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# Above 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 θ. 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 θ that allows total internal reflection at A?

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#### When examining figure in question, it is important to note that the angle (measured from the central axis) for the light ray in air, θ, is not the angle for the ray in the glass core, which we denote θ′. The law of refraction leads to,sinθ′=1n1sinθassuming nair=1. The angle of incidence for the light ray striking the coating is the complement of θ′, which we denote as θ′comp, and recall thatsinθ′comp=cosθ′=√1−sin2θ′.In the critical case, θ′comp must equal θc specified by equation n1sinθB=n2sinθr. Therefore,n2n1=sinθ′comp=√1−sin2θ′=√1−(1n1sinθ)2.which leads to the result: sinθ=√n21−n22. With n1=1.58 and n2=1.53, we obtainθ=sin−1(1.582−1.532)=23.20.

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