Question

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.550 and is submerged in ethyl alcohol, which has an index of refraction of 1.361. What is the critical angle (in degrees) for the light ray to remain inside the fiber?

Answer 1

To solve this exercise it is necessary to apply the concepts related to the Snells law.

The law defines that,

`n_1 sin\theta_1 = n_2 sin\theta_2`

`n_1 =` Incident index

`n_2 =` Refracted index

`\theta_1` = Incident angle

`\theta_2 =` Refracted angle

Our values are given by

`n_1 = 1.550`

`n_2 = 1.361`

`\theta_2 =90\° \rightarrow`Refractory angle generated when light passes through the fiber.

Replacing we have,

`(1.55)sin \theta_1 = (1.361) sin90`

`sin \theta_1 = ((1.361) sin90)/((1.55))`

`\theta_1 =sin^(-1) ((1.361) sin90)/((1.55))`

`\theta_1 =61.4\°`

Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,

`\theta_(max) = 90-\theta \\\theta_(max) =90-61.4\\\theta_(max)=28.6\°`

Therefore the critical angle for the light ray to remain insider the fiber is 28.6°