Comparison of Speed of Light Across Two Media

Based on the principles of wave optics and refraction at a plane interface:

1. Snell's Law of Refraction:
   When a light wave travels from medium-1 into medium-2, the angles of incidence ($i$) and refraction ($r$) are related to the speeds of light $v_1$ and $v_2$ in the respective media by:
   $\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$
   where $n_1$ and $n_2$ are the refractive indices of medium-1 and medium-2, respectively.

2. Analysis of Refraction:
   From the given passage and wavefront diagram, the ray of light bends towards the normal upon entering medium-2 from medium-1. This implies that the angle of incidence is greater than the angle of refraction:
   $i > r$

   Since $\sin \theta$ is a strictly increasing function for $0^\circ < \theta < 90^\circ$:
   $\sin i > \sin r \implies \frac{\sin i}{\sin r} > 1$

3. Comparison of Speeds:
   Substituting this into Snell's law:
   $\frac{v_1}{v_2} > 1 \implies v_1 > v_2$

   Hence, the speed of light is larger in medium-1 than in medium-2. Furthermore, since both media are individually homogeneous and isotropic, the speed of light is uniform throughout medium-1 (including point $b$) and uniform throughout medium-2 (including point $d$).

Therefore, the correct statement is that the speed of light is larger in medium-1 than in medium-2.

Correct Option: (B)