Question

. A light signal is travelling through a fiber. What is the delay in the signal if the length of the fiber-optic cable is 10 m, 100 m, and 1 Km (assume a propagation speed of2x 108 ill)?

Answer 1

Answer:

The delay=distance/ (propagation speed).

therefore, we have:

Explanation:

a. Delay = 10/ (2×108)       = 0.05ms

b. Delay = 100/ (2×108)     = 0. 5ms

c. Delay = 1000/ (2×108)   = 5ms

Answer 2

The delay in the signal for the length of the fiber-optic cable 10 m, 100 m, and 1 Km is 100 ns, 1 micro sec, and 10 micro sec respectively

Explanation:

• Here the signal is traveling through optic fiber.

Also given the propagation speed of the cable $s=2\times10^{8} m/s$.

• Here we need to derive out the delay in signal for different lengths of cables.
• The delay in signal is also generally referred to as transmission delay.
• The transmission time is twice the propagation time of the network.

Here transmission time $t_{t}=2\times t_{p}$.

Also propagation time $t_{p}=\frac{Distance}{speed}$

As a result, $t_{t}=2\times \frac{distance}{2 \times 10^{8}}$

• For distance=10 m,

$t_{t}=2\times \frac{10}{2 \times 10^{8}}$

$t_{t}=100 \hspace{0.1 cm}ns$

• For distance=100 m,

$t_{t}=2\times \frac{100}{2 \times 10^{8}}$

$t_{t}=1 \hspace{0.1 cm} \mu s$

• For distance=1  km

$t_{t}=2\times \frac{1000}{2 \times 10^{8}}$

$t_{t}=10 \hspace{0.1 cm} \mu s$

Hence, For $distance=10 m$ ==>  $t_{t}=100 \hspace{0.1 cm}ns$,

            for $distance=100 m$ ==>  $t_{t}=1 \hspace{0.1 cm}\mu s$ &

           for $distance=1 km$ ==>  $t_{t}=10 \hspace{0.1 cm}\mu s$.