Answer to Question #168544 in Optics for Tushar Singhal

Question

A multimode step index fiber with a core diameter of 80 μm and a relative index difference of 1.5 % is

operating at a wavelength of 0.85 μm. If the core refractive index is 1.48, estimate the normalized

frequency for the fiber and number of guided modes.

Answer 1

Answer

Here wavelegth

$\lambda=085\mu m$

Radius of core

$a=D/2=40\mu m$

$n_1=$ core refractive index=1.48

$\Delta=0.015$

Putting all value in below equation

normalised frequency of fiber

$f= (2π/λ) a n_1 (2Δ)^{0.5}\\=\frac{2\times3.14\times40\mu m\times1.48}{0.085\mu m}\times(2\times0.015)^{0.5}\\= 75.8Hz$

Now new parameters

$V=\frac{2\pi a}{\lambda}(\sqrt{n_1^2-n_2^2})\\= \frac{2\times3.14\times0.00004}{0.85\mu m}(\sqrt{0.015}) \\ =75.80$

the number of guided modes.

$M ≈\frac{ V^2} { 2 }= 2873$

(i.e. nearly 3000 guided modes!)