A pulse of light has numbers in optic fibers?? a) 0 b) 1 c) 2 d) 3
Answers
option C
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Answer:
Consider the optical fiber from Sample Problems 1 and 2 . The index of refraction of the inner core is 1.480 , and the index of refraction of the outer cladding is 1.44.
A.
What is the critical angle for the core-cladding interface?
B.
For what range of angles in the core at the entrance of the fiber (2) will the light be completely internally reflected at the core-cladding interface?
C.
What range of incidence angles in air does this correspond to?
D.
If light is totally internally reflected at the upper edge of the fiber, will it necessarily be totally internally reflected at the lower edge of the fiber (assuming edges are parallel)?
1.
As before, we will need to combine Snell's Law with geometrical arguments.
2.
We have the same illustration as for Sample Problem 2, with the addition of 5.
3.
The normals are the same as before, and
4.
we will label the angles in the same manner. The only addition is due to part (d) of this problem. We will call the angle of incidence on the bottom edge of the fiber 5. We know n1= 1.00, n2= 1.480, and n4= 1.44.
5.
We want C for the core-cladding interface, values of 2 for which 3 is greater than C, and then the corresponding values for 1 and 5.
6.
The critical angle is found from
C= sin -1(n4/n2).
Geometrical arguments (same as Sample Problem 2) give
2= 3'= 90o- 3.
Snell's Law at the entrance gives
n1sin 1= n2sin 2.
Finally, 3 and 5 are alternate angles for parallel lines. Therefore the two angles are equal:
5= 3.
7.
(a)
Using the definition of the critical angle,
C= sin-1(n4/n2)= sin-1(1.44/1.480)= 76.65o.
The critical angle for the core-cladding interface is 76.7o.
(b)
For total internal reflection to occur, 3 must be greater than the critical angle. Therefore,
3= 90o- 2> C, so
90o- C> 2, and
90o-76.65o= 13.35o> 2.
The angle of refraction at the entrance must be less than 13.4o if total internal reflection is to occur at the upper edge of the fiber.
(c)
If 2< 13.4o, then sin 2< sin (13.4o). (As always in this module, angles are kept between 0o and 90o. Otherwise, this inequality might not hold). So, using Snell's Law,
sin 1= (n2/n1)sin 2< (n2/n1)sin (13.40 o).
Therefore,
1= sin-1[(n2/n1)sin 2] < sin-1[ (n2/n1)sin (13.40o)] .
1< sin -1[ (1.480/1.00)sin (13.40o)] , so
1< 20.0o.
Light will experience total internal reflection if light enters the fiber at an angle less than 20.0o. This maximum value for the angle of incidence on the fiber is called the cut-off angle
(d)
Yes. Since 5= 3, if 3 is greater than the critical angle, then so is 5.
8.
First, let's check the appropriateness of our answers. As light travels from air into the fiber, it should bend toward the normal, so 2 should be smaller than 1. The allowed values calculated for these angles match this expectation. The smaller 2 is, the larger 3 should be, so a minimum value for 3 (the critical angle) should correspond to a maximum value for 2, which it does. Since the values of n for the core and cladding are not very different, total internal reflection should not occur unless the light strikes the upper edge of the fiber at a very large angle. This too is consistent with our calculated values.
Now, let's work backwards and check our math. If 1< 20.0o, then
2= sin-1[ (n1/n2)sin 1] < sin-1[ (1.00/1.480)sin 20.06o], so
2< 13.40o.
Then,
3= 90o- 2>90o-13.40o.
3>76.60o.
The the critical angle must be 76.6o:
To the three digits of accuracy given in our problem, these agree. Our solution appears to be correct.
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