What is the Numerical Aperture of an Optical Fiber?The optical fiber system shown below has a core of refractive index n_{1} and a cladding of refractive index n_{2} such that n_{1} > n_{2}. For a light ray to be internally reflected at the core  cladding interface, the angle on incidence θ must be greater than the critical angle θ_{c} given by
\theta_c = \sin^{1} \left(\dfrac{n_2}{n_1} \right)
When a light ray is incident from the outside (left side of the diagram) with refractive index n at an angle α at the outside  core interface, it will be refracted at an angle β inside the core of the fiber and both angles are related by Snell's law as follows
n \sin\alpha = n_1 \sin\beta
But angles θ and β are complementary; hence
\cos\theta = \sin\beta
which gives the equation
n \sin\alpha = n_1 \cos\theta
Solve the above equation for θ
\theta = cos^{1} \left( \dfrac{n \sin \alpha}{n_1} \right)
To have total internal internal reflection at the core cladding interface, angle θ must be greater than the critical angle θ_{c} given above; hence the inequality
\cos^{1} \left( \dfrac{n \sin \alpha}{n_1} \right) > \sin^{1} \left(\dfrac{n_2}{n_1} \right)
Take the cosine of both sides of the above inequality and change the symbol of the inequality because cos(x) is a decreasing function on the interval [0 , π/2]. Hence
\cos \left(\cos^{1} \left( \dfrac{n \sin \alpha}{n_1} \right) \right) < \cos \left( \sin^{1} \left(\dfrac{n_2}{n_1} \right) \right)
Use the following properties of trigonometric functions and their inverses in the above inequality
\cos(\cos^{1} x ) = x \text{ and } \cos(\sin^{1} x ) = \sqrt{1  x^2}
to simplify and obtain
\dfrac{n \sin \alpha}{n_1} < \sqrt{1\dfrac{n_2^2}{n_1^2}}
Multiply both sides of the inequality by n_{1} / n and simplify
\sin \alpha < \dfrac{1}{n} \sqrt{n_1^2n_2^2}
Take the sin^{1} of both sides to obtain
\alpha < \sin^{1} \left (\dfrac{1}{n} \sqrt{n_1^2n_2^2} \right)
The quantity
N.A = \sqrt{n_1^2n_2^2}
is called the numerical aperture and
\alpha_{max} = \sin^{1} \left (\dfrac{1}{n} \sqrt{n_1^2n_2^2} \right)
is called the angle of acceptance which is the largest angle α for which light is totally reflected at the corecladding interface and hence guided along the fiber.
Numerical Aperture of Optical Fiber Sytems Problems with Solutions
A numerical aperture of optical fibers calculator is included in this site and may be used to check the calculations in the following problems.
Problem 1
let n = 1, n_{1} = 1.46 and n_{2} = 1.45 in the diagram of the optical fiber system above. Find
a) the critical angle θ_{c} at the core  cladding interface.
b) the numerical aperture N.A. of the optical fiber
c) the angle of acceptance α_{max} of the the optical fiber system.
Solution to Problem 1
a) θ_{c} = sin^{1} (n_{2} / n_{1}) = sin^{1} (1.45 / 1.46) = 83.29 °
b) N.A. = √(n^{2}_{1}  n^{2}_{2}) = √(1.46^{2}  1.45^{2}) = 0.17
c) α_{max} = sin^{1}√(1.46^{2}  1.45^{2}) = 9.82 °
Problem 2
We use the same values for n , n_{1} and n_{2} in the diagram of the optical fiber system above as those used in problem 1. Let the angle of incidence of a light ray on the outside  core interface be α = 5°. Find
a) angle of refraction β at the outside  core interface.
b) angle θ
c) and explain why this light ray will be reflected at the core  cladding interface and hence guided along the fiber.
Solution to Problem 2
a) Angle β is found using Snell's law at the outside  core interface as follows
n sin(α) = n_{1} sin(β)
Substitute the given parameters from problem 1 to obtain
β = sin^{1} ( sin(5°) / n_{1}) = 3.42 °
b) Angle θ is complementary to angle β hence
θ = 90  3.42 = 86.58 °
c) The angle of incidence θ = 86.58 ° at the core  cladding interface is larger that the critical angle θ_{c} = 83.29 ° calculated in problem 1 above and will therefore be totally reflected at this interface and hence guided along the fiber.
Of course we could have answered this question by stating that since angle α = 5° is smaller to α_{max} = 9.82 ° calculated in problem 1, the given light ray will be reflected at the core  cladding interface, but the idea behind this problem is to gain better understanding of the concept of numerical aperture and angle of acceptance of an optical fiber systems using numerical values and calculations at each interface.
Problem 3
let n = 1 and n_{1} = 1.48 in the diagram of the optical fiber system above. Find n_{2} such that light rays incident at an angle α greater than 12 ° are not reflected at the core  cladding and therefore not guided along the optical fiber system.
Solution to Problem 3
In this problem, we are given α_{max} = 12 ° given by the formula
α_{max} = sin^{1}(√(n_{1}^{2}  n_{2}^{2})/ n)
The above is equivalent to
sin(α_{max}) = √(n_{1}^{2}  n_{2}^{2})/ n
Substitute α_{max}, n and n_{1} by the given values and calculate n_{2}
n_{2}^{2} = n_{1}^{2}  sin^{2}(α_{max})
n_{2} = √(1.48^{2}  sin^{2}(12°)) = 1.465
Problem 4
let n = 1 in the diagram of the optical fiber system above. Find n_{1} and n_{2} such that the acceptance α_{max} = 10° and the critical angle at the core  cladding interface θ_{c} = 82 ° .
Solution to Problem 4
In this problem, we are given α_{max} and θ_{c} whose formulas are given by \alpha_{max} = \sin^{1} \left (\dfrac{1}{n} \sqrt{n_1^2n_2^2} \right)
and
\theta_c = \sin^{1} \left(\dfrac{n_2}{n_1} \right)
Take the sine of the first formula, simplify and square both sides to obtain
\sin^2\alpha_{max} = \left (\dfrac {n_1^2n_2^2} {n^2} \right)
The second equation is equivalent to
\sin \theta_c = \left(\dfrac{n_2}{n_1} \right)
Substitute the known values to obtain the equations
n_1^2n_2^2 = n^2 \sin^2 \alpha_{max} \quad\quad (equation 1) \dfrac{n_2}{n_1} = \sin \theta_c \quad\quad (equation (2)
The last equation gives
n_2 = n_1 \sin \theta_c
Substitute the above in equation (1) and solve for n_{1} and n_{2}n_1^2 (n_1 \sin \theta_c )^2 = n^2 \sin^2 \alpha_{max} n_1^2(1  \sin^2 (\theta_c)) = n^2 \sin^2 \alpha_{max} n_1^2 = \dfrac{\sin^2 \alpha_{max} }{1  \sin^2 \theta_c} = n^2 \dfrac{\sin^2 \alpha_{max}}{ \cos^2 \theta_c } n_1 = \dfrac{n\sin \alpha_{max} }{ \cos (\theta_c)} n_2 = n_1 \sin \theta_c = \dfrac{n \sin(\alpha_{max})}{ \cos \theta_c} \sin \theta_c = n \sin \alpha_{max} \tan \theta_c
Substitute n , θ_{c} and α_{max} by their values to obtain numerical values for n_{1} and n_{2}n_1 = \dfrac{\sin 10^{\circ} }{ \cos 82^{\circ}} = 1.2477
and
n_2 = \tan 82^{\circ} \sin 10^{\circ} = 1.2355 More References and LinksOptical Fibers.
numerical aperture of optical fibers calculator
Total Internal Reflection of Light Rays at an Interface.
Refraction of Light Rays, Examples and Solutions.
