Answer to Question #167449 in Optics for Rajeev

Question #167449

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%​

1
Expert's answer
2021-03-08T07:27:55-0500

Given,

Core index =1.68

Radius =35μm=35\mu m

Index difference =3.75%=3.75\%

wavelength (λ)=1150nm=1150×109m(\lambda) =1150nm = 1150\times 10^{-9}m

0.0375=1.68×I10.0375=1.68 \times I_1


I1=0.03751.68=0.02232I_1=\frac{0.0375}{1.68}= 0.02232


Index =IoI1=I_o-I_1

=1.680.02232=1.68-0.02232

=1.65=1.65

P=12Io(πW2)P=\frac{1}{2} I_o(\pi W^2)


P=12×1.65×3.14×0.1150P=\frac{1}{2} \times 1.65 \times 3.14 \times 0.1150

=0.0297=0.0297

P=2.5%P=2.5\%


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