## What is the Numerical Aperture of an Optical Fiber?The optical fiber system shown below has a core of refractive index n \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_1^2-n_2^2}
Take the sin^{-1} of both sides to obtain
\alpha < \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right)
The quantity
N.A = \sqrt{n_1^2-n_2^2}
is called the numerical aperture and
\alpha_{max} = \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right)
is called the angle of acceptance which is the largest angle α for which light is totally reflected at the core-cladding interface and hence guided along the fiber.
## Numerical Aperture of Optical Fiber Sytems Problems with SolutionsA 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. Finda) 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}) = 0.17c) α _{max} = sin^{-1}√(1.46^{2} - 1.45^{2}) = 9.82 °
Problem 2 Problem 3 Problem 4 \alpha_{max} = \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_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_1^2-n_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_1^2-n_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. |