A multimode step-index fiber has a core refractive index of 1.5 and relative refractive index difference of 1%. The length of the optical link is 6 km. Estimate the RMS pulse broadening due to intermodal dispersion on the link.a. a.86.7 nsb. b.69.3 nsc. c.92.6 nsd. d.68.32 ns

The RMS pulse broadening due to intermodal dispersion in an optical fiber can be estimated using the formula:

Δτ = L * (n1 * Δn/c)

where:

• Δτ is the pulse broadening (in seconds)
• L is the length of the fiber (in meters)
• n1 is the core refractive index
• Δn is the refractive index difference
• c is the speed of light (approximately 3*10^8 m/s)

Given in the problem:

• n1 = 1.5
• Δn/n1 = 1% = 0.01, so Δn = 0.01 * n1 = 0.01 * 1.5 = 0.015
• L = 6 km = 6000 m

Substituting these values into the formula, we get:

Δτ = 6000 * (1.5 * 0.015 / 3*10^8) = 0.00000003 seconds = 0.03 microseconds = 30 nanoseconds

So, none of the options given (a. 86.7 ns, b. 69.3 ns, c. 92.6 ns, d. 68.32 ns) match the calculated value. There might be a mistake in the problem or the provided options.

The RMS pulse broadening due to intermodal dispersion in an optical fiber can be estimated using the formula:

Δτ = L * (n1 * Δn/c)

where:

• Δτ is the RMS pulse broadening (in seconds)
• L is the length of the optical link (in meters)
• n1 is the core refractive index
• Δn is the relative refractive index difference (n1 - n2)
• c is the speed of light in vacuum (approximately 3 * 10^8 m/s)

Given:

• n1 = 1.5
• Δn = 1% of n1 = 0.01 * 1.5 = 0.015
• L = 6 km = 6000 m

we can substitute these values into the formula to get:

Δτ = 6000 * (1.5 * 0.015 / 3 * 10^8) = 0.00000003 seconds = 0.03 microseconds = 30 nanoseconds

So, the RMS pulse broadening due to intermodal dispersion on the link is approximately 30 ns. This is not an option in the given choices, so there may be a mistake in the question or the provided options.