(5%) Problem 8: Fiber optics are an important part of our modern internet. In th

Question

(5%) Problem 8: Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore = 1.511) and the cladding (ncladding = 1.461). "cladding- 'core ©theexpertta.com * 50% Part (a) Numerically, what is the largest angle (in degrees) a ray will make with respect to the interface of the fiber Omar, and still experience total internal reflection? A 50% Part (b) Suppose you wanted the largest angle at which total internal reflection occurred to be o'max = 5 degrees. What index of refraction does the cladding need if the core is unchanged?

Explanation / Answer

given,

refractive index of core n_core = 1.511

refractive index of cladding n_cladding = 1.461

by the condition of total internal reflection

n_core * sin(critical angle) = n_cladding * sin(90)

1.511 * sin(critical angle) = 1.461 * sin(90)

critical angle = 75.22 degree

theta_max = 90 - 75.22

theta_max = 14.78 degree

b) theta_i = 90 - 5 = 85°

n_cladding = 1.511 x sin85°

n_cladding = 1.505

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