Question

3- Consider a bare fiber consisting of core of refractive index 1.48 and having air as cladding What is its NA? What is the maximum incident angle up to which light can be guided by the fiber? 4- Consider a fiber from which cladding is removed over a short length as shown below. Assume that the core and cladding refractive indices are 1.5 and 1.4 respectively, What ill happen to the output power if the bare portion of the fiber is covered by a liquid whose refractive index is varied from 1.4 to 1.5 CLADDINO S A link of optical fiber connects Tripoli with Gharian. It is used to send TV signal to Gharian. Assume the loss of transmitter and the recelver are 2 and 3dB respectively, Also assume the fiber optic loss is 1dB/Km 6- A TV channel needed to be transmitted from Tripoli to Gharian through fiber optic (93 Km distance). Assume the transmitter's power is 17mw, wave length is 870nm and its loss is 2dB. Also assume the loss coming from the fiber is 1dB/Km. Libyan markets have two types of receivers and two types of optical repeaters such as 7 Repeater 1 (Gain 17de) Repeater 2 Receiver price 90 LD 10 LD 30 LD S5LD Design the link that gives us the lowest system cost Dr. Khaled Due date is 30/1/2019

Answer 1

As you have not mentioned which question i have to solve, so i am solving the first one only. If you need solution for others, then please upload them separately.

(3) Numerical aperture (NA) is a light gathering property of optical fibre, which gives the quantity of light that brought into the center of optical fibre in terms of incidence angle.

Let the refractive index of outside medium, cladding ,and core are and $n_{2}$ respectively. Then the numerical aperture is given by

As you have not mentioned which question i have to solve, so i am solving the first one only. If you need solution for others, then please upload them separately. (3) Numerical aperture (NA) is a light gathering property of optical fibre, which gives the quantity of light that brought into the center of optical fibre in terms of incidence angle. Let the refractive index of outside medium, cladding, and core are n_0, n_1andn_2 respectively. Then the numerical aperture is given by NA = n_1/n_0 squareroot 1 - (n_2/n_1)^2 Given, n_1 = 1.48 n_2 = 1 (refractive index of air) so, in this case NA = 1.48/n_0 squareroot 1 - (1/1.48)^2 = 1.48/n_0 times squareroot 0.5435 = 1.48/n_0 times 0.7372 = 1.09/n_0 If the outside medium is air then n_0 = 1, Therefore, N A = 1.09 This value of numerical aperture is greater than one, which is not acceptable

Given,

  

   (refractive index of air)

So, in this case

  

If the outside medium is air then ,

Therefore,

This value of numerical aperture is greater than one, which is not acceptable. Hence, you can't use air as cladding for the fibre if the outside medium is air too.

The maximum incident angle up to which light can be guided by , is known as acceptance angle. The acceptance angle is given by

Here you can see that , if you put , you will get an undefined quantity.

For any doubt please comment.