Light travels in two media $$M_1$$ and $$M_2$$ with speeds $$1.5 \times 10^8 \text{ m s}^{-1}$$ and $$2.0 \times 10^8 \text{ m s}^{-1}$$ respectively. The critical angle between them is

Light travels in medium $$M_1$$ with speed $$v_1 = 1.5 \times 10^8 \text{ m s}^{-1}$$ and in medium $$M_2$$ with speed $$v_2 = 2.0 \times 10^8 \text{ m s}^{-1}$$. To determine the critical angle between them, we begin by finding the refractive indices.

Since the speed of light in vacuum is $$c = 3 \times 10^8 \text{ m s}^{-1}$$, we have $$n_1 = \dfrac{c}{v_1} = \dfrac{3 \times 10^8}{1.5 \times 10^8} = 2$$ and $$n_2 = \dfrac{c}{v_2} = \dfrac{3 \times 10^8}{2.0 \times 10^8} = 1.5$$. Because $$n_1 > n_2$$, medium $$M_1$$ is denser than medium $$M_2$$ and total internal reflection occurs when light travels from $$M_1$$ to $$M_2$$.

At the critical angle $$\theta_c$$, the refracted ray emerges along the interface, so $$\sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1.5}{2} = \dfrac{3}{4}$$. This gives $$\sin \theta_c = \dfrac{3}{4}$$.

Next, we compute $$\cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \dfrac{9}{16}} = \sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{4}$$, which leads to $$\tan \theta_c = \dfrac{\sin \theta_c}{\cos \theta_c} = \dfrac{3/4}{\sqrt{7}/4} = \dfrac{3}{\sqrt{7}}$$. Therefore, $$\theta_c = \tan^{-1}\dfrac{3}{\sqrt{7}}$$.

Hence, the correct answer is Option A: $$\tan^{-1}\dfrac{3}{\sqrt{7}}$$.