Angle Between Reflected and Refracted Rays in Water

To determine the behavior of the light ray, consider the processes occurring at the boundary between water (optically denser medium, refractive index $\mu > 1$) and air (optically rarer medium, refractive index $\mu_{\text{air}} = 1$):

1. Existence of Reflected and Refracted Rays:

   • The ray travels from water to air with an angle of incidence $\theta$.
   • It is given that the angle of incidence is less than the critical angle ($\theta < \theta_c$).
   • Since total internal reflection does not take place, partial reflection occurs at the interface. Thus, both a reflected ray and a refracted ray are present.

2. Angle of Refraction:

   • According to Snell's law:
     $\mu \sin \theta = 1 \cdot \sin r \implies \sin r = \mu \sin \theta$
   • Since water is denser than air ($\mu > 1$), it follows that:
     $\sin r > \sin \theta \implies r > \theta$
     where $r$ is the angle of refraction.

3. Angle Between Reflected and Refracted Rays:

   • By the law of reflection, the angle of reflection is equal to the angle of incidence:
     $r' = \theta$
   • The normal is a straight line through the point of incidence subtending an angle of $180^\circ$.
   • The reflected ray lies in the water at an angle $\theta$ to the normal, while the refracted ray lies in the air on the same side of the normal at an angle $r$ to the upward normal.
   • Therefore, the angle $\phi$ between the reflected ray and the refracted ray is given by:
     $\phi = 180^\circ - (r' + r) = 180^\circ - (\theta + r)$
   • Since $r > \theta$, we have:
     $\theta + r > 2\theta$
   • Substituting this inequality gives:
     $\phi < 180^\circ - 2\theta$

Thus, there will be a reflected ray and a refracted ray, and the angle between them is less than $180^{\circ} - 2\theta$.

Therefore, the correct option is C.