1. Given situation

A light ray goes from glass to air and is incident at angle $\theta$ on a glass-air interface. We need the qualitative variation of:

• reflected intensity $R$ with $\theta$
• transmitted intensity $T$ with $\theta$

2. Key physical facts

When light goes from a denser medium (glass) to a rarer medium (air):

• For small angle of incidence, some part is reflected and some part is transmitted.
• As $\theta$ increases, reflectance generally increases.
• At the critical angle $\theta_c$, the refracted ray grazes the interface.
• For $\theta > \theta_c$, total internal reflection occurs.

So, $\theta_c = \sin^{-1}\!\left(\frac{n_{\text{air}}}{n_{\text{glass}}}\right)$

3. Behavior of transmitted intensity $T$

• For $0 < \theta < \theta_c$, transmission exists, so $T>0$.
• As $\theta \to \theta_c^{-}$, transmission decreases.
• For $\theta > \theta_c$, no transmission occurs because of total internal reflection.

Hence, $T=0 \quad \text{for } \theta \ge \theta_c$

So the graph of $T$ must fall to zero at $\theta_c$ and remain zero afterward.

4. Behavior of reflected intensity $R$

• For small $\theta$, only partial reflection occurs, so $R<1$.
• As $\theta$ increases, reflected intensity increases.
• At $\theta=\theta_c$, total internal reflection starts.
• For $\theta > \theta_c$, all the light is reflected.

Hence, $R=1 \quad \text{for } \theta \ge \theta_c$

So the graph of $R$ must rise and become unity at $\theta_c$, then stay at unity.

5. Energy conservation check

Neglecting absorption, $R+T=1 \qquad (\theta<\theta_c)$

and for total internal reflection, $R=1,\quad T=0 \qquad (\theta>\theta_c)$

6. Correct sketch

Therefore the correct sketch must show:

• $R$ increasing with $\theta$
• $T$ decreasing with $\theta$
• At the critical angle, $R \to 1$ and $T \to 0$
• Beyond critical angle, $R=1$ and $T=0$

This corresponds to Option C.

7. Comparison with stored answer

Stored correct answer = C. Our derived answer = C. So they agree.