Index of refraction of an optical fiber

[rayhan619]

Homework Statement

The glass core of an optical fiber has an index of refraction 1.62. The index of refraction of the cladding is 1.47.
What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

Homework Equations

theta = sin ^-1 (n2/n1)

The Attempt at a Solution

i used the equation and got angel 65.15 which was wrong

 

[Delphi51]

The critical angle is 65 all right, but that is the angle of incidence - measured from a line perpendicular to the surface. The question asks for the angle "with the wall of the core". So you want 90-65.15.

 

[nlightner]

, but then i realized that i had to use the inverse sine function (sin^-1) to find the angle. So, the correct answer is theta = 41.88 degrees. This means that the maximum angle a light ray can make with the wall of the core is 41.88 degrees if it is to remain inside the fiber. This is known as the critical angle and is an important factor in determining the efficiency of an optical fiber. If the angle is greater than the critical angle, the light ray will undergo total internal reflection and remain inside the fiber, allowing for efficient transmission of light signals. Understanding the index of refraction of the core and cladding is crucial in designing and optimizing optical fibers for various applications such as telecommunication, medical imaging, and data transmission.