A 4 km optical link consists of multimode step index fiber with core refractive index of 1.3 and a relative refractive index difference of 1%. Find the delay difference between the slowest and fastest modes at the fiber output.

Question

Quizplus · Accepted Answer

The delay difference between the slowest and fastest modes in an optical fiber is given by the formula: $\Delta \tau = \frac{L \cdot \Delta n}{c}$, where $L$ is the length of the fiber, $\Delta n$ is the difference in refractive indices between the core and cladding, and $c$ is the speed of light in vacuum ($3 \times 10^8$ m/s). The relative refractive index difference ($\Delta$) is given by $\frac{\Delta n}{n_1}$, where $n_1$ is the refractive index of the core. Therefore, $\Delta n = n_1 \cdot \Delta$.Given $n_1 = 1.3$, $\Delta = 1\% = 0.01$, and $L = 4$ km = $4000$ m, we find:$\Delta n = 1.3 \times 0.01 = 0.013$.Substituting these values into the delay difference formula:$\Delta \tau = \frac{4000 \cdot 0.013}{3 \times 10^8} = \frac{52}{3 \times 10^8}$ s.Converting to microseconds ($\mu$s) by multiplying by $10^6$, we get:$\Delta \tau = \frac{52}{300} \mu s \approx 0.173 \mu s$.

A 4 Km Optical Link Consists of Multimode Step Index