Question 24: 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 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 sin im.

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.

Answer

NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index $${{16} \over {3\sqrt {15} }}$$

Answer

NA of S1 immersed in liquid of refractive index $${6 \over {\sqrt {15} }}$$ is the same as that of S2 immersed in water

Answer

NA of S1 placed in air is the same as that S2 immersed in liquid of refractive index $${4 \over {\sqrt {15} }}$$

Answer

NA of S1 placed in air is the same as that of S2 placed in water

Ray Optics

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