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    an optical fiber has core-index of 1.480 and a cladding index of 1.478. what should be the core size for single mode operation at 1310nm?

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    [Solved] Optical fiber has a core index of 1.480 and a cladding index of 1.478. What should be the core size for single

    Optical fiber has a core index of 1.480 and a cladding index of 1.478. What should be the core size for single-mode operation at 1310nm?

    Optical fiber has a core index of 1.480 and a cladding index of 1.478. What should be the core size for single-mode operation at 1310nm?

    Optical fiber has a core index of 1.480 and a cladding index of 1.478. What should be the core size for single-mode operation at 1310nm?

    Right Answer is:

    6.50μm

    SOLUTION

    The normalized frequency for the fiber can be adjusted within the range by 0 ≤ V≤ 2.405 reducing core radius and refractive index difference < 1%.

    Normalized frequency V≤ 2.405 is the value at which the lowest order Bessel function J=0. Core size(radius).

    $a=\frac{v\lambda }{2\pi }\frac{1}{\sqrt{n_{1}^{2}-n_{2}^{2}}}$

    $a = \frac{2.405\times 1.55}{2\Pi }\frac{1}{\sqrt{(1.480)^{2}-}(1.478)^{2}}$

    a = 6.50

    If this fiber also should be single-mode at 1310 nm, then the core radius must be 6.50

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    Q.

    An optical fiber has core-index of 1.480 and a cladding index of 1.478. What should be the core size for single mode operation at 1310nm?

    A. 7.31μm B. 8.71μm C. 5.26μm D. 6.50μm Answer» D. 6.50μm

    Explanation: normalized frequency v<=2.405 is the value at which the lowest order bessel function j=0. core size(radius)

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