Question

calculate the numerical aperture and acceptance angle of optical fibre with core of refractive index 1.62 and cladding of refractive index 1.52.

Answer 1

Answer:

Numerical Aperture = $\sqrt{n1^{2} - n2^{2} }$

Hence; Numerical Aperture = 0.56

Acceptance Angle = $Sin^{-1}\alpha$

So, its $34.05^{0}$.

Answer 2

Given: refractive index of core = 1.62

           refractive index of cladding = 1.52

To find: (a) Numerical aperture

             (b) Acceptance angle of optical fiber

Solution:

(a) Numerical aperture is defined as the measurement of an optical fiber's ability to collect the occurrence light ray in it is known as the numerical aperture.

Numerical aperture can be calculated by the formula = $\sqrt{n1^2 - n2^2$

Where n₁ is the refractive index of the core, and n₂ is the refractive index of the cladding.

= $\sqrt{1.62^2 - 1.52^2$

= $\sqrt{2.62 - 2.31$

= 0.56

Therefore, the natural aperture is 0.56

(b) The acceptance angle of an optical fiber 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.

It is calcultaed as = Sin⁻¹α

So, it is 34.05°

Therefore, the acceptance angle is 34.05°