Sample Problems for Numerical Aperture

Sample Problem for Numerical Aperture

Important Information

Light entering a fiber at angles less than the cut-off angle θ_0max will be trapped in the fiber. This cut-off angle is related to the numerical aperture, NA, which is a property of the fiber:

n₀ sin θ_0max = (n₁² - n₂²)^1/2 = NA.

Sample Problem 1:

A step-index fiber has a core index of refraction of n₁ = 1.425. The cut-off angle for light entering the fiber from air is found to be 8.50^o. (a) What is the numerical aperture of the fiber? (b) What is the index of refraction of the cladding of this fiber? (c) If the fiber were submersed in water, what would be the new numerical aperture and cut-off angle?

Solution:

(a)	From the Table of indices of refraction, we see that n₀ = n_air = 1.0003. The numerical aperture is therefore NA = n₀ sin θ_0max = (1.0003) sin (8.50^o) = 0.148.
(b)	The index of refraction of the cladding can be found from the numerical aperture: n₁² - n₂² = NA². n₂² = n₁² - NA² = (1.425)² - (0.1479)² = 2.0088 n₂ = 1.417.
(c)	From the Table, we see that the n₀ = n_water = 1.33. Since the numerical aperture is a property of the fiber and only depends upon n₁ and n₂, it will not change when the medium outside the fiber changes. The cut-off angle, however, will have to change if the numerical aperture is to be unaffected by a change in n₀: NA = 0.148. sin θ_0max = NA/n₀ θ_0max = sin^-1(NA/n₀) = sin^-1(0.1479/1.33) = = sin^-1(0.1112) = 6.38^o.

Practice Exercises

On-Line Reading

At-Home Assignments

Click on the Prev button to go to sample problems for other topics.