QJEEM19Ph0ee2: JEE Main 2019 Physics - Class 12 JEEM19Ph0ee2 Solution: In figure, the optical fiber is = 2m long and has a diameter of …

If we approximate the angle $\theta _2$ as 30° 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 as following.

EXACT SOLUTION
By Snell’s law 1.sin 40° = (1.31) sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

Now tan $\theta _2$
= ${{64} \over {\sqrt {{{\left( {131} \right)}^2} - {{\left( {64} \right)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$

Now number of reflections

$= {{2 \times 64} \over {114.3 \times 20 \times {{10}^{ - 6}}}} = {{64 \times {{10}^5}} \over {114.3}}$

$\approx 55991 \approx 55000$

APPROXIMATE SOLUTION
By Snell’s law 1.sin 40° = (1.31)sin $\theta _2$

$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$

If assume $\Rightarrow$ $\theta _2$ $\approx$ 30°

tan 30° = ${d \over x} \Rightarrow x = {\sqrt3} d$

Now number of reflections
$= {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3 \times 20 \times {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$

$\approx 57735 \approx 57000$