Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of a glass slab. In the vacuum the beams makes an angle of `37^(@)` with the normal to the surface, while in the glass it makes an angle of `23.5^(@)` with the normal. What is the index of refraction of the glass ?

Answer

To find the index of refraction of the glass slab, we can use Snell's Law, which states:

\[ \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \]

Where:
- \( i \) is the angle of incidence (in vacuum),
- \( r \) is the angle of refraction (in glass),
- \( n_1 \) is the refractive index of the first medium (vacuum, which is 1),
- \( n_2 \) is the refractive index of the second medium (glass, which we want to find).

### Step-by-Step Solution:

1. **Identify the angles**:
- Angle of incidence, \( i = 37^\circ \)
- Angle of refraction, \( r = 23.5^\circ \)

2. **Apply Snell's Law**:
\[ \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \]
Since \( n_1 = 1 \) (for vacuum), the equation simplifies to:
\[ \sin i = n_2 \cdot \sin r \]

3. **Calculate \( \sin i \) and \( \sin r \)**:
- Calculate \( \sin 37^\circ \) and \( \sin 23.5^\circ \):
- \( \sin 37^\circ \approx 0.6018 \)
- \( \sin 23.5^\circ \approx 0.3987 \)

4. **Substitute the values into Snell's Law**:
\[ \frac{\sin 37^\circ}{\sin 23.5^\circ} = n_2 \]
\[ n_2 = \frac{0.6018}{0.3987} \]

5. **Calculate \( n_2 \)**:
\[ n_2 \approx 1.51 \]

### Final Answer:
The index of refraction of the glass is approximately \( 1.51 \).

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