The speed of light in vacuum and in the core of a step-index fiber is 3x108 ms–1 and 2x108 ms–1, respectively. When the fiber is placed in air, the critical angle at the core–cladding interface is 75°. Calculate the NA of the fiber
Answers
Answer:
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.
Explanation:
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