Question

In figure, the optical fiber is $\ell =2m$ long and has a diameter of $d=20pm$. If a ray of light is incident on one end of the fiber at angle ${\theta }_1=40^{\circ }$, the number of reflection it makes before emerging from the other end is close to: (refractive index o f fibre is $1.31$ and $\sin 40^{\circ }=0.64)$

• A. 55000
• B. 57000
• C. 66000
• D. 45000

Answer 1

Answer: (B)

Exact solution
by Snells' law $1.\sin 40^{\circ }=\left(1.31\right)\sin {\theta }_2$
$\sin {\theta }_2=\frac{.64}{1.31}=\frac{64}{131}\approx .49$
Now $\tan {\theta }_2=\frac{64}{\sqrt{{\left(131\right)}^2-{\left(64\right)}^2}}=\frac{64}{\sqrt{13065}}\approx \frac{64}{114.3}=\frac{d}{x}$
Now no. o f reflections
$=\frac{2\times 64}{114.3\times 20\times 10^{-6}}=\frac{64\times 10^5}{114.3}$
$\approx 55991\approx 55000$
Approximate solution
By Snells' law $1.\sin 40^{\circ }=(1.31)\sin {\theta }_2$
$\sin {\theta }_2=\frac{.64}{1.31}=\frac{64}{131}\approx .49$
If assume $\Rightarrow {\theta }_2\approx 30^{\circ }$
$\tan 30^{\circ }=\frac{d}{x}\Rightarrow x=\sqrt{3}d$
Now number of reflections
$=\frac{\ell }{\sqrt{3}d}=\frac{2}{\sqrt{3}\times 20\times 10^{-6}}=\frac{10^5}{\sqrt{3}}$
$\approx 57735\approx 57000$