In figure, the optical fiber is l = 2m long and has a diameter of d = 20 μ m. If a ray of light is incident on one end of the fiber at angle
θ₁ = 40º , the number of reflection it makes before emerging from the other end is close to:
(A) 57000 (B) 45000
(C) 66000 (D) 55000
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The number of reflection it makes before emerging from the other end is close to:
(A) 57000
This can be calculated as follows:
- Given, l = 2m and θ₁ = 40º
- Thus, As per the question,
l sin 40º = 1.31 sin θ
sin θ = 0.64 / 1.31
≈ 1/2
- ∴ θ = 30º
- Now, a = D / ( 1 / √3 )
= 20√3 x 10⁻⁶ m
- Thus, number of reflections = 20 / (20√3 x 10⁻⁶)
≈ 57000
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