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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鈥搒heath boundary. (Thus there is no loss of light through that boundary.) What is the maximum value of that allows total internal reflection at A?

Figure:

Short Answer

Expert verified

The maximum value of that 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鈥檚 law, we can find the value of that allows the total internal reflection.

Formula:

Critical angle is

c=sin-1n2n1

Snell鈥檚 law is :

n1蝉颈苍胃1=n2蝉颈苍胃2

03

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

The critical angle is given by

c=sin-1n2n1c=sin-11.531.58=75.547

So, by geometry, the incident angle is

1=90-c1=90-75.547o=14.45

Now, we have to apply Snell鈥檚 law,

nairsin=n1sin1

Since nair=1

sin=1.58sin14.45sim=0.3942=23.2

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