The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability \( \displaystyle\frac{4}{3} \) for this wavelength, will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30°
Light travels in a medium with speed \( v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_r\varepsilon_r}} \), so the refractive index of the medium (relative to vacuum) is \( n = c/v = \sqrt{\mu_r\varepsilon_r} \).
\( \displaystyle n = \sqrt{3\times\frac{4}{3}} = \sqrt{4} = 2 \)
The critical angle is the angle of incidence in the medium for which the refracted ray grazes the surface, \( \sin\theta_c = 1/n \) (the other medium being vacuum or air).
\( \displaystyle \sin\theta_c = \frac{1}{2} \Rightarrow \theta_c = 30^\circ \)