A beam of light is travelling from region II to region III (see the figure) . The refractive index in the region I,II and III are `n_0,n_0/sqrt2, and n_0/2` respectively . The angle of incidence `theta` for which the beam just misses entering region III is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `sin ^(-1) sqrt(2)`

Answer

To solve the problem, we need to determine the angle of incidence \( \theta \) for which a beam of light traveling from region II to region III just misses entering region III. This situation corresponds to the condition for total internal reflection at the boundary between regions II and III.

### Step-by-Step Solution:

1. **Identify the refractive indices**:
- Let the refractive index of region I be \( n_0 \).
- The refractive index of region II is \( \frac{n_0}{\sqrt{2}} \).
- The refractive index of region III is \( \frac{n_0}{2} \).

2. **Determine the critical angle**:
- The critical angle \( \theta_c \) for total internal reflection occurs when light travels from a medium with a higher refractive index (region II) to a medium with a lower refractive index (region III).
- The formula for the critical angle is given by:
\[ \sin \theta_c = \frac{n_3}{n_2} \]
- Substituting the values:
\[ \sin \theta_c = \frac{\frac{n_0}{2}}{\frac{n_0}{\sqrt{2}}} = \frac{1}{\sqrt{2}} \]
- Therefore, \( \theta_c = 45^\circ \).

3. **Apply Snell's Law at the boundary between region I and region II**:
- At the boundary between region I (with refractive index \( n_0 \)) and region II (with refractive index \( \frac{n_0}{\sqrt{2}} \)), we apply Snell's Law:
\[ n_1 \sin \theta = n_2 \sin \theta_c \]
- Substituting the known values:
\[ n_0 \sin \theta = \frac{n_0}{\sqrt{2}} \sin 45^\circ \]
- Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we have:
\[ n_0 \sin \theta = \frac{n_0}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{n_0}{2} \]

4. **Solve for \( \sin \theta \)**:
- Dividing both sides by \( n_0 \):
\[ \sin \theta = \frac{1}{2} \]

5. **Calculate \( \theta \)**:
- The angle \( \theta \) can be found using the inverse sine function:
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

Thus, the angle of incidence \( \theta \) for which the beam just misses entering region III is \( 30^\circ \).