First slide

Refraction of light

Question

The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $\frac{4}{3}$ for this wavelength, will be:

Easy

A

$45^{0}$
B

$15^{0}$
C

$60^{0}$
D

$30^{0}$

Solution

$\begin{array}{l} R e f r a c t i v e i n d e x o f a m e d i u m = n = \sqrt{μ_{r} ε_{r}} = \sqrt{\frac{4}{3} \times 3} = 2 \\ L e t s a y c = c r i t i c a l a n g l e, \\ \sin c = \frac{1}{n} = \frac{1}{2} \\ \Rightarrow c = 30^{0} \end{array}$

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