• D. $60°$

Answer 1

The correct option is B

$30°$

Step 1: Given

Relative permittivity $=3$

Relative permeability $=\frac{4}{3}$

Step 2: Formula used

Note that if the speed of light in the given medium is $V$then,

$V=\frac{1}{\sqrt{\left(\mu \in \right)}}$, where $\mu =$permittivity, and $\in =$epsilon

Critical Angle, $\sin {\theta }_c=\frac{n_r}{n_i}$, where $n_r=$refractive index, and $n_i=$incident index

Step 3: Determine the critical angle

We know that $n=\frac{c}{v}$ , where $n=$is the index of refraction, $c=$speed of light in vacuum, and $v=$speed of light in that medium and

$\begin{array}{l}\Rightarrow n=\sqrt{\left({\mu }_r{\in }_r\right)}\end{array}\dots \dots \dots \dots \dots \left(1\right)$

Substitute the known values in equation $\left(1\right)$.

$$\begin{gathered} n=\sqrt{3\times \frac{4}{3}} \\ n=2 \end{gathered}$$

As we know critical angle, $\sin {\theta }_c=\frac{n_r}{n_i}$. So.

$\begin{array}{rcl}\sin {\theta }_c& =& \frac{1}{2}\\ & =& {30}^{°}\end{array}$

Therefore the critical angle for wavelength is ${\theta }_c=30°$.

Hence, option B is correct.