Question 65 of 295

In figure, the optical fiber is $l=2m$ 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 ${\theta }_{1}=40^{\circ}$ , the number of reflections it makes before emerging from the other end is close to:
(refractive index of fiber is $1.31$ , $sin40^{\circ}=0.64$ and ${\sin }^{-1}⁡0.49=30^{\circ}$ .)

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

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Answer: (B) $57000$

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:
${n}_{1}\sin {\theta }_{1}={n}_{2}\sin {\theta }_{2}$
$\sin {40}^{^{\circ}}=1.31\times \sin {\theta }_{2}$
$\Rightarrow \sin {\theta }_{2}=\dfrac{0.64}{1.31}=0.4885\approx 0.5$
$\Rightarrow {\theta }_{2}={30}^{^{\circ}}$
The distance covered along the length of the optical fibre after one reflection:
$QN=RN{cot\theta }_{2}$
$=20\times {10}^{-6}\times \cot {30}^{^{\circ}}=20\sqrt{3}\times {10}^{-6}m$
The reflections take place throughout the length of the optical fibre.
Length of the wire$=2m$
Total number of reflections along its length$=\dfrac{Lengthofthewire}{distancecoveredinonereflectionalongthelength}$
$=\dfrac{2\times {10}^{6}}{20\sqrt{3}}=\dfrac{100000}{\sqrt{3}}=57735.02\approx 57000$