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602.


Problem 43.48 (RHK)


A particular optical fibre consists of a non-graded glass core (index of refraction ) surrounded by a cladding (index of refraction ). Suppose that a beam of light enters the fibre from air at angle with the fibre axis as shown in the figure. (a) We have to show that the greatest possible value of for which a ray can be propagated down the fibre is given by

(b) Assuming that the indices of refraction of the glass and the coating are 1.58 and 1.53, respectively, we have to calculate the value of this angle.


Solution:           Click For PDF Version

(a)

As shown in the figure, let a beam of light enter the optical fibre from air at incident angle . The angle of refraction in the glass is related to by the Snell’s law. Let the index of refraction of the glass be . We have

.

This beam hits the glass-coating interface at angle of incidence, . For greater than the critical angle , determined by the condition that

,

beam that enters the optical fibre from air will travel by undergoing total internal reflection. Therefore, the maximum possible value of the angle with which the beam can enter the optical fibre from air so that it travels inside the optical fibre through process of total internal reflection will be given by

And, therefore,

(b)

For and , we find