Question

A ray of light is incident at \(30^\circ\) from a medium of refractive index \(2\) into a medium of refractive index \(1\). Then the angle of refraction is:

• A. \(30^\circ\)
• B. \(60^\circ\)
• C. \(45^\circ\)
• D. \(90^\circ\)

Hint:

If the refracted angle becomes \[ 90^\circ, \] the angle of incidence is equal to the critical angle.

Answer 1

The Correct Option is D

Step 1: Recall the condition for total internal reflection.
When light travels from a denser medium (higher $n$) to a rarer medium (lower $n$), total internal reflection occurs if the angle of incidence exceeds the critical angle $\theta_c$.

Step 2: Calculate the critical angle for this interface.
The critical angle is given by:
\[ \sin\theta_c = \frac{n_2}{n_1} = \frac{1}{2} \] \[ \theta_c = 30^\circ \]

Step 3: Compare the angle of incidence with the critical angle.
The given angle of incidence is $i = 30^\circ$, which equals the critical angle.

Step 4: Understand what happens exactly at the critical angle.
At the critical angle, the refracted ray travels exactly along the interface. This means the angle of refraction is $90^\circ$.

Step 5: Verify using Snell's law.
\[ n_1 \sin i = n_2 \sin r \] \[ 2 \times \sin 30^\circ = 1 \times \sin r \] \[ 2 \times \frac{1}{2} = \sin r \implies \sin r = 1 \implies r = 90^\circ \]

Step 6: State the answer.
\[ \boxed{90^\circ} \]