Question

Derive the equation for acceptance angle and numerical aperture, of optical fiber.

Answer 1

1. To ensure the critical angle incident in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering into it. This angle is called an acceptance angle.
   
   acceptance angle
   
   acceptance cone
2. By Snell’s law
   n₃ sin i_a = n₁ sin r_a
   To have the internal reflection inside optical fibre,
   n₁ sin i₁ = n₂ sin 90°
   n₁ sin i_c = n₂ sin 90° = 1
   ∴ sin i_c = `"n"_2/"n"_1`
3. From the right angle triangle AABC,
   i_c = 90° – r_a
   Now, equation becomes
   sin (90° – r_a) = `"n"_2/"n"_1`
4. Using trigonometry,
   cos r_a = `"n"_2/"n"_1`
   sin r_a = `sqrt(1 - cos^2 "r"_"a")`
   Substituting for cos r_a
   sin r_a = `sqrt(1 - ("n"_2/"n"_1)^2) = sqrt(("n"_1^2 - "n"_2^2)/"n"_1^2)`
   Substituting this in equation (1)
   n₃ sin i_a = `"n"_1 sqrt(("n"_1^2 - "n"_2^2)/"n"_1^2) = sqrt("n"_1^2 - "n"_2^2)`
   On further simplification,
   sin i_a = `sqrt(("n"_1^2 - "n"_2^2)/"n"_3)` or sin i_a = `sqrt(("n"_1^2 - "n"_2^2)/"n"_3)`
   `"i"_"a" = sin^-1 (sqrt(("n"_1^2 - "n"_2^2)/"n"_3))`
   If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes,
   `"i"_"a" = sin^-1 (sqrt("n"_1^2 - "n"_2^2))`
5. Light can have any angle of incidence from o to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone.
   The term (n₃ sin i_a) is called numerical aperture NA of the optical fibre
   NA = n₃ sin i_a = `sqrt("n"_1^2 - "n"_2^2)`
6. 6. If outer medium is air, then n₃ = 1. The numeric aperture NA becomes,
   NA = sin i_a = `sqrt("n"_1^2 - "n"_2^2)`