Question

A step-index has a numerical aperture of 0.17 and a cladding refractive index of 1.46. Determine (i) the acceptance angle of the fiber when it is placed in water (refractive index =1.33) and (ii) the critical angle at the core-cladding interface.

Answer 1

The Acceptance angle is 0 and critical angle is 90°.

Explanation: We have N.A = $\frac{1}{no} \sqrt{n1^{2} -n2^{2}$

When the fibre is in air , n₀ = 1

then N.A = $\sqrt{(n1) ^{2} -(n2)^{2}$  = 0.17

n₁ = $\sqrt{NA}^{2} + (n1)^{2}$

n₁ = $\sqrt{(0.17)^{2} + (1.46)^{2}$  = 1.46

When the fibre is in water n₀ = 1.33

N.A = $\frac{1}{1.33}\sqrt{(n1) ^{2} -(n2)^{2}$  = $\frac{1}{1.33}\sqrt{(1.46)^{2} - (1.46)^{2}$

N.A = 0

Now Acceptance angle i₀= sin-¹(0) = 0

Critical angle Sin∅c = n₂/n₁ = 1.46/1.46 = 1

∅c = 90°