The core refractive index and a relative refractive index difference of a multimode step-index fiber are specified as 1.5 and 2%, respectively. At operating wavelength of 1300 nm, the approximate number of propagating modes is 1000. The diameter of the fiber core is :

Concept:

The number of propagating modes in a multimode step-index fiber is approximately given by, \( M = \frac{V^2}{2} \)

Where V is the normalized frequency (V-number), and is given by, \( V = \frac{2\pi a}{\lambda} \cdot NA \)

NA is the numerical aperture and is defined as, \( NA = n_1 \cdot \sqrt{2\Delta} \)

Where, n₁ = core refractive index, Δ = relative refractive index difference, λ = operating wavelength, a = core radius

Calculation:

Given:

n₁ = 1.5, Δ = 2% = 0.02, λ = 1300 nm = 1.3 × 10^-6 m, M = 1000

Calculate NA: \( NA = 1.5 \cdot \sqrt{2 \cdot 0.02} = 1.5 \cdot \sqrt{0.04} = 1.5 \cdot 0.2 = 0.3 \)

Now, \( M = \frac{V^2}{2} \Rightarrow V^2 = 2M = 2000 \Rightarrow V = \sqrt{2000} \approx 44.72 \)

From, \( V = \frac{2\pi a}{\lambda} \cdot NA \)

\( 44.72 = \frac{2\pi a}{1.3 \times 10^{-6}} \cdot 0.3 \Rightarrow a = \frac{44.72 \cdot 1.3 \times 10^{-6}}{2\pi \cdot 0.3} \approx \frac{58.136 \times 10^{-6}}{1.884} \approx 30.85 \times 10^{-6}~m \)

Core diameter = 2a ≈ \( 61.7~\mu m \)

Correct Option:

The closest answer is: 3) 62 µm