A ray of light is incident on a glass slab at an angle of `45^(@)`. If refractive index of glass be 1.6, what is the angle of refraction ?

Options

A. `45.2^(@)`
B. `30^(@)`
C. `26.2^(@)`
D. `15.2^(@)`

Explanation

To find the angle of refraction when a ray of light passes from air into a glass slab, we can use Snell's Law, which states:

\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]

Where:
- \( n_1 \) is the refractive index of the first medium (air in this case, which is approximately 1),
- \( \theta_1 \) is the angle of incidence,
- \( n_2 \) is the refractive index of the second medium (glass in this case, which is 1.6),
- \( \theta_2 \) is the angle of refraction.

### Step-by-Step Solution:

1. **Identify the given values**:
- Refractive index of air, \( n_1 = 1 \)
- Refractive index of glass, \( n_2 = 1.6 \)
- Angle of incidence, \( \theta_1 = 45^\circ \)

2. **Apply Snell's Law**:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
Substituting the known values:
\[ 1 \cdot \sin(45^\circ) = 1.6 \cdot \sin(\theta_2) \]

3. **Calculate \( \sin(45^\circ) \)**:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \]
So the equation becomes:
\[ 0.707 = 1.6 \cdot \sin(\theta_2) \]

4. **Solve for \( \sin(\theta_2) \)**:
\[ \sin(\theta_2) = \frac{0.707}{1.6} \]
\[ \sin(\theta_2) \approx 0.44125 \]

5. **Calculate \( \theta_2 \)**:
To find \( \theta_2 \), we take the inverse sine:
\[ \theta_2 = \sin^{-1}(0.44125) \approx 26.2^\circ \]

### Final Answer:
The angle of refraction \( \theta_2 \) is approximately \( 26.2^\circ \).

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