Find the minimum angle${\mathbf{\theta }}_{\mathbf{i}}$for total reflection in the optical fiber in Figure 20-21 if the index of refraction of the cladding is 1.400 and the index of refraction of the core is

(a) 1.600 or

(b) 1.800.

• Fiber optics, often known as optical fibre, is a technology that uses light pulses to convey data along a glass or plastic fibre.
• A fibre optic cable can have anywhere from a few to several hundred of these glass fibres.
• The glass fibre core is surrounded by another glass layer called cladding.

(a)

For entire reflection, the minimal angle ${\mathrm{\theta }}_{\mathrm{i}}$is

${\mathrm{sin\theta }}_{\mathrm{i}}=\frac{{\mathrm{n}}_{\mathrm{cladding}}}{{\mathrm{n}}_{\mathrm{core}}}$

Simplify fortheminimum angle${\mathrm{\theta }}_{\mathrm{i}}$for total reflection in the optical fiber in 1.600.

For${\mathrm{n}}_{\mathrm{core}}=1.6$,the${\mathrm{\theta }}_{\mathrm{i}}$is:

$$\begin{gathered} {\mathrm{sin\theta }}_{\mathrm{i}}=\frac{1.4}{1.6} \\ {\mathrm{sin\theta }}_{\mathrm{i}}=0.875 \\ {\mathrm{\theta }}_{\mathrm{i}}=61.04° \end{gathered}$$

The entire reflection in an optical fibre, the minimum angle ${\mathrm{\theta }}_{\mathrm{i}}$is $\mathrm{\theta i}\ge 61.04°$

(b)

Simplify forthe minimum angle ${\mathrm{\theta }}_{\mathrm{i}}$for total reflection in the optical fiber in 1.800

For ${\mathrm{n}}_{\mathrm{core}}=1.8$, the ${\mathrm{\theta }}_{\mathrm{i}}$is:

$$\begin{gathered} {\mathrm{sin\theta }}_{\mathrm{i}}=\frac{1.4}{1.8} \\ {\mathrm{sin\theta }}_{\mathrm{i}}=0.778 \\ {\mathrm{\theta }}_{\mathrm{i}}=51.08° \end{gathered}$$

The entire reflection in an optical fibre, the minimum angle ${\mathrm{\theta }}_{\mathrm{i}}$is ${\mathrm{\theta }}_{\mathrm{i}}\ge 51.06°$