Let the angle of incidence of a light ray on the outside - core interface be α = 5°. 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. calculations, etc... This calc is accurate to ~2 significant figures. 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 θc given by. 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. Solution to Problem 2 All rights reserved. where λ \lambda λ is the wavelength of the signal, a a a is the core radius of the fiber and N A \mathrm{NA} N A is the numerical aperture of its core. Problem 1 let n = 1, n 1 = 1.46 and n 2 = 1.45 in the diagram of the optical fiber system above. 2). sin(αmax) = √(n12 - n22)/ n where Î»\lambdaÎ» 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. Numerical aperture is the ability to gather light otherwise an optical fiber capacity. 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. Your feedback is important for improving our service and meet your expectations. (1), where a ray making an angle θi with the fiber axis is incident at the core center. n2 = √(1.482 - sin2(12°)) = 1.465, Problem 4 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. 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. n22 = n12 - sin2(αmax) n sin(α) = n1 sin(β) The V parameter is defined as. αmax = sin-1(√(n12 - n22)/ n) contact : contact@optical-calculation.com - website : www.optical-calculation.com Find b) Angle θ is complementary to angle β hence For applications, it is most commonly expressed as. Size of a Collimated Beam: NA : Bundle Dia. We rely on you to give us your comments and suggestions. β = sin-1 ( sin(5°) / n1) = 3.42 ° Solution to Problem 4 larger calculation portfolio that we hope to upload as soon as possible. In fiber optics, it describes the angles range where light is occurring on the fiber optic will be broadcasted along with it. 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]. At shorter wavelengths, at least two LP modes can propagate; at longer wavelengths the fiber guides only a single transverse mode. MFD is determined by the numerical aperture (NA) and cut-off wavelength of the fiber and is related to the diameter of the fiber core. Known ratios, together with the light wavelength, allow spot size calculations. Problem 2 What is the application of numerical aperture? photometry and optoelectronic. Find a) the critical angle θ c at the core - cladding interface. = 1.45 in the diagram of the optical fiber system above. All rights reserved, contact : contact@optical-calculation.com - website : www.optical-calculation.com, Copyright © 2009 CLAVIS S.A.R.L. "optical-calculation.com" includes basic calculations as well as more advanced one like coupling efficiency in a fiber optic, laser beam shaping, aberration

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