The critical angle of medium for specific wavelength, if the medium has relative permittivity $$3$$ and relative permeability $$\dfrac{4}{3}$$ for this wavelength, will be:
Given that $$\in_e = 3$$ and $$\mu_r = \dfrac{4}{3}$$
Let $$\mu_1, \mu_2$$ be refractive indices of force space and the medium.
We know $$\mu_r = \dfrac{\mu}{\mu_0}$$ and $$\epsilon_r = \dfrac{\epsilon}{\epsilon_0}$$
and $$C = \dfrac{1}{\sqrt{\nu_0 \in_0}}$$, $$V = \dfrac{1}{\sqrt{\mu \epsilon_e}}$$
Where $$C$$ is the speed of light in vacuum and $$v$$ is the speed of of light in the medium.
Refractive index of the medium $$\mu_2 = \dfrac{c}{v} = \sqrt{\dfrac{\mu \epsilon}{\mu_0 \epsilon_0}}$$
$$= \sqrt{\mu_r\epsilon_r} = \sqrt{3\times \dfrac{4}{3}}$$
$$=2$$
From Sneii's Law,
$$\mu_2 \sin \theta_i = \mu_1\sin \theta_r$$
At critical angle $$\theta_r = 90^o$$
Let $$\theta_c$$ be the incident angle at that moment.
$$\therefore \mu_2 \sin \theta_c= 1 \times 1 = 1$$
$$2 \sin \theta_c = 1$$, $$\sin \theta_c = \dfrac{1}{2}$$
$$\theta_c = \sin^{-1}\left(\dfrac{1}{2}\right) = 30^o$$
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