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The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer.The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer.

The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer.The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer.

Grade:12th pass

4 Answers

Amit kumar
15 Points
2 years ago
  • \Delta=n1-n2/n1
n12_n22=(n1+n2)=(n1+n2/2)(n1_n2/n1)2n1
n1+n2/2\approxn1
and 
NA=\sqrt{}n122\Delta
therfore
na=n1\sqrt{}2\Delta
v-no. or normalized frequence
v=\pi d/\lambda \sqrt{}}n12_n22
No. of fibre traveling
for step index fibre
N=v2/2
for graded index fibre
N=v2/4
abhinab
13 Points
one year ago
The core diameter of multi-mode step index fiber is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Determine the number of guided modes when the operating wavelength is 0.75 micrometer
vivek
13 Points
6 months ago
The core diameter of multimode step index fibre is 60 micrometer. The difference in refractive index is 0.013. The core refractive index is 1.46. Predict the number of guided modes when the operating wavelength is 0.75 micrometer
Mohit Yadav
26 Points
one month ago

Given,

Core diameter of multi mode step index fiber =60 micrometer

Difference in refractive index=0.013

Core refractive index=1.46

To Find,

The number of guided modes when the wavelength is 0.75

Solution,

n1=1.46

n2=1.46+0.013

   =1.473

D=60micrometer

wavelngth=0.75micrometer

NA=√n2²-n1²

  =√1.473²-1.46²

 =0.19

Number of modes=1/2×((π×D×NA)/lambda))²

                              =1/2×((3.14×60×10^-6×0.19)/0.75×10^-6)²

                              =1200

 

Hence, the number of guided modes are 1200.

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