Physics - Geometrical Optics Question with Solution | TestHub

Let $n_1$ and $n_2$ be the refractive indices of the air and optical fibre respectively. Applying Snell's law at Q, we get:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

$\sin {40}^{\circ }=1.31\times \sin {\theta }_2$

$\Rightarrow \sin {\theta }_2=\frac{0.64}{1.31}=0.4885\approx 0.5$

$\Rightarrow {\theta }_2={30}^{\circ }$

The distance covered along the length of the optical fibre after one reflection:

$\mathrm{QN}=\mathrm{RN} {\mathrm{cot\theta }}_2$

$=20\times {10}^{-6}\times \cot {30}^{\circ }=20\sqrt{3}\times {10}^{-6} \mathrm{m}$

The reflections take place throughout the length of the optical fibre.

Length of the wire$=2 \mathrm{m}$

Total number of reflections along its length$=\frac{\mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{wire}}{\mathrm{distance} \mathrm{covered} \mathrm{in} \mathrm{one} \mathrm{reflection} \mathrm{along} \mathrm{the} \mathrm{length}}$

$=\frac{2\times {10}^6}{20\sqrt{3}}=\frac{100000}{\sqrt{3}}=57735.02\approx 57000$