On a glass plate, a light wave is incident at an angle of `60^(@)`. If the reflected and the refracted waves are mutully perpendicular, the refractive index of material is

Options

A. `(sqrt3)/2`
B. `sqrt3`
C. `3/2`
D. zero

Explanation

To solve the problem, we need to find the refractive index of the glass plate given that the reflected and refracted waves are mutually perpendicular when a light wave is incident at an angle of \(60^\circ\).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Angle of incidence (\(i\)) = \(60^\circ\)
- The reflected and refracted waves are mutually perpendicular. This means that the angle between the reflected ray and the refracted ray is \(90^\circ\).

2. **Use the Relationship Between Angles:**
- According to the law of reflection, the angle of reflection (\(r\)) is equal to the angle of incidence. Therefore, \(r = i = 60^\circ\).
- If the reflected and refracted rays are perpendicular, we can express this relationship as:
\[ r + r' = 90^\circ \]
where \(r'\) is the angle of refraction. Thus:
\[ r' = 90^\circ - r = 90^\circ - 60^\circ = 30^\circ \]

3. **Apply Snell's Law:**
- Snell's Law states:
\[ n_1 \sin(i) = n_2 \sin(r') \]
- Here, \(n_1\) is the refractive index of air (approximately \(1\)), \(n_2\) is the refractive index of the glass plate, \(i = 60^\circ\), and \(r' = 30^\circ\).

4. **Substitute the Values into Snell's Law:**
- Plugging in the values:
\[ 1 \cdot \sin(60^\circ) = n_2 \cdot \sin(30^\circ) \]

5. **Calculate the Sine Values:**
- We know:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
- Substitute these values into the equation:
\[ 1 \cdot \frac{\sqrt{3}}{2} = n_2 \cdot \frac{1}{2} \]

6. **Solve for \(n_2\):**
- Rearranging the equation gives:
\[ n_2 = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]

7. **Final Answer:**
- The refractive index of the material (glass plate) is:
\[ n_2 = \sqrt{3} \]