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:

First, we recall the basic connection between the optical refractive index of a non-conducting medium and its electrical properties. The refractive index $$n$$ of a dielectric, when the wavelength is such that absorption is negligible, is given by the relation

$$n=\sqrt{\varepsilon_r\,\mu_r}$$

where $$\varepsilon_r$$ is the relative permittivity (also called the dielectric constant) and $$\mu_r$$ is the relative permeability of the medium.

For the given medium we are told that

$$\varepsilon_r = 3, \qquad\mu_r = \dfrac{4}{3}.$$

We now substitute these numerical values into the formula for $$n$$. We have

$$n = \sqrt{\varepsilon_r \,\mu_r} = \sqrt{ \;3 \times \dfrac{4}{3}\;}.$$

Inside the square root, the factor $$3$$ in the numerator and the factor $$3$$ in the denominator cancel out, leaving

$$n = \sqrt{4}.$$

The square root of $$4$$ is $$2$$, so

$$n = 2.$$

Next, we recall the definition of the critical angle. When light passes from a denser medium (refractive index $$n_1$$) to a rarer medium (refractive index $$n_2$$) and the angle of incidence inside the denser medium is such that the angle of refraction in the rarer medium becomes $$90^\circ$$, that angle of incidence is called the critical angle $$C$$. The mathematical statement of Snell’s law for this limiting case is

$$n_1 \,\sin C = n_2 \,\sin 90^\circ.$$

Because $$\sin 90^\circ = 1$$, we can write

$$n_1 \,\sin C = n_2.$$

If the rarer medium is air (or vacuum) we take $$n_2 \approx 1$$. Here the given medium of refractive index $$n = 2$$ is acting as the denser medium, so $$n_1 = 2$$ and $$n_2 = 1$$. Substituting these values we obtain

$$2 \,\sin C = 1.$$

Dividing both sides by $$2$$ gives

$$\sin C = \dfrac{1}{2}.$$

We know from basic trigonometry that

$$\sin 30^\circ = \dfrac{1}{2}.$$

Therefore

$$C = 30^\circ.$$

Hence, the correct answer is Option B.