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Cal the numerical aperture of the fiber with core index μ1=1.61 and cladding index μ2=1.55

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

Marks : 3M

Year : Dec13

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Given: μ1=1.61 , μ2=1.55

soln: NA= μ12μ22

= 1.6121.552

= 0.1896

=0.4354

Hence, the numerical aperture of fibre is 0.4354

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