Hence, Use the following properties of trigonometric functions and their inverses in the above inequality, Multiply both sides of the inequality by n. 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. Of course we could have answered this question by stating that since angle α = 5° is smaller to αmax = 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. Aim of experiment: In this experiment, we measure the numerical aperture. Fiber NA : Distance : Circle Size : NA = Sin of Half Angle = Numerical Aperture is an angle measurement and is specified by the fiber manufacturer. Numerical Aperture. let n = 1 and n1 = 1.48 in the diagram of the optical fiber system above. at Focal Length : Diver-gence Full Angle (radians) Diver-gence Full Angle (degrees) Distance From Lens Focus : Beam Dia. Numerical aperture is the ability to gather light otherwise an optical fiber capacity. What is the application of numerical aperture? The above is equivalent to In this problem, we are given αmax = 12 ° given by the formula This calculator computes the V parameter/number (also known as normalized frequency) of a step-index optical fiber. Because of refraction at the fiber–air interface, the ray bends toward the normal. Problem 1 let n = 1, n 1 = 1.46 and n 2 = 1.45 in the diagram of the optical fiber system above. 2). Numerical Aperture (NA), a measure of the acceptance angle of a fiber, is a dimensionless quantity. Numerical aperture is commonly used in microscopy to describe the acceptance cone of an objective (and hence its light-gathering ability and resolution), and in fiber optics, in which it describes the range of angles within which light that is incident on the fiber will be transmitted along it. n22 = n12 - sin2(αmax) where λ\lambdaλ is the wavelength of the signal, aaa is the core radius of the fiber and NA\mathrm{NA}NA is the numerical aperture of its core. c) and explain why this light ray will be reflected at the core - cladding interface and hence guided along the fiber. Rays launched outside the angle specified by a fiber's NA will excite radiation modes of the fiber. Solution to Problem 4 Passes are free and working for 30 days. Measuring Numerical Aperture of Optical Fiber. β = sin-1 ( sin(5°) / n1) = 3.42 ° V parameter value at the cutoff wavelength is Vcutoff=2.405V_{cutoff}=2.405Vcutoff​=2.405 so the cutoff wavelength can be calculated from the V parameter as follows. Your feedback is important for improving our service and meet your expectations. Find n1 and n2 such that the acceptance αmax = 10° and the critical angle at the core - cladding interface θc = 82 ° . let n = 1 in the diagram of the optical fiber system above. larger calculation portfolio that we hope to upload as soon as possible. 3). Numerical aperture is abbreviated as NA and shows the efficiency with which light is collected inside the fiber in order to get propagated.. We know light through an optical fiber is propagated through total internal reflection.Or we can say multiple TIR takes place inside the optical fiber for the light ray to get transmitted from an end to another through an optical fiber. n sin(α) = n1 sin(β) We use the same values for n , n1 and n2 in the diagram of the optical fiber system above as those used in problem 1. "optical-calculation.com" has been developped to make you save time and money. c) The angle of incidence θ = 86.58 ° at the core - cladding interface is larger that the critical angle θc = 83.29 ° calculated in problem 1 above and will therefore be totally reflected at this interface and hence guided along the fiber. The Numerical Aperture (NA) of a fiber is defined as the sine of the largest angle an incident ray can have for total internal reflectance in the core. θ = 90 - 3.42 = 86.58 ° Frequently, Glim is determined by the input slit dimensions and F/# of a monochromator, or the small diameter and numerical aperture (NA) of an optical fiber. In this problem, we are given αmax and θc whose formulas are given by, \theta_c = \sin^{-1} \left(\dfrac{n_2}{n_1} \right), \theta = cos^{-1} \left( \dfrac{n \sin \alpha}{n_1} \right), \cos^{-1} \left( \dfrac{n \sin \alpha}{n_1} \right) > \sin^{-1} \left(\dfrac{n_2}{n_1} \right), \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), \cos(\cos^{-1} x ) = x \text{ and } \cos(\sin^{-1} x ) = \sqrt{1 - x^2}, \dfrac{n \sin \alpha}{n_1} < \sqrt{1-\dfrac{n_2^2}{n_1^2}}, \sin \alpha < \dfrac{1}{n} \sqrt{n_1^2-n_2^2}, \alpha < \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right), \alpha_{max} = \sin^{-1} \left (\dfrac{1}{n} \sqrt{n_1^2-n_2^2} \right), \sin^2\alpha_{max} = \left (\dfrac {n_1^2-n_2^2} {n^2} \right), \sin \theta_c = \left(\dfrac{n_2}{n_1} \right), 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), 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, n_1 = \dfrac{\sin 10^{\circ} }{ \cos 82^{\circ}} = 1.2477, n_2 = \tan 82^{\circ} \sin 10^{\circ} = 1.2355, 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, But angles θ and β are complementary; hence, internal reflection at the core cladding interface, angle θ must be greater than the critical angle θ, 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]. Problem 3 The optical fiber system shown below has a core of refractive index n1 and a cladding of refractive index n2 such that n1 > n2. Problem 2 online optical computation "optical-calculation.com" is a useful online tool enabling quick computation in the main fields of photonics : geometrical optics, physical optics, laser, fiber optic, photometry and optoelectronic. n2 = √(1.482 - sin2(12°)) = 1.465, Problem 4 b) angle θ "optical-calculation.com" is the most complete online service dedicated to experts as well as contact : contact@optical-calculation.com - website : www.optical-calculation.com This calculator also computes the cutoff wavelength, which determines when the fiber becomes single-mode. a) angle of refraction β at the outside - core interface. Find a) the critical angle θ c at the core - cladding interface. For applications, it is most commonly expressed as. a) Angle β is found using Snell's law at the outside - core interface as follows All rights reserved, contact : contact@optical-calculation.com - website : www.optical-calculation.com, Copyright © 2009 CLAVIS S.A.R.L. This calculator also computes the cutoff wavelength, which determines when the fiber becomes single-mode. Let the angle of incidence of a light ray on the outside - core interface be α = 5°. Known ratios, together with the light wavelength, allow spot size calculations. Light coming from a larger étendue component will be partially accepted by the limiting component with the lowest étendue. calculations, etc... Find, Take the sine of the first formula, simplify and square both sides to obtain, Substitute the known values to obtain the equations, Substitute the above in equation (1) and solve for n, by their values to obtain numerical values for n, numerical aperture of optical fibers calculator, Total Internal Reflection of Light Rays at an Interface, Refraction of Light Rays, Examples and Solutions.

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