Question

Optical engineers need to know the cone of acceptance of an optical fiber. This is the maximum angle that an entering light ray can make with the axis of the fiber if it is to be guided down the fiber.

What is the cone of acceptance of an optical fiber for which the index of refraction of the core is 1.75 while that of the cladding is 1.55? You can model the fiber as a cylinder with a flat entrance face.(Show all steps please)

Answer 1

The minimum angle of incidence to observed total internal reflection is critical angle.

$$ \begin{aligned} i_{c} &=\sin ^{-1}\left(\frac{n_{c \text { iad }}}{n_{\text {cors }}}\right) \\ &=\sin ^{-1}\left(\frac{1.55}{1.75}\right) \\ &=62.3^{\circ} \end{aligned} $$

Therefore, the angle of reflection at the air-core boundary is,

$$ \begin{aligned} r &=180-90-i_{c} \\ &=90-i_{c} \\ &=90-62.3 \\ &=27.7^{\circ} \end{aligned} $$

Therefore, the acceptance angle is,

$$ A=\sin ^{-1}\left(\frac{n_{\text {core }} \sin r}{n_{\text {air }}}\right) $$

$$ \begin{array}{l} =\sin ^{-1}\left(\frac{1.75 \sin 27.7}{1.0}\right) \\ =54.4^{\circ} \end{array} $$