In figure, the optical fiber is $$l = 2$$ m long and has a diameter of $$d = 20 \; \mu$$m. If a ray of light is incident on one end of the fiber at angle $$\theta_1 = 40°$$, the number of reflections it makes before emerging from the other end is close to: (refractive index of fiber is 1.31, sin 40° = 0.64 and sin$$^{-1}$$(0.49) = 30°.)

Using Snell’s law at the entrance of the optical fiber,

$$1\times \sin40^\circ=(1.31)\sin\theta_2$$

Given,

$$\sin40^\circ=0.64$$

So,

$$\sin\theta_2=\frac{0.64}{1.31}$$

$$\sin\theta_2\approx0.49$$

$$\theta_2=\sin^{-1}(0.49)\approx30^\circ$$

Inside the fiber,

$$\tan\theta_2=\frac{d}{x}$$

where

$$d=20\ \mu m=20\times10^{-6}\ m$$

x = distance between two successive reflections

Using,

$$\tan30^\circ=\frac{1}{\sqrt3}$$

$$\frac{1}{\sqrt3}=\frac{d}{x}$$

$$x=\sqrt3 d$$

Number of reflections:

$$N=\frac{\ell}{x}$$

Given,

$$\ell=2\ m$$

So,

$$N=\frac{2}{\sqrt3\times20\times10^{-6}}$$

$$N=\frac{10^5}{\sqrt3}$$

$$N\approx57735$$

Hence,

$$\boxed{N\approx57000}$$