An optical fibre (mu = 1.72) is surrounded by a glass coating (mu= 1.50). Find the critical angle for total internal reflection at the fibre-glass interface.

Explanation

To find the critical angle for total internal reflection at the fibre-glass interface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Critical Angle**:
The critical angle (C) is the angle of incidence above which total internal reflection occurs when light travels from a denser medium to a rarer medium.

2. **Identify the Refractive Indices**:
- The refractive index of the optical fiber (denser medium) is \( \mu_1 = 1.72 \).
- The refractive index of the glass coating (rarer medium) is \( \mu_2 = 1.50 \).

3. **Use Snell's Law**:
According to Snell's Law, at the critical angle, the relationship can be expressed as:
\[ \mu_1 \sin(C) = \mu_2 \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), this simplifies to:
\[ \mu_1 \sin(C) = \mu_2 \]

4. **Substitute the Values**:
Substitute the values of the refractive indices into the equation:
\[ 1.72 \sin(C) = 1.50 \]

5. **Solve for \( \sin(C) \)**:
Rearranging the equation gives:
\[ \sin(C) = \frac{1.50}{1.72} \]

6. **Calculate \( \sin(C) \)**:
Performing the division:
\[ \sin(C) = \frac{1.50}{1.72} \approx 0.8721 \]

7. **Find the Critical Angle \( C \)**:
To find the critical angle, take the inverse sine:
\[ C = \sin^{-1}(0.8721) \]

8. **Calculate the Angle**:
Using a calculator, we find:
\[ C \approx 61.0^\circ \]

### Final Answer:
The critical angle for total internal reflection at the fibre-glass interface is approximately \( 61.0^\circ \).

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