question_answer

In figure, the optical fiber is \[\ell =2m\]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 \[{{\theta }_{1}}={{40}^{o}},\]the number of reflection it makes before emerging from the other end is close to: (refractive index of fibre is 1.31 and sin \[40{}^\circ \text{ }=\text{ }0.64\]) [JEE Main 8-4-2019 Morning]

A) 55000             

B) 57000

C) 66000             

D) 45000

Correct Answer: B

Solution :

Note : If we approximate the angle \[{{\theta }_{2}}\,\text{as}\,{{30}^{o}}\]as initially then answer will be closer to 57000. But if we solve thoroughly, answer will be close to 55000. So both the answers must be awarded. Detailed solution is as following. Exact solution by Snells' law 1.\[sin40{}^\circ =\left( 1.31 \right)sin{{\theta }_{2}}\] \[sin{{\theta }_{2}}=\frac{.64}{1.31}=\frac{64}{131}\approx .49\] 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 Snells' law 1.\[sin40{}^\circ =\left( 1.31 \right)sin{{\theta }_{2}}\] \[sin{{\theta }_{2}}=\frac{.64}{1.31}=\frac{64}{131}\approx .49\] If assume \[\Rightarrow {{\theta }_{2}}\approx {{30}^{o}}\] \[\tan {{30}^{o}}=\frac{d}{x}\Rightarrow x=\sqrt{3}d\] Now number of reflections \[=\frac{\ell }{\sqrt{3}d}=\frac{2}{\sqrt{3}\times 20\times {{10}^{-6}}}=\frac{{{10}^{5}}}{\sqrt{3}}\] \[\approx 57735\approx 57000\]