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 \[=\frac{\ell }{\sqrt{3}d}=\frac{2}{\sqrt{3}\times 20\times {{10}^{-6}}}=\frac{{{10}^{5}}}{\sqrt{3}}\] \[\approx 57735\approx 57000\]You need to login to perform this action.
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