A step-index Fiber has a core index of refraction = 1:25. The cut-off angle for light entering the Fiberfrom air is found to be 8.50º.-(1) What is the numerical aperture of the Fiber?(2) If the Fiber were submersed in water witi same NA, what would be the cut-off anglewhennwater = 1.33.(3) Find the power of the Fiber after (2) km, when te power losses is (1.5) dB/km.

Approach to solving the question:Numerical Aperture formula and Power loss formulaDetailed…

Answered: A step-index Fiber has a core index of…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Determine the transfer function, of the rotational mechanical system shown in T(s) Figure Q2. The variables 6,(t) and 02(t) refer to angular displacement of motion, while T(t) is a torque applied to the system. Given the value of spring, damping coefficient and inertia as; J: 5 kg-m? Di: 5 N-m-s/rad J2: 10 kg-m? K : 6 N-m/rad K2 : 5 N-m/rad D::4 N-m-s/rad D3:2 N-m-s/rad T(t) e,(1) D2 K2 0000 D1 D3 Figure Q2
Problem (1) For the circuit shown in fig. find the current supplied by the battery by using delta/star transformation 400 www 2002 500 30 V 2002 www 30Ω www 5Ω ww
19
A uniform Simply-supported beam is found to vibrate in its first mode with the amplitude of 12 mm at its center. Assume E=200 GPa, I = 1000 mm“ , p=7500 kg/m², A=50mm² and L=2 m. Determine the maximum bending moment in the beam. Also find the first three natural frequencies.
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where is it demonstrated that the Hamiltonian circuit's tour length is, at most, 4/3 times that of the ideal TSP trip.
19. For the network of Fig. 6.82 D, find: a. The voltage V. b. The current /2. c. The current I,. d. The power to the 12 kQ resistor. 12 kΩ 18 kN 48 V Is + v - 3 kN 12
2. calculates the trajectory r(t) and stores the coordinates for time steps At as a nested list trajectory that contains [[xe, ye, ze], [x1, y1, z1], [x2, y2, z2], ...]. Start from time t = 0 and use a time step At = 0.01; the last data point in the trajectory should be the time when the oscillator "hits the ground", i.e., when z(t) ≤ 0; 3. stores the time for hitting the ground (i.e., the first time t when z(t) ≤ 0) in the variable t_contact and the corresponding positions in the variables x_contact, y_contact, and z_contact. Print t_contact = 1.430 X_contact = 0.755 y contact = -0.380 z_contact = (Output floating point numbers with 3 decimals using format (), e.g., "t_contact = {:.3f}" .format(t_contact).) The partial example output above is for ze = 10. 4. calculates the average x- and y-coordinates 1 y = Yi N where the x, y, are the x(t), y(t) in the trajectory and N is the number of data points that you calculated. Store the result as a list in the variable center = [x_avg, y_avg]…

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