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

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Cal the numerical aperture of the fiber with core index

$\mu_1 =1.61$ and cladding index

$\mu_2=1.55$

written 8.8 years ago by

teamques10 ★ 69k

modified 3.2 years ago by

pedsangini276 • 4.8k

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

Marks : 3M

Year : Dec13

engineering physics

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written 8.8 years ago by

teamques10 ★ 69k

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

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