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Question 20298
JEE Physics · Hard

Principles of Communication: Optical Fibre MCQ

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Optical FibreHardQuestion 20298

Question

In a step-index optical fibre, core n₁ = 1.50 and cladding n₂ = 1.47. The critical angle θ_c and numerical aperture NA are:
  1. A
    θ_c ≈ 78.5°, NA ≈ 0.298
    Correct
  2. B
    θ_c ≈ 11.5°, NA ≈ 0.5
  3. C
    θ_c ≈ 60°, NA ≈ 0.866
  4. D
    θ_c ≈ 80°, NA ≈ 0.1

Correct answer

θ_c ≈ 78.5°, NA ≈ 0.298

Explanation

sin θc = n₂/n₁ = 1.47/1.50 = 0.98 → θc = sin⁻¹(0.98) ≈ 78.5°. NA = √(n₁²−n₂²) = √(2.25−2.1609) = √0.0891 ≈ 0.298. Acceptance angle in air = sin⁻¹(NA) ≈ 17.3°.

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