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Optical Path Length

in an optical...

Question

in an optical fibre refractive index of inner part is 1.68 and outer part is 1.68. The numerical aperture of fibre is

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Solution

$N u m e r i c a l a p e r t u r e o f f i b r e = \sqrt{{μ_{1}}^{2} - {μ_{2}}^{2}} = \sqrt{1 . 68^{2} - 1 . 68^{2}} . N u m e r i c a l a p e r t u r e o f f i b r e = 0$

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