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Standard XII

Physics

Optical Fiber

In optical fi...

Question

In optical fiber, refractive index of inner part is $1.68$ and refractive index of outer part is $1.44$ . The numerical aperture of the fibre is

0.5653

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0.6653

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0.7653

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0.8653

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Solution

The correct option is C 0.8653
$N u m e r i c a l$ $a p e r t u r e$ $o f$ $f i b r e$ $=$ $\sqrt{μ_{1}^{2} - μ_{2}^{2}}$ $= \sqrt{{1.68}^{2} - {1.44}^{2}}$
$= \sqrt{2.8224 - 2.0736}$ $= \sqrt{0.7488} = 0.8653$

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In optical fiber, refractive index of inner part is 1.68 and refractive index of outer part is 1.44. The numerical aperture of the fibre is

The correct option is C 0.8653Numerical aperture of fibre = √μ21−μ22 =√1.682−1.442=√2.8224−2.0736 =√0.7488=0.8653

In optical fiber, refractive index of inner part is $1.68$ and refractive index of outer part is $1.44$ . The numerical aperture of the fibre is

The correct option is C 0.8653
$N u m e r i c a l$ $a p e r t u r e$ $o f$ $f i b r e$ $=$ $\sqrt{μ_{1}^{2} - μ_{2}^{2}}$ $= \sqrt{{1.68}^{2} - {1.44}^{2}}$
$= \sqrt{2.8224 - 2.0736}$ $= \sqrt{0.7488} = 0.8653$