Answer to Question #167449 in Optics for Rajeev

Question

For multimode step-index fiber with core index of 1.68 and radius of 35 um. The indexdifference is 3.75%. For a wavelength of 1150 nm, the optical power propagating incladding is what percentage of total power?(a) 2.1%(b) 2.5%(c) 1.8%(d) 1.4%​

Answer 1

Given,

Core index =1.68

Radius $=35\mu m$

Index difference $=3.75\%$

wavelength $(\lambda) =1150nm = 1150\times 10^{-9}m$

$0.0375=1.68 \times I_1$

$I_1=\frac{0.0375}{1.68}= 0.02232$

Index $=I_o-I_1$

$=1.68-0.02232$

$=1.65$

$P=\frac{1}{2} I_o(\pi W^2)$

$P=\frac{1}{2} \times 1.65 \times 3.14 \times 0.1150$

$=0.0297$

$P=2.5\%$