Cal the numerical aperture of the fiber with core index $\mu_1 =1.61$ and cladding index $\mu

1.3kviews

Cal the numerical aperture of the fiber with core index $\mu_1 =1.61$ and cladding index $\mu_2=1.55$

written 8.4 years ago by

teamques10 ★ 68k

modified 2.8 years ago by

pedsangini276 • 4.8k

Mumbai University > First year engineering > sem2 > Applied physics 2

Marks : 3M

Year : Dec13

engineering physics

ADD COMMENT EDIT

1 Answer

2views

written 8.4 years ago by

teamques10 ★ 68k

Given: μ1=1.61 , μ2=1.55

soln: NA= $\surd\mu1^2-\mu2^2$

= $\surd1.61^2-1.55^2$

= $\surd0.1896$

=0.4354

Hence, the numerical aperture of fibre is 0.4354

ADD COMMENT EDIT

Please log in to add an answer.