1. Given Data

• Refractive Index of Glass Fibre (n₁): 1.68
• Refractive Index of Outer Pipe (n₂): 1.44
• Refractive Index of Air (n_air): 1.00

2. Part (a): With Outer Covering (Cladding)

For Total Internal Reflection (TIR) to occur inside the pipe, the angle of incidence at the core-cladding interface (i_c) must be greater than the critical angle (C).

Step A: Find Critical Angle (C)

sin(C) = n₂ / n₁

sin(C) = 1.44 / 1.68 = 0.8571

C = sin^-1(0.8571)

C ≈ 59°

Step B: Find Maximum Angle of Refraction (r)

Inside the fibre, the ray hits the wall at angle i_c. From geometry: r + i_c = 90°.

For TIR, minimum i_c = 59°, so maximum r is:

r_max = 90° - 59° = 31°

Step C: Find Maximum Incident Angle (i)

Using Snell's Law at the entry point:

sin(i) = n₁ × sin(r_max)

sin(i) = 1.68 × sin(31°)

sin(i) = 1.68 × 0.515

sin(i) = 0.8652

i = sin^-1(0.8652) ≈ 60°

3. Part (b): Without Outer Covering

If there is no outer covering, the outside medium is Air (n₂ = 1.00).

Step A: New Critical Angle (C')

sin(C') = 1.00 / 1.68 = 0.5952

C' = sin^-1(0.5952) ≈ 36.5°

Step B: New Maximum Refraction (r')

r'_max = 90° - 36.5° = 53.5°

Step C: New Incident Angle (i')

sin(i') = 1.68 × sin(53.5°)

sin(i') = 1.68 × 0.8038

sin(i') = 1.35

Since the value of sine cannot be greater than 1, this means that for all possible angles of incidence (0° to 90°), the condition for total internal reflection is satisfied.