Question 163: Ray Optics

Question

In figure, the optical fiber is $l$ = 2m long and
has a diameter of d = 20 μm. If a ray of light
is incident on one end of the fiber at angle
$	heta _1$ = 40°, the number of reflection it makes
before emerging from the other end is close to:
(refractive index of fibre is 1.31 and
sin 40° = 0.64)
![](https://imagex.cdn.examgoal.net/oP5gfCsOhjqLifh5t/dgwekGnyYdptBPFizU64yMfXN7VIG/C7kOCUvhBN5P5TyLy3kw7G/image.png)

Accepted Answer

If we approximate the angle $	heta _2$ as 30ÃÂ° 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 SnellÃ¢ÂÂs law 1.sin 40ÃÂ° = (1.31) sin $	heta _2$
$\sin {	heta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$
Now tan$	heta _2$
= ${{64} \over {\sqrt {{{\left( {131} ight)}^2} - {{\left( {64} ight)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$
Now number of reflections
$ = {{2 	imes 64} \over {114.3 	imes 20 	imes {{10}^{ - 6}}}} = {{64 	imes {{10}^5}} \over {114.3}}$
$ \approx 55991 \approx 55000$
APPROXIMATE SOLUTION
By SnellÃ¢ÂÂs law 1.sin 40ÃÂ° = (1.31)sin $	heta _2$
$\sin {	heta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$
If assume $ \Rightarrow $ $	heta _2$ $\approx $ 30ÃÂ°
tan 30ÃÂ° = ${d \over x} \Rightarrow x = {\sqrt3} d$
Now number of reflections
$ = {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3  	imes 20 	imes {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$
$ \approx 57735 \approx 57000$

Answer

57000

Answer

55000

Answer

66000

Answer

45000

Ray Optics

In figure, the optical fiber is l = 2m long and has a diameter of d = 20 μm. If a ray of light is incident on one end of the fiber at angle \theta _1 = 40°, the number of reflection it makes before emerging from the other end is close to: (refractive index of fibre is 1.31 and sin 40° = 0.64)

Choose the correct answer: