A ray of light incident at an angle `theta` on a refracting face of a prism emerges from the other face normally. If the angle of the prism is `5^@` and the prism is made of a material of refractive index `1.5`, the angle of incidence is.

Options

A. `7.5^(@)`
B. `5^(@)`
C. `15^(@)`
D. `2.5^(@)`

Explanation

To solve the problem, we need to find the angle of incidence (θ) for a ray of light incident on a prism with an angle of 5 degrees and a refractive index of 1.5, where the ray emerges normally from the other face of the prism.

### Step-by-Step Solution:

1. **Understand the Geometry of the Prism:**
- The prism has an angle \( A = 5^\circ \).
- The ray of light enters the prism at an angle of incidence \( \theta \) and emerges normally from the opposite face.

2. **Identify Angles:**
- Let \( r_1 \) be the angle of refraction at the first face of the prism.
- Since the ray emerges normally from the second face, the angle of refraction \( r_2 = 0^\circ \).

3. **Apply the Prism Formula:**
- For a prism, the relationship between the angles of refraction is given by:
\[ r_1 + r_2 = A \]
- Substituting \( r_2 = 0^\circ \) into the equation:
\[ r_1 + 0 = 5^\circ \implies r_1 = 5^\circ \]

4. **Use Snell's Law:**
- Snell's law states:
\[ \mu = \frac{\sin \theta}{\sin r_1} \]
- Here, \( \mu = 1.5 \) (the refractive index of the prism) and \( r_1 = 5^\circ \).

5. **Substitute Values into Snell's Law:**
- Rearranging Snell's law gives:
\[ \sin \theta = \mu \cdot \sin r_1 \]
- Substitute the known values:
\[ \sin \theta = 1.5 \cdot \sin(5^\circ) \]

6. **Calculate \( \sin(5^\circ) \):**
- Using a calculator, we find:
\[ \sin(5^\circ) \approx 0.0872 \]
- Therefore:
\[ \sin \theta = 1.5 \cdot 0.0872 \approx 0.1308 \]

7. **Find \( \theta \):**
- To find \( \theta \), take the inverse sine:
\[ \theta = \sin^{-1}(0.1308) \approx 7.5^\circ \]

### Conclusion:
The angle of incidence \( \theta \) is approximately \( 7.5^\circ \).