In the figure,the optical fiber is L = 2,m lo | vedclass

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(A) Given: Length $L = 2\,m$,Diameter $d = 20 	imes 10^{-6}\,m$,Incident angle $	heta_1 = 40^\circ$. Assuming the refractive index of the fiber core

Vedclass · Accepted Answer

$57000$

Vedclass · Answer

(A) Given: Length $L = 2\,m$,Diameter $d = 20 	imes 10^{-6}\,m$,Incident angle $	heta_1 = 40^\circ$. Assuming the refractive index of the fiber core is $n = 1.31$.By Snell's law at the entrance:$1 \cdot \sin(40^\circ) = n \cdot \sin(	heta_2)$$\sin(	heta_2) = \frac{\sin(40^\circ)}{1.31} \approx \frac{0.6428}{1.31} \approx 0.4907$Now,$\cos(	heta_2) = \sqrt{1 - \sin^2(	heta_2)} = \sqrt{1 - (0.4907)^2} \approx \sqrt{1 - 0.2408} \approx \sqrt{0.7592} \approx 0.8713$$	an(	heta_2) = \frac{\sin(	heta_2)}{\cos(	heta_2)} \approx \frac{0.4907}{0.8713} \approx 0.5632$The horizontal distance $x$ covered by the ray between two consecutive reflections is given by $	an(	heta_2) = \frac{d}{x}$,so $x = \frac{d}{	an(	heta_2)}$.The number of reflections $N$ is given by $N = \frac{L}{x} = \frac{L \cdot 	an(	heta_2)}{d}$.$N = \frac{2 	imes 0.5632}{20 	imes 10^{-6}} = \frac{1.1264}{20 	imes 10^{-6}} = 0.05632 	imes 10^6 = 56320$.This value is closest to $57000$.

In the figure,the optical fiber is $L = 2\,m$ long and has a diameter of $d = 20\,\mu m$. If a ray of light is incident on one end of the fiber at an angle $\theta_1 = 40^\circ$,the number of reflections it makes before emerging from the other end is close to:

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