[amp_mcq option1=”12.5ns km-1″ option2=”9.6ns km-1″ option3=”9.0ns km-1″ option4=”10.2ns km-1″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B) 9.6ns km-1}}$.<\/p>\n
The rms pulse broadening due to material dispersion is given by:<\/p>\n
$$\\Delta \\tau = \\frac{L}{\\nu_0} \\cdot \\frac{D_m}{\\lambda_0^2}$$<\/p>\n
where $L$ is the fiber length, $\\nu_0$ is the central frequency of the light, $D_m$ is the material dispersion coefficient, and $\\lambda_0$ is the central wavelength of the light.<\/p>\n
In this case, we are given that $L = 1 \\text{ km}$, $\\nu_0 = 299,792,458 \\text{ m s}^{-1}$, $D_m = 150 \\text{ ps nm}^{-1} \\text{ km}^{-1}$, and $\\lambda_0 = 600 \\text{ nm}$. Substituting these values into the equation, we get:<\/p>\n
$$\\Delta \\tau = \\frac{1 \\text{ km}} {299,792,458 \\text{ m s}^{-1}} \\cdot \\frac{150 \\text{ ps nm}^{-1} \\text{ km}^{-1}} {600 \\text{ nm}^2} = 9.6 \\text{ ns km}^{-1}$$<\/p>\n
Therefore, the rms pulse broadening due to material dispersion is $\\boxed{\\text{9.6ns km-1}}$.<\/p>\n
The other options are incorrect because they do not take into account the correct values of the parameters.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”12.5ns km-1″ option2=”9.6ns km-1″ option3=”9.0ns km-1″ option4=”10.2ns km-1″ correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[684],"tags":[],"class_list":["post-51435","post","type-post","status-publish","format-standard","hentry","category-optical-communication","no-featured-image-padding"],"yoast_head":"\n