An optical fiber consists of a glass core (index of refraction 1.53) surrounded

Question

An optical fiber consists of a glass core (index of refraction 1.53) surrounded by a cladding (index of refraction 1.52). Suppose a beam of light enters the fiber from air at an angle theta with the fiber axis as shown in Fig. 34-57. What is the greatest possible angle for theta that will still give total internal reflection at the core/cladding interface. (Note from Dr. Mike: This problem requires that you know the angle of the light compared with the normal of the core/cladding interface before you use the equation for total internal reflection. Use your knowledge of trigonometry.)

Explanation / Answer

As we know that

Snell's Law: n1sin(x1) = n2sin(x2)
where n1 is the refractive index of the fist material (the core), n2 is the refractive index of the second material (the cladding), x1 is the angle of incidence and x2 is the angle of refraction.

The mimimum angle that causes total internal reflection is the value of x1 when x2, the angle of refraction, is 90 degrees. So putting in the refractive indices and this data fact into the equayion, gives

1.53sin(x1) = 1.52sin(90), and since sin(90)=1, this reduces to just

1.53sin(x1) = 1.52, or

sin(x1) = 1.52/1.53 and performing the calculation gives 83.44566337 degrees, and subtracting this from 90 degrees gives the angle with the wall of the core, which is 6.554336625 degrees.

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