Correct Answer - Option 3 : 3 μm
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 mathematically given by:
\({\lambda _c} = \frac{{2\pi a}}{{{V_c}}}\times NA\) ---(1)
λc = 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.
NA = Numerical aperture calculated as:
\(NA=\sqrt{n_1^2-n_2^2}\)
Calculation:
Given n1 = 1.484, n2 = 1.402 and diameter = 5 μm
The radius of the core will be 2.5 μm.
\(NA=\sqrt{(1.484^2-1.402^2}\)
Δ = 0.486
Putting on the respective values in Equation-(1), we get:
\({\lambda _c} = \frac{{2\pi \left( {2.5 \times {{10}^{ - 6}}} \right)}}{{2.4}}\times 0.486\)
λc = 3.18 μm ≈ 3 μm
