In figure, the optical fiber is l = 2 m long and has a diameter of d = 20 μ m . If a ray of light is incident on one end of the fiber at angle θ 1 = 40 ° , the number of reflections it makes before emerging from the other end is close to: (refractive index of fiber is 1.31 , s i n 40 ° = 0.64 and sin - 1 ⁡ 0.49 = 30 ° .)

Step-by-step solution

Let n 1 and n 2 be the refractive indices of the air and optical fibre respectively. Applying Snell's law at Q, we get: n 1 sin θ 1 = n 2 sin θ 2 sin 40 ∘ = 1 . 31 × sin θ 2 ⇒ sin θ 2 = 0 . 64 1 . 31 = 0 . 4885 ≈ 0 . 5 ⇒ θ 2 = 30 ∘ The distance covered along the length of the optical fibre after one reflection: QN = RN   cotθ 2 = 20 × 10 - 6 × cot 30 ∘ = 20 3 × 10 - 6   m The reflections take place throughout the len