Question

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 $\theta_1 = 40^{\circ}$, the number of reflections it makes before emerging from the other end is : (refractive index of fiber is $1.31$ and $sin \,40^{\circ} = 0.64)$

Answer 1

Using Snell’s law of refraction,
$ 1\times sin\,40^{\circ} = 1.31 \,sin \,\theta$
$\Rightarrow sin \,\theta = \frac{0.64}{1.31} = 0.49 \approx 0.5$
$\Rightarrow \theta = 30^{\circ}$

$ x = 20\,\mu m \times cot\,\theta$
$\therefore $ Number of reflections $ = \frac{2}{20 \times 10^{-6} \times cot\,\theta}$
$ = \frac{2 \times 10^6}{20 \times \sqrt{3}} = 57735 \approx 57000$