A ray of light is travelling is travelling from air to water. What is the angle of incidence in air, if angle of refraction in water is `41^(@)` ? Take refractive index of water `= 1.32`.

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `90^(@)`

Explanation

To solve the problem of finding the angle of incidence in air when a ray of light travels from air to water, we can use Snell's Law, which states:

\[ n_1 \sin(i) = n_2 \sin(r) \]

Where:
- \( n_1 \) is the refractive index of the first medium (air),
- \( n_2 \) is the refractive index of the second medium (water),
- \( i \) is the angle of incidence,
- \( r \) is the angle of refraction.

### Step-by-Step Solution:

1. **Identify the given values:**
- Refractive index of air, \( n_1 = 1 \) (approximately),
- Refractive index of water, \( n_2 = 1.32 \),
- Angle of refraction, \( r = 41^\circ \).

2. **Apply Snell's Law:**
\[ n_1 \sin(i) = n_2 \sin(r) \]
Substituting the known values:
\[ 1 \cdot \sin(i) = 1.32 \cdot \sin(41^\circ) \]

3. **Calculate \( \sin(41^\circ) \):**
Using a calculator or trigonometric tables, we find:
\[ \sin(41^\circ) \approx 0.6561 \]

4. **Substitute \( \sin(41^\circ) \) into the equation:**
\[ \sin(i) = 1.32 \cdot 0.6561 \]
\[ \sin(i) \approx 0.8660 \]

5. **Determine the angle \( i \):**
We know that:
\[ \sin(i) = 0.8660 \]
The value of \( \sin(i) = 0.8660 \) corresponds to:
\[ i \approx 60^\circ \]

6. **Conclusion:**
The angle of incidence in air is:
\[ i = 60^\circ \]

### Final Answer:
The angle of incidence in air is \( 60^\circ \).