Question

In figure, the optical fibrer is $l=2n$ 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 reflection it makes before emerging from the other end is close to: (refractive index of fibre is $1.31$ and $\sin {40}^{\circ }=0.64)$.

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

Answer 1

The correct option is B $57000$

By Snell's 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 $\tan {\theta }_2=\frac{64}{\sqrt{(131{)}^2-(64{)}^2}}=\frac{64}{\sqrt{13065}}\approx \frac{64}{114.3}=\frac{d}{x}$
Now no. of 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 Snell's law $1.\sin {40}^{\circ }=\left(1.31\right)\sin {\theta }_2$
$\sin {\theta }_2=\frac{.64}{1.31}=\frac{64}{131}\approx .49$
If assume $\Rightarrow {\theta }_2\approx {30}^{\circ }$
$\tan {30}^{\circ }=\frac{d}{x}\Rightarrow x=\sqrt{3}d$
Now number of reflections
$=\frac{l}{\sqrt{3}d}=\frac{2}{\sqrt{3}\times 20\times {10}^{-6}}=\frac{{10}^5}{\sqrt{3}}$
$\approx 57735\approx 57000$.