Question

Above figure depicts a simplistic optical fiber: a plastic core ($n_1=1.58$) is surrounded by a plastic sheath ($n_2=1.53$). A light ray is incident on one end of the fiber at angle $\theta$. 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 $\theta$ that allows total internal reflection at $A$?

Answer 1

When examining figure in question, it is important to note that the angle (measured from the central axis) for the light ray in air, $\theta$, is not the angle for the ray in the glass core, which we denote ${\theta }^{\prime }$. The law of refraction leads to,

$\sin {\theta }^{\prime }=\frac{1}{n_1}\sin \theta$

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

$\sin {\theta }_{comp}^{\prime }=\cos {\theta }^{\prime }=\sqrt{1-{\sin }^2{\theta }^{\prime }}$.

In the critical case, ${\theta }_{comp}^{\prime }$ must equal ${\theta }_c$ specified by equation $n_1\sin {\theta }_B=n_2\sin {\theta }_r$. Therefore,

$\frac{n_2}{n_1}=\sin {\theta }_{comp}^{\prime }=\sqrt{1-{\sin }^2{\theta }^{\prime }}=\sqrt{1-{\left(\frac{1}{n_1}\sin \theta \right)}^2}$.

which leads to the result: $\sin \theta =\sqrt{n_1^2-n_2^2}$. With $n_1=1.58$ and $n_2=1.53$, we obtain

$\theta ={\sin }^{-1}({1.58}^2-{1.53}^2)={23.2}^0$.