Compute the maximum radius allowed for a fiber having core refractive index 1.5 and 1.48.

Compute the maximum radius allowed for a fiber having core refractive index 1.5 and 1.48. the fiber is to support only one mode at a wavelength of 1500 nm

Maximum core radius ≈ 2.35 µm.

A step-index fibre supports only one mode when its V-number is below 2.405:

\[V = \frac{2\pi a}{\lambda}\,\text{NA} \le 2.405\]

Step 1: Numerical aperture (\(n_1 = 1.5,\ n_2 = 1.48\)):

\[\text{NA} = \sqrt{n_1^2 - n_2^2}\]

\[= \sqrt{2.25 - 2.1904} = \sqrt{0.0596}\]

\[\text{NA} = 0.2441\]

Step 2: Maximum radius (\(\lambda = 1500\ \text{nm}\)):

\[a_{max} = \frac{2.405\,\lambda}{2\pi\,\text{NA}}\]

\[= \frac{2.405 \times 1500\times10^{-9}}{2\pi \times 0.2441}\]

\[= \frac{3.6075\times10^{-6}}{1.5339}\]

\[a_{max} \approx 2.35\times10^{-6}\ \text{m} = 2.35\ \mu\text{m}\]

So the core radius must not exceed about 2.35 µm (core diameter about 4.7 µm) for single-mode operation at 1500 nm.