Question

For a 3-um-diameter optical fiber with core and cladding indexes of refraction of 1.545 and 1.510, respectively. The cut off wavelength is.
1. 2.3 um
2. 1.29 um
3. 1.5 um
4. 3.24 um

Answer 1

Correct Answer - Option 2 : 1.29 um

Concept:

Assume it is a single-mode step-index fiber.

For single-mode step index fiber normalized frequency number (V) ranges:

0 < V ≤ 2.405

∴ \({V_{max}} = \frac{{2\pi \cdot a \cdot \left( {N.A} \right)}}{{{\lambda _c}}}\)

a ⇒ Core radius

For max V, the wavelength is taken as the critical wavelength.

Given:

Core diameter (d) = 3 μm

Core indexes (n₁) = 1.545

Calculation:

\(N.A = \sqrt {n_1^2 - n_2^2} = 0.326\)

\({\lambda _c} = \frac{{2\pi \cdot a \cdot \left( {N.A.} \right)}}{{{V_{max}}}} = \frac{{2\pi \; \times \;1.5\; \times\; 0.326}}{{2.405}}\)

λ_c = 1.29 μm