00:01
In this question, it is said that the core of an optical fiber is made out of a transparent material with a refractive index of 1 .50, which is n -core.
00:11
And the cladding is made out of material with a refractive index of 1 .44, which is, and cladding is equals to 1 .44, which is as shown in the speaker.
00:22
A monochromatic light rate travels in air incident on the one side of the optical mirror and refracted.
00:30
That the reflective ray will enter the core cladding interference with its corresponding critical angle c.
00:37
In the first part of the question, we need to calculate the value of theta.
00:42
We need to calculate solution we are able to draw this diagram in which adb is a triangle.
00:51
Let c be the critical angle.
00:54
So angle bad equals to pi divided by 2 minus c and angle eab is equal to to c.
01:07
So from snar's law we get encode multiply sine c, sine c is equals to enclading multiply by sine 90 degrees, so we get sine c is 2 .4 .4 divided by 1 .55 which is 73 .7 7 .7 4 degree and again when we apply snitch law we get n -a sine theta n -er sine theta is equals to n -core sine 5 y2, pi divided by 2 minus c this is equal to we have 1 .5 g.
02:38
Sine 16 .26 degree thus theta is is equal to 24 .83 degree.
02:51
This is the answer for the first part of the question...