A ray of light strikes an air-glass interface at an angle of incidence `(i=60^@)` and gets refracted at an angle of refraction r. On increasing the angle of incidence `(i gt 60^@)`, the angle of refraction r

Options

A. decreases
B. remains same
C. is equal to `60^@`
D. increases

Answer

To solve the problem, we need to analyze the behavior of light as it passes from air (a rarer medium) into glass (a denser medium) at different angles of incidence. We will use Snell's Law, which states that:

\[ \mu = \frac{\sin i}{\sin r} \]

where:
- \( \mu \) is the refractive index of the medium,
- \( i \) is the angle of incidence,
- \( r \) is the angle of refraction.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Angle of incidence, \( i = 60^\circ \)
- Refractive index of air, \( \mu_{air} = 1 \)
- Refractive index of glass, \( \mu_{glass} = \frac{3}{2} \)

2. **Apply Snell's Law**:
Using Snell's Law at the air-glass interface:
\[ \mu_{air} \cdot \sin i = \mu_{glass} \cdot \sin r \]
Substituting the known values:
\[ 1 \cdot \sin(60^\circ) = \frac{3}{2} \cdot \sin r \]

3. **Calculate \( \sin(60^\circ) \)**:
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
So the equation becomes:
\[ \frac{\sqrt{3}}{2} = \frac{3}{2} \cdot \sin r \]

4. **Solve for \( \sin r \)**:
Rearranging the equation:
\[ \sin r = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

5. **Increasing the Angle of Incidence**:
Now, if we increase the angle of incidence \( i \) beyond \( 60^\circ \), we need to analyze the effect on \( r \):
- As \( i \) increases, \( \sin i \) also increases.
- Since \( \mu \) is constant, \( \sin r \) must also increase to maintain the equality in Snell's Law.

6. **Behavior of the Angle of Refraction**:
- When the angle of incidence is increased, the angle of refraction \( r \) will also increase.
- However, as the angle of incidence approaches \( 90^\circ \), the angle of refraction approaches a maximum value, which is \( 90^\circ \) (total internal reflection occurs beyond a certain critical angle).

### Conclusion:
Thus, as the angle of incidence \( i \) increases beyond \( 60^\circ \), the angle of refraction \( r \) also increases.