Paragraph: Light guidance in an optical fiber can be understood by considering a...
Paragraph:
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sinim.

Question:
For two structures namely S1 with n1=√45/4 and n2=3/2, and S2 with n1=8/5 and n2=7/5 and taking the refractive index of water to be 4/3 and that of air to be 1 , the correct option(s) is(are)
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sinim.

Question:
For two structures namely S1 with n1=√45/4 and n2=3/2, and S2 with n1=8/5 and n2=7/5 and taking the refractive index of water to be 4/3 and that of air to be 1 , the correct option(s) is(are)
Solution:
Let the whole structure is placed in a medium of refractive index n′ , then
n′sini=n1cos(90-θ)
n′sini=n1 cosθ .....(i)
Here for im; θ=C and sinC=n2n1
From equation (i), n′sinim=n1 √1-n22n21=√n21 -n22
⇒sinim=√n21-n22n′
Now, for (i) (NA)s1=34 √4516-94=34×34=916
(NA)s2=3√1516 √6425-4925=3√151615√15=916
For (ii) (NA)s1=√156×34=√158
(NA)s2=34=√155 Not equal
For (iii) (NA)s1=1×34=34
(NA)s2=√154×√155=154×5=34
For (iv) (NA)s1=34
(NA)s2=34√155 Not equal
n′sini=n1cos(90-θ)
n′sini=n1 cosθ .....(i)
Here for im; θ=C and sinC=n2n1
From equation (i), n′sinim=n1 √1-n22n21=√n21 -n22
⇒sinim=√n21-n22n′
Now, for (i) (NA)s1=34 √4516-94=34×34=916
(NA)s2=3√1516 √6425-4925=3√151615√15=916
For (ii) (NA)s1=√156×34=√158
(NA)s2=34=√155 Not equal
For (iii) (NA)s1=1×34=34
(NA)s2=√154×√155=154×5=34
For (iv) (NA)s1=34
(NA)s2=34√155 Not equal
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