An optical fibre is $l = 2\ \text{m}$ long and has a diameter of $d = 20\ \mu\text{m}$. If a ray of light is incident on one end of the fibre at angle $\theta_1 = 40^\circ$ (with the axis of the fibre), the number of reflections it makes before emerging from the other end is close to (refractive index of fibre is $1.31$, $\sin40^\circ = 0.64$ and $\sin^{-1}0.49 = 30^\circ$)
Answer: (B) $57000$
Refraction at the end face: $\sin\theta_2 = \dfrac{\sin40^\circ}{1.31} = \dfrac{0.64}{1.31} = 0.49 \Rightarrow \theta_2 = 30^\circ$ with the axis.
Between two successive reflections the ray crosses the diameter $d$ while moving along the axis a distance
$$x = \frac{d}{\tan30^\circ} = 20\sqrt3\ \mu\text{m} \approx 34.6\ \mu\text{m}$$
Number of reflections:
$$N = \frac{l}{x} = \frac{2}{34.6\times10^{-6}} \approx 5.8\times10^4 \approx 57000$$
Solution by Sreeraj P, M.Sc Physics