In figure, the optical fiber |GEOMETRICALOPTICSJEEMain2019

JEE MAIN 2019

08-04-2019 S1

Question

In figure, the optical fiber is I = 2 m long and has a diameter of $\mathrm{d}=20 \mu \mathrm{~m}$. 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 and $\sin 40^{\circ}=0.64$ )

Select the correct option:

66000

55000

45000

57000

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & 1 \times \sin 40^{\circ}=1.31 \sin \theta \\ & \Rightarrow \quad \sin \theta=\frac{0.64}{1.31} \Rightarrow \theta \simeq 30^{\circ} \\ & \Rightarrow \quad I=20 \mu \mathrm{~m} \times \cot \theta \\ & \therefore \quad N=\frac{2}{20 \times 10^{-6} \times \cot \theta} \\ & \quad=\frac{2 \times 10^6}{20 \times \sqrt{3}}=57735 \\ & N \simeq 57000\end{aligned}$

Question Tags

JEE Main

Physics

Hard

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