How is the critical angle related to the refractive index of a medium ?

Answer

To understand how the critical angle is related to the refractive index of a medium, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Consider a boundary between two media: a denser medium (like glass) and a rarer medium (like air).

2. **Defining the Critical Angle**:
- The critical angle (denoted as \( C \)) is the angle of incidence in the denser medium at which the angle of refraction in the rarer medium is \( 90^\circ \). This means that the light ray travels along the boundary.

3. **Applying Snell's Law**:
- According to Snell's Law, the relationship between the angles of incidence and refraction is given by:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) is the refractive index of the denser medium (glass), \( n_2 \) is the refractive index of the rarer medium (air), \( i \) is the angle of incidence, and \( r \) is the angle of refraction.

4. **Setting Up for Critical Angle**:
- At the critical angle, the angle of refraction \( r \) is \( 90^\circ \). Therefore, we can substitute \( r = 90^\circ \) into Snell's Law:
\[ n_{\text{glass}} \sin(C) = n_{\text{air}} \sin(90^\circ) \]

5. **Simplifying the Equation**:
- Since \( \sin(90^\circ) = 1 \), the equation simplifies to:
\[ n_{\text{glass}} \sin(C) = n_{\text{air}} \]

6. **Finding the Relationship**:
- Rearranging the equation gives:
\[ \sin(C) = \frac{n_{\text{air}}}{n_{\text{glass}}} \]
- The refractive index of air is approximately \( 1 \) (since it is the reference medium), thus:
\[ \sin(C) = \frac{1}{n_{\text{glass}}} \]

7. **Conclusion**:
- Therefore, the critical angle \( C \) is related to the refractive index \( n \) of the denser medium (glass) by the formula:
\[ C = \sin^{-1}\left(\frac{1}{n_{\text{glass}}}\right) \]