A fiber optic wire uses total internal reflection to transmit light down the wir

Question

A fiber optic wire uses total internal reflection to transmit light down the wire. It is constructed from two types of glass, with indices of refraction n = 1.45 and n_2 = 1.52. A glass filament forms the core of the wire, with a cladding of the other glass type wrapped around it. Which glass forms the core and which forms the cladding? Briefly justify your answer. What is the minimum incident angle a light ray originating from within the core can strike the core/cladding boundary and not be transmitted through it al all?

Explanation / Answer

(a)
Light travelling in the core reflects from the core-cladding boundary due to total internal reflection, This is only possible when cladding has lower index of refraction. Therefore,
Core is made of Material With , n2 = 1.52
Cladding is made of Material With , n1 =  1.45

(b)
The minimum incident angle is known as, Critical Angle.
The Angle of Refraction is 90o for Critical Angle.

n2 * sin(c) = n1 * sin(90)
1.52 * sin(c) = 1.45 * 1
c = 72.5o
Minimum Angle, c = 72.5o

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