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

A numerical aperture of optical fibers calculator is included in this site and may be used to check the calculations in the following problems.

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 °

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.

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

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

\( \alpha_{max} = \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right) \)
and
\(\theta_c = \sin^{-1} \left(\dfrac{n_2}{n_1} \right) \)

Take the sine of both sides in the formula of \( \alpha_{max} \)

\( \sin \left(\alpha_{max}\right) = \sin \left(\sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right) \right) \)

Simplify using the identity \( \sin (\sin^{-1} (x)) = x \)

\( \sin \left(\alpha_{max}\right) = \dfrac{1}{n} \sqrt{n_1^2-n_2^2} \)

square both sides to obtain

\(\sin^2\alpha_{max} = \left (\dfrac {n_1^2-n_2^2} {n^2} \right) \)

Take the sine of both sides in \(\theta_c = \sin^{-1} \left(\dfrac{n_2}{n_1} \right) \) and simplify to obtain

\(\sin \theta_c = \left(\dfrac{n_2}{n_1} \right) \)

Substitute the known values to obtain the equations

\(n_1^2-n_2^2 = n^2 \sin^2 \alpha_{max} \quad\quad (equation 1) \)

\(\dfrac{n_2}{n_1} = \sin \theta_c \quad\quad (equation (2) \)

Equation (2) gives

\(n_2 = n_1 \sin \theta_c \)

Substitute the above in equation (1) and solve for n_{1} and n_{2}

\(n_1^2- (n_1 \sin \theta_c )^2 = n^2 \sin^2 \alpha_{max} \)

Rewrite the above as

\(n_1^2(1 - \sin^2 (\theta_c)) = n^2 \sin^2 \alpha_{max} \)

Solve for \(n_1^2 \)

\(n_1^2 = \dfrac{\sin^2 \alpha_{max} }{1 - \sin^2 \theta_c} = n^2 \dfrac{\sin^2 \alpha_{max}}{ \cos^2 \theta_c } \)

Take the square root of both sides in the above

\(n_1 = \dfrac{n\sin \alpha_{max} }{ \cos (\theta_c)} \)

Use equation 2 to find

\(n_2 = n_1 \sin \theta_c = \dfrac{n \sin(\alpha_{max})}{ \cos \theta_c} \sin \theta_c = n \sin \alpha_{max} \tan \theta_c \)

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

\(n_2 = \tan 82^{\circ} \sin 10^{\circ} = 1.2355 \)

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