Question

A single mode fiber with radius of 4.2um with core refractive index =1.48 and that of cladding=1.475, the cutoff wavelength is given by

Answer 1

Answer:

(viii) If you take away 6 from 6 times y, you get 60.

(ix) If you add 3 to one-third of z, you get 30.

Solution:

Question 5.

Write the following equations in statement forms.

Explanation:

please give brainliest answer

Answer 2

Concept:

The CUTOFF WAVELENGTH of a single-mode

fiber is the wavelength above which the fiber propagates only the fundamental mode.

Below cut-off, the fiber will transmit more than one mode. An optical fiber that is single-mode at a particular wavelength may have two or more modes at wavelengths lower than the cutoff wavelength. It is given by:

de = 2πani (24) ³ 2 kan Ve

Ac = Theoretical Cut-Off Wavelength

a = radius of the core

n₁ = Refractive index of the core

Where Vc = Normalized Frequency, which is typically 2.4 for single-mode fiber.

A = Relative Refractive Index Difference (Between the core and the cladding)Calculation:

Given n₁ = 1.48, n₂ =1.475 and a = 4.2 μm

A (n²-n²) 2n² n₁

Putting on respective values in Equation-(1), we get;

Ac = 2TT (4.2x10-6) (1.48) 2.4 (2A) =/

λο Ξ λ = 1334 nm( please follow)