Question 374: Ray Optics

Question

Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n
1
surrounded by a medium of lower refractive index n
2
. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n
1
and n
2
as shown in the figure. All rays with the angle of incidence i less than a particular value i
m
are confined in the medium of refractive index n
1
. The numerical aperture (NA) of the structure is defined as sin i
m
.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0wl450o/9fa571d0-0c9d-45e9-8987-2801d260e167/45908e80-a6d1-11ec-9cd8-49eccebff69a/file-1l0wl450p.png?format=png)

Accepted Answer

Let the whole structure be placed in the medium of refractive index n
0
. From geometry, if the angle of incidence at Q is $	heta$ then the angle of refraction at P is 90$^\circ$ $-$ $	heta$.
The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e.,
$	heta$ $\ge$ $	heta$
c
= sin
$-$1
(n
2
/ n
1
). ........ (1)
Apply Snell's law for refraction at P to get
n
0
sin i = n
1
sin(90$^\circ$ $-$ $	heta$) = n
1
cos$	heta$
= ${n_1}\sqrt {1 - {{\sin }^2}	heta } $. ...... (2)
From equation (2), the angle of incidence is maximum (i = i
m
) when $	heta$ is minimum i.e., when $	heta$ = $	heta$
c
(from equation (1)). Thus, the numerical aperture is given by
$NA = \sin {i_m} = {{{n_1}\sqrt {1 - {{\sin }^2}{	heta _c}} } \over {{n_0}}} = {{\sqrt {n_1^2 - n_2^2} } \over {{n_0}}}$ ........ (3)
Substitute ${n_1} = \sqrt {45} /4$ and ${n_2} = 3/2$ in equation (3) to get the numerical aperture for the structure S
1
is
$N{A_1} = {{\sqrt {45/16 - 9/4} } \over {{n_0}}} = {3 \over {4{n_0}}}$ ...... (4)
Similarly, substitute n
1
= 8/4 and n
2
= 7/5 in equation (3) to get the numerical aperture for the structure S
2
as
$N{A_2} = {{\sqrt {64/25 - 49/25} } \over {{n_0}}} = {{\sqrt {15} } \over {5{n_0}}}$ ...... (5)
In case (A), substitute n
0
= 4/3 in equation (4) to get NA
1
= 9/16 and substitute n
0
= 16/3$\sqrt15$ in equation (5) to get NA
2
= 9/16.
In case (B), substitute n
0
= 6/$\sqrt15$ in equation (4) to get NA
1
= $\sqrt15$/8 and substitute n
0
= 4/3 in equation (5) to get NA
2
= 3$\sqrt15$/20.
In case (C), substitute n
0
= 1 in equation (4) to get NA
1
= 3/4 and substitute n
0
= 4/$\sqrt15$ equation (5) to get NA
2
= 3/4.
In case (D), substitute n
0
= 1 in equation (4) to get NA
1
= 3/4 and substitute n
0
= 4/3 equation (5) to get NA
2
= 3$\sqrt15$/20.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l0ym6ht0/c18c24e8-91e1-4fbb-b328-94c06a5239b7/017d3d50-a7ef-11ec-82ec-9768dcc201cf/file-1l0ym6ht1.png?format=png)

Answer

NA of S
1
immersed in water is the same as that of S
2
immersed in a liquid of refractive index ${{16} \over {3\sqrt {15} }}$

Answer

NA of S
1
immersed in liquid of refractive index ${6 \over {\sqrt {15} }}$ is the same as that of S
2
immersed in water

Answer

NA of S
1
placed in air is the same as that S
2
immersed in liquid of refractive index ${4 \over {\sqrt {15} }}$

Answer

NA of S
1
placed in air is the same as that of S
2
placed in water

Ray Optics

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