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 a given angle and refractive index. Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Angle of the prism (A) = 5°
- Refractive index (μ) = 1.5

2. **Use the Formula for Angle of Deviation (δ):**
The angle of deviation (δ) can be calculated using the formula:
\[ δ = (μ - 1) \times A \]
Substituting the values:
\[ δ = (1.5 - 1) \times 5° = 0.5 \times 5° = 2.5° \]

3. **Understand the Emergence Condition:**
The ray emerges normally from the second face of the prism, which means the angle of refraction (r2) at the second face is 0°.

4. **Apply the Prism Formula:**
The relationship between the angles of the prism and the angles of incidence and refraction is given by:
\[ A = r_1 + r_2 \]
Since \( r_2 = 0° \), we have:
\[ A = r_1 + 0° \implies r_1 = A = 5° \]

5. **Use Snell's Law:**
According to Snell's law:
\[ \frac{\sin i_1}{\sin r_1} = μ \]
Since \( r_1 = 5° \), we can write:
\[ \sin i_1 = μ \cdot \sin r_1 \]
Substituting the values:
\[ \sin i_1 = 1.5 \cdot \sin(5°) \]

6. **Calculate \( \sin(5°) \):**
Using a calculator or trigonometric tables:
\[ \sin(5°) \approx 0.0872 \]
Therefore:
\[ \sin i_1 = 1.5 \cdot 0.0872 \approx 0.1308 \]

7. **Find \( i_1 \):**
To find the angle of incidence \( i_1 \), we take the inverse sine:
\[ i_1 = \sin^{-1}(0.1308) \approx 7.5° \]

### Final Answer:
The angle of incidence \( i_1 \) is approximately **7.5°**.

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