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Updated on : 2022-09-05

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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,

$sinθ_{′}=n_{1}1 sinθ$

assuming $n_{air}=1$. The angle of incidence for the light ray striking the coating is the complement of $θ_{′}$, which we denote as $θ_{comp}$, and recall that

$sinθ_{comp}=cosθ_{′}=1−sin_{2}θ_{′} $.

In the critical case, $θ_{comp}$ must equal $θ_{c}$ specified by equation $n_{1}sinθ_{B}=n_{2}sinθ_{r}$. Therefore,

$n_{1}n_{2} =sinθ_{comp}=1−sin_{2}θ_{′} =1−(n_{1}1 sinθ)_{2} $.

which leads to the result: $sinθ=n_{1}−n_{2} $. With $n_{1}=1.58$ and $n_{2}=1.53$, we obtain

$θ=sin_{−1}(1.58_{2}−1.53_{2})=23.2_{0}$.

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