Answer to Question #188736 in Optics for Shashank

Question #188736

 Consider a step-index fiber for which n1 = 1.475, n2 = 1.460 and a = 25µm. What is the maximum value of θ for which the ray ill be guided by the fiber. 

I need this answer sir



1
Expert's answer
2021-05-04T18:47:13-0400

Given,

n1=1.475n2=1.460n_1=1.475\\n_2=1.460

a=25 μma=25 \ \mu m

Numerical aperture (NA) =n12n22=(1.475)2(1.460)2=2.252.19=0.210\sqrt{n_1^2-n_2^2}=\sqrt{(1.475)^2-(1.460)^2}=\sqrt{2.25-2.19}=0.210

θmax=sin1(0.210)=12.12°\theta_{max}=sin^{-1}(0.210)=12.12\degree


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment