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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)
NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16315
NA of S1 immersed in liquid of refractive index 615 is the same as that of S2 immersed in water
NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 415
NA of S1 placed in air is the same as that of S2 placed in water
Solution:
Let the whole structure is placed in a medium of refractive index n , then
nsini=n1cos(90-θ)
nsini=n1 cosθ .....(i)
Here for im; θ=C and sinC=n2n1 
From equation (i), nsinim=n1 1-n22n21=n21 -n22
sinim=n21-n22n
Now, for (i) (NA)s1=34 4516-94=34×34=916
(NA)s2=31516 6425-4925=315161515=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=34155 Not equal