The critical angle for the material of a prism is `45^(@)` and its refractive angle is `30^(@)`. A monochromatic ray goes out perpendicular to the surface of emergence from the prism. Then the angle of incidence on the prism will be

Options

A. `60^(@)`
B. `75^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle of incidence (A) on the prism given the critical angle (C) and the refractive angle (R). Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Critical angle (C) = 45°
- Refractive angle (R) = 30°
- The ray emerges perpendicular to the surface of the prism.

2. **Using the Concept of Critical Angle:**
- The critical angle is defined as the angle of incidence in the denser medium (prism) for which the angle of refraction in the less dense medium (air) is 90°.
- The refractive index (μ) of the prism material can be calculated using the formula:
\[ \mu = \frac{1}{\sin C} \]
- Substituting the critical angle:
\[ \mu = \frac{1}{\sin 45°} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

3. **Applying Snell's Law:**
- According to Snell's Law:
\[ \frac{\sin A}{\sin R} = \mu \]
- Here, A is the angle of incidence we need to find, and R is the angle of refraction (30°).
- Substituting the known values:
\[ \frac{\sin A}{\sin 30°} = \sqrt{2} \]
- We know that \(\sin 30° = \frac{1}{2}\), so:
\[ \frac{\sin A}{\frac{1}{2}} = \sqrt{2} \]

4. **Solving for \(\sin A\):**
- Rearranging gives:
\[ \sin A = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} \]

5. **Finding the Angle A:**
- The value of \(\sin A = \frac{\sqrt{2}}{2}\) corresponds to:
\[ A = 45° \]

6. **Conclusion:**
- Therefore, the angle of incidence (A) on the prism is **45°**.

### Final Answer:
The angle of incidence on the prism is **45°**.

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