An optical fiber has index of refraction n and diameter d....

An optical fiber has index of refraction $n$ and diameter $d$. It is surrounded by air. Light is sent into the fiber along its axis, as shown in Figure P35.40. (a) Find the smallest outside radius $R$ permitted for a bend in the fiber if no light is to escape. (b) What If? Does the result for part (a) predict reasonable behavior as $d$ approaches zero? As $n$ increases? As $n$ approaches $1 ?$ (c) Evaluate $R$ assuming the fiber diameter is $100 \mu \mathrm{m}$ and its index of refraction is 1.40. (FIGURE CANT COPY)

Key Concepts

Total Internal Reflection

Total internal reflection is the phenomenon in which light traveling within a medium is completely reflected at the boundary with a less dense medium if the angle of incidence is greater than a specific critical angle. This principle is essential in optical fibers, as it confines light within the fiber core, preventing it from escaping into the surrounding environment and ensuring efficient transmission of signals.

Critical Angle

The critical angle is the minimum angle of incidence at which light is totally internally reflected at the interface between two media with different refractive indices. It is determined by the refractive indices of the fiber material and the surrounding medium. Understanding the critical angle is key to analyzing how bending or other perturbations can cause light to exceed the critical incidence conditions, potentially leading to leakage.

Optical Fiber Bending and Losses

Bending an optical fiber alters the propagation path of light, potentially causing some light rays to hit the fiber boundary at angles less than the critical angle. This can result in light leaking out of the fiber, an effect known as bend loss. Determining the smallest bend radius that prevents leakage is crucial in fiber design and installation, ensuring minimal signal loss and optimal performance.

Geometrical Optics in Waveguides

The analysis based on geometrical optics considers light as rays and predicts how these rays propagate within optical fibers. This approach simplifies the problem by relating fiber geometry, such as diameter and bend radius, to the incidence angles of the light rays. It provides a practical framework for designing optical fiber paths that maintain total internal reflection even when the fiber is bent.

Transcript

0:00 Hi there.

00:01 So for this problem, we have an optical fiber with index of refraption and a diameter t that is surrounded by earth, as is shown in the figure.

00:10 So light is sent into the fiber along its axis.

00:14 And for part a, what we need to find is the smallest upside radius art permitted for a bend in the fiber if no light is to escape.

00:29 So the values they are given are the index of refraption, the diameter of this, and where we need to determine the expression for the radius.

00:42 So with that said, we can observe in the sketch at this that a ray originally traveling along the inner edge will have the smallest angle of incidence.

00:55 So if we do this in here, if we draw in here, something like this, we're going to have this one, this path in here.

01:05 Well, we're going to have like a little more to the left.

01:11 So we're going to just draw in here like this.

01:14 Do, do, two, two, too, too, too.

01:15 So in here we're going to have a reflection that we know is with the normal is going to be equal to this.

01:27 This angle and just goes there.

01:35 Something like this, so that is going to happen.

01:38 This angle in here is the angle theta.

01:42 And the separation, so we will have that this in here, this distance is r minus d, as you can see.

01:55 So the smallest angle of incidence when it is tries the outer edge of the fiber in the curve.

02:02 So if this race is totally internally reflected, all of the others are also totally reflected.

02:10 So for this ray to be totally internally reflected, it is necessary that the angle is greater or equal to the critical angle.

02:23 And we also can see, can do in here that the sign of theta should be greater or equal to the sign of the critical angle.

02:36 And we know that the sign of the critical angle is equal to the ratio between the index of reflection of the earth, because that is the first medium divided by the index of reflection of the pipe, which is just given, and we name it n.

02:56 So we know that the index of refruption of earth is just equal to 1.

03:00 So we can say that this is sign of theta c is equal to 1 over the index of refraction and given...

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