1. Given situation

A ray travels from water to air and is incident at angle $\theta$, with $\theta < \theta_c$ where $\theta_c$ is the critical angle.

Since the angle of incidence is less than the critical angle, total internal reflection does not occur.

So, at the boundary:

• there is a reflected ray in water,
• and a refracted ray in air.

Hence, options A and B are immediately false.

2. Reflection and refraction angles

Let:

• angle of incidence $= \theta$,
• angle of reflection $= \theta$,
• angle of refraction $= r$.

By Snell's law, $n_w \sin\theta = n_a \sin r$ Since light goes from denser medium (water) to rarer medium (air), we have $n_w > n_a$ Therefore, $\sin r > \sin\theta \quad \Rightarrow \quad r > \theta$ (for acute angles involved here).

3. Angle between reflected and refracted rays

The reflected ray makes angle $\theta$ with the normal on the water side. The refracted ray makes angle $r$ with the normal on the air side.

Therefore, the angle between the reflected and refracted rays is $\phi = 180^\circ - (\theta + r)$

Now since $r > \theta$, $\theta + r > 2\theta$ So, $180^\circ - (\theta + r) < 180^\circ - 2\theta$ That is, $\phi < 180^\circ - 2\theta$

Thus the angle between the reflected and refracted rays is less than $180^\circ - 2\theta$.

4. Check options

• A: only reflected ray and no refracted ray → false
• B: only refracted ray and no reflected ray → false
• C: reflected ray and refracted ray, and angle between them is less than $180^\circ - 2\theta$ → true
• D: angle greater than $180^\circ - 2\theta$ → false

5. Final answer

The correct option is: $\boxed{\text{C}}$

This matches the stored correct answer.