Question

In optical fiber, the refractive index of the inner part is 1.68 and the refractive index of outer part is 1.44. The numerical aperture of the fiber is:
A. 0.5653
B. 0.6653
C. 0.7653
D. 0.8653

Answer 1

Hint: An optical fiber makes use of the principle of total internal reflection which states that while light travels from an optically denser to an optically rarer medium, if the angle of incidence is higher than a certain critical angle, it is completely reflected back into the optically denser medium from the interface between the two media.
Numerical aperture is a measure of the ability of the optical fiber to keep the light travelling inside it and not let it escape to the outside.

Formula used:
$\text{Numerical aperture of an optical fiber = }\sqrt{{{\mu }_{1}}^{2}-{{\mu }_{2}}^{2}}$
Where ${{\mu }_{1}}$ and ${{\mu }_{2}}$ are the refractive indices of the optically denser inner material of the optical fiber and the optically rarer outer coating material.

Complete Step-by-Step solution:
An optical fiber makes use of the principle of total internal reflection which states that while light travels from an optically denser to an optically rarer medium, if the angle of incidence is higher than a certain critical angle, it is completely reflected back into the optically denser medium from the interface between the two media.
It allows the travel of information at the speed of light by continuous total internal reflections throughout the length of the fiber.
The numerical aperture is a measure of the ability of the optical fiber to keep the light travelling, inside it and not let it escape to the outside. The higher the numerical aperture, the higher is the efficiency of light or information retention in the optical fiber.
$\text{Numerical aperture of an optical fiber = }\sqrt{{{\mu }_{1}}^{2}-{{\mu }_{2}}^{2}}$ --(1)
Where ${{\mu }_{1}}$ and ${{\mu }_{2}}$ are the refractive indices of the optically denser inner material of the optical fiber and the optically rarer outer coating material.
Now, let us analyse the information given to us.
The refractive index of the inner part is ${{\mu }_{1}}=1.68$ --(2)
The refractive index of the outer part is ${{\mu }_{2}}=1.44$ --(3)
For finding out the numerical aperture, we will put (2) and (3) in (1),
$\therefore \text{Numerical aperture = }\sqrt{{{1.68}^{2}}-{{1.44}^{2}}}=\sqrt{2.8224-2.0736}=\sqrt{0.7488}=0.8653$
Hence, the numerical aperture is 0.8653.
So, the correct option is D) 0.8653.

Note: An optical fiber with a high numerical aperture is always preferred, since it allows for more light retention even at lower angles of incidence. Thus, manufacturers always try to keep the numerical aperture as high as possible either by keeping a high refractive index material inside or a material with a much lower refractive index as the outside.
Optical fibers have changed the world and become the fastest carriers of information since they literally carry information at the speed of light.