An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher

1. We know that total internal reflection occurs when the angle of incidence is greater than the

An optical fiber consists of an outer "cladding" layer and...

An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher index of refraction. Light rays entering the core are trapped inside by total internal reflection and forced to travel along the fiber (see Figure 23.56 ). Suppose the cladding has an index of refraction of $n_{b}=1.44$ and the core has an index of refraction of $n_{a}=1.46 .$ Calculate the largest angle $\theta$ between a light ray and the longitudinal axis of the fiber (see the figure) for which the ray will be totally internally reflected at the core/cladding boundary.

Transcript

00:02 We have a optical fiber.

00:04 We have a optical fiber and we will first of all we will draw the optical fiber and the draw the all of the angles all of the data that is given.

00:15 So let's just say that this is the optical fiber that we have second.

00:21 This is optical fiber and this is the outer cladding and this is the inner.

00:27 This is inner core.

00:28 This is outer cladding.

00:29 So outer cladding has a refractive index.

00:32 Let's just say first of all we will denote n1 and one is the refractive index of air which is one and inner core has a refractive index n2 and it is given by 1 .46 and cladding has a refractive index n3 which is given by 1 .44.

00:49 Now it is given that light enters this is the longitudinal axis.

00:55 This is the longitudinal axis.

01:00 Now in the question it is given that the light enters at an angle of theta 1 we can say and after going through a refraction it will travel something like this and let's just say this is the normal this is theta 3 and this is theta 2 and after that it will go here and this is the refraction.

01:21 This is the path that the light will follow.

01:23 Now we know and sorry this place this let us say that this is pointed this point is the interface of the air and optical fiber boundary.

01:33 So now we can see that for tir we can say for tir for total internal refraction we can say for total internal reflection total internal reflection we say that theta 3 must be greater than the critical angle c.

01:51 So if theta 3 is greater than critical angle c then we know that theta 3 can be written as 90 minus of theta 2.

01:59 So it will be greater than critical angle c...

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