To solve this exercise it is necessary to apply the concepts related to the Snells law.
The law defines that,
[tex]n_1 sin\theta_1 = n_2 sin\theta_2[/tex]
[tex]n_1 =[/tex] Incident index
[tex]n_2 =[/tex] Refracted index
[tex]\theta_1[/tex] = Incident angle
[tex]\theta_2 =[/tex] Refracted angle
Our values are given by
[tex]n_1 = 1.550[/tex]
[tex]n_2 = 1.361[/tex]
[tex]\theta_2 =90\° \rightarrow[/tex]Refractory angle generated when light passes through the fiber.
Replacing we have,
[tex](1.55)sin \theta_1 = (1.361) sin90[/tex]
[tex]sin \theta_1 = \frac{(1.361) sin90}{(1.55)}[/tex]
[tex]\theta_1 =sin^{-1} \frac{(1.361) sin90}{(1.55)}[/tex]
[tex]\theta_1 =61.4\°[/tex]
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,
[tex]\theta_{max} = 90-\theta \\\theta_{max} =90-61.4\\\theta_{max}=28.6\°[/tex]
Therefore the critical angle for the light ray to remain insider the fiber is 28.6°
