A ray of light passes from air to glass `(n = 1.5)` at an angle of `30^(@)`. Calculate the angle of refraction.

Explanation

To solve the problem of finding the angle of refraction when a ray of light passes from air to glass, we can use Snell's Law. Here is the step-by-step solution:

### Step-by-Step Solution:

1. **Identify the given values:**
- Angle of incidence, \( i = 30^\circ \)
- Refractive index of air, \( n_1 = 1 \)
- Refractive index of glass, \( n_2 = 1.5 \)

2. **Write down Snell's Law:**
\[ n_1 \sin i = n_2 \sin r \]
where \( r \) is the angle of refraction.

3. **Substitute the given values into Snell's Law:**
\[ 1 \cdot \sin 30^\circ = 1.5 \cdot \sin r \]

4. **Calculate \(\sin 30^\circ\):**
\[ \sin 30^\circ = \frac{1}{2} \]

5. **Substitute \(\sin 30^\circ\) into the equation:**
\[ 1 \cdot \frac{1}{2} = 1.5 \cdot \sin r \]
\[ \frac{1}{2} = 1.5 \cdot \sin r \]

6. **Solve for \(\sin r\):**
\[ \sin r = \frac{\frac{1}{2}}{1.5} \]
\[ \sin r = \frac{1}{2} \cdot \frac{1}{1.5} \]
\[ \sin r = \frac{1}{2} \cdot \frac{2}{3} \]
\[ \sin r = \frac{1}{3} \]
\[ \sin r = 0.333 \]

7. **Find the angle \( r \) by taking the inverse sine (arcsin) of 0.333:**
\[ r = \sin^{-1}(0.333) \]

8. **Calculate the angle of refraction:**
\[ r \approx 19.47^\circ \]

So, the angle of refraction is approximately \( 19.47^\circ \).