In figure, the optical fiber is \(I = 2\,m\) 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

So both the answers must be awarded. Detailed solution as following.

Exact solution

By Snell's law  \(1.\sin {40^o} = (1.31)\sin {\theta _2}\)

\(\sin \theta_{2}=\frac{64}{1.31}=\frac{64}{131} \approx .49\)

Now \(\tan {\theta _2} = \frac{{64}}{{\sqrt {{{(131)}^2} - {{(64)}^2}} }}\) \( = \frac{{64}}{{\sqrt {13065} }} \approx \frac{{64}}{{114.3}} = \frac{{\text{d}}}{x}\)

Now number of reflections

\(=\frac{2 \times 64}{114.3 \times 20 \times 10^{-6}}=\frac{64 \times 10^{5}}{114.3}\)

\(=55991 \approx 55000\)

Approximate solution

By Snell's law \(1 . \sin 40^{\circ}=(1.31) \sin \theta_{2}\)

\({\sin \theta_{2}=\frac{0.64}{1.31}=\frac{64}{131} \approx 0.49}\)

If assume  \( \Rightarrow {\theta _2} \approx {30^o}\)

\(\tan {30^o} = \frac{{\text{d}}}{{\text{x}}} \Rightarrow {\text{x}} = \sqrt 3 {\text{d}}\)

Now number of reflections

\(=\frac{\ell}{\sqrt{3} \mathrm{d}}=\frac{2}{\sqrt{3} \times 20 \times 10^{-6}}=\frac{10^{5}}{\sqrt{3}}\)

\(\approx 57735 \approx 57000\)