Question

2. The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

Answer 1

The maximum angle a light ray can make with the wall of the core of an optical fiber is approximately 63.47 degrees.

Step-by-step explanation:

To calculate the maximum angle a light ray can make with the wall of the core of an optical fiber, we can use the concept of the critical angle. The critical angle is the incident angle that produces an angle of refraction of 90°. When the angle of incidence is greater than the critical angle, total internal reflection occurs, and the light ray remains inside the fiber. In this case, the glass core has an index of refraction of 1.60, and the cladding has an index of refraction of 1.48. We can use the formula:

critical angle = arcsin(n2/n1)

where n1 is the index of refraction of the core and n2 is the index of refraction of the cladding. Plugging in the values, we get:

critical angle = arcsin(1.48/1.60)

Using a calculator, the critical angle is approximately 63.47 degrees.

Answer 2

We know that the maximum angle that a light ray can wake with the wall of the core is equipment to the minimum angle with the normal of the core that will give rise in total internal reflection. so using Snell's law the angle is subtracted from 90° to get the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber.

So using

n1sinစ1. = n2sinစ2

1.6sin(x1) = 1.48sin(90),

But sin(90)=1

1.6sin(စ1) = 1.48,

sin(စ1) = 1.48/1.6

စ = 68°

Step-by-step explanation: