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Question: Figure depicts a simplistic optical fiber: a plastic core(n1=1.58)is surrounded by a plastic sheath (n2=1.53). A light ray is incident on one end of the fiber at angle.The ray is to undergo total internal reflection at point A, where it encounters the core–sheath boundary. (Thus there is no loss of light through that boundary.) What is the maximum value of θthat allows total internal reflection at A?

Figure:

Short Answer

Expert verified

The maximum value of θthat allows total internal reflection at A is 23.2∘.

Step by step solution

01

Given

  1. Refractive index of the plastic core,n1=1.58
  2. Refractive index of the plastic sheath,n2=1.53
02

Understanding the concept

By using the concept of the critical angle and Snell’s law, we can find the value of θthat allows the total internal reflection.

Formula:

Critical angle is

θc=sin-1n2n1

Snell’s law is :

n1²õ¾±²Ôθ1=n2²õ¾±²Ôθ2

03

Calculate the maximum value of θ that allows total internal refection at A.

The critical angle is given by

θc=sin-1n2n1θc=sin-11.531.58=75.547∘

So, by geometry, the incident angle is

θ1=90-θcθ1=90-75.547o=14.45∘

Now, we have to apply Snell’s law,

nairsinθ=n1sinθ1

Since nair=1

sinθ=1.58sin14.45∘simθ=0.3942θ=23.2∘

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