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प्रश्न
Derive the equation for acceptance angle and numerical aperture, of optical fiber.
उत्तर
- 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 - By Snell’s law
n3 sin ia = n1 sin ra
To have the internal reflection inside optical fibre,
n1 sin i1 = n2 sin 90°
n1 sin ic = n2 sin 90° = 1
∴ sin ic = `"n"_2/"n"_1` - From the right angle triangle AABC,
ic = 90° – ra
Now, equation becomes
sin (90° – ra) = `"n"_2/"n"_1` - Using trigonometry,
cos ra = `"n"_2/"n"_1`
sin ra = `sqrt(1 - cos^2 "r"_"a")`
Substituting for cos ra
sin ra = `sqrt(1 - ("n"_2/"n"_1)^2) = sqrt(("n"_1^2 - "n"_2^2)/"n"_1^2)`
Substituting this in equation (1)
n3 sin ia = `"n"_1 sqrt(("n"_1^2 - "n"_2^2)/"n"_1^2) = sqrt("n"_1^2 - "n"_2^2)`
On further simplification,
sin ia = `sqrt(("n"_1^2 - "n"_2^2)/"n"_3)` or sin ia = `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 ia becomes,
`"i"_"a" = sin^-1 (sqrt("n"_1^2 - "n"_2^2))` - Light can have any angle of incidence from o to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone.
The term (n3 sin ia) is called numerical aperture NA of the optical fibre
NA = n3 sin ia = `sqrt("n"_1^2 - "n"_2^2)` - 6. If outer medium is air, then n3 = 1. The numeric aperture NA becomes,
NA = sin ia = `sqrt("n"_1^2 - "n"_2^2)`
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