Question

An optical fiber made of glass with an index of refraction 1.50 is coated with a plastic with index of refraction 1.30. What is the critical angle of this fiber at the glass-plastic interface? 61.1° 60.1° 90.0° 50.2° 41.8°

Answers

  1. Answer:

    option B

    Explanation:

    Given,

    Refractive index of the glass, n₂ = 1.50

    Refractive index of plastic, n₁ = 1.30

    critical angle = ?

    \theta_{critical}=\sin^{-1} (\dfrac{n_1}{n_2})

    n₁ is the refractive index of the rare medium

    n₂ is the refractive index of the denser medium.

    \theta_{critical}=\sin^{-1} (\dfrac{1.30}{1.50})

    \theta_{critical}=\sin^{-1} (0.8667)

    \theta_{critical}=60.1^\circ

    The correct answer is option B

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