Question

In figure, the optical fiber is $l=2m$ 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 $sin40^{\circ }=0.64)$

Answer 1

Answer: 57000

Using Snell’s law of refraction,
$1\times sin40^{\circ }=1.31sin\theta$
$\Rightarrow sin\theta =\frac{0.64}{1.31}=0.49\approx 0.5$
$\Rightarrow \theta =30^{\circ }$

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