Question

The velocity of light in the core of a step index fibre is $2 \times 10^{8} \mathrm{m} / \mathrm{s}$ and the critical angle at the core-cladding interfere is $80^{\circ}$. Find the numerical aperture and the acceptance angle for the fibre in air. The velocity of light in vacuum is $3 \times 10^{8} \mathrm{m} / \mathrm{s}$
(A) $0.264;75.3{}^\circ$
(B) $0.464;45.3{}^\circ$
(C) $0.364;25.3{}^\circ$
(D) $0.264;15.3{}^\circ$

Answer 1

Hint We know that step-index fibres have a uniform core with one index of refraction, and a uniform cladding with a smaller index of refraction. (Air serves as the cladding in the simple glass tube example.) When plotted on a graph as a function of distance from the center of the fibre, the index of refraction resembles a step-function. Refractive Index (Index of Refraction) is a value calculated from the ratio of the speed of light in a vacuum to that in a second medium of greater density. The refractive index variable is most commonly symbolized by the letter n or n' in descriptive text and mathematical equations.

Complete step by step answer
We know that numerical aperture is defined as the sine of half of the angle of fibre's light acceptance cone.
The formula to find numerical aperture is given as:
$\sqrt{{{n}^{2}}_{core}-{{n}^{2}}_{cladding}(1+\dfrac{{{r}_{00}}}{2R})}$
Once we put the values where the step index fibre is $2 \times 10^{8} \mathrm{m} / \mathrm{s}$ and the velocity of light in vacuum is $3 \times 10^{8} \mathrm{m} / \mathrm{s}$.
After the evaluation we get that:
The value of numerical aperture 0.264.
We know that the acceptance angle of an optical fiber is defined based on a purely geometrical consideration ( ray optics ): it is the maximum angle of a ray (against the fiber axis) hitting the fiber core which allows the incident light to be guided by the core.
The above statement can be written in the formula format in the following way:
$NA=\operatorname{Sin}{{\theta }_{a}}$
In the above expression ${{\theta }_{a}}$ is the acceptance angle.
Here the value will be after we put the mentioned values is ${{15.3}^{{}^\circ }}$.

Hence the correct answer is option D.

Note We can conclude that the numerical aperture (NA) of an optical system (e.g. an imaging system) is a measure for its angular acceptance for incoming light. It is defined based on geometrical considerations and is thus a theoretical parameter which is calculated from the optical design. Numerical aperture is an important consideration when trying to distinguish detail in a specimen viewed down the microscope. NA is a number without units and is related to the angles of light which are collected by a lens. n optics, the numerical aperture (NA) of an optical system is a dimensionless number that characterizes the range of angles over which the system can accept or emit light.