Question

(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?

Answer 1

(a) Let's first derive the condition for total internal reflection.

see figure, critical angle for the interface of medium 1 to medium 2 is given by, $\mu_1 SinC=\mu_2sin90^{\circ}$

or, $sinC=\frac{\mu_2}{\mu_1}$

given, $\mu_1=1.68$ and $\mu_2$ = 1.44

so, SinC = 1.44/1.68 = 0.8571

so, C = sin^-1(0.8571) ≈ 59°

condition for total internal reflection,

$i_2$ must be greater than C.

i.e., $i_2>59^{\circ}$

see triangle ∆ABC,

so, $i_2$ = 90° - r

so, r ≤ 90° - 59° ⇒r ≤ 31°

Now , for refraction at first surface of air to core.

Snell's law, $\frac{sini_1}{sinr}=\frac{\mu_1}{\mu_a}$

sini1/sin31°= 1/1.68

or, sini1 = 1.68sin31° ⇒$i_1$ = 60°

Thus all incident rays which makes angle of incidence between 0° to 60° will suffer total internal reflection in the optical fibre.

(b) When there is no outer covering critical angle from core to surface.

sinC = $\frac{\mu_{air}}{\mu_1}$

= 1/1.68

C = sin^-1(1/1.68) ≈ 36.5°

so, condition for total internal reflection from core to surface.

$i_2$ > 36.5°

or, r ≤ 90° - 36.5° ⇒r ≤ 53.5°

so, incident angle at first surface air to core, $i_1$

or, $\frac{\mu_1}{\mu_a}=\frac{sini_1}{sinr}$

$i_1$ = sin^-1(1.68 sin53.5°) ≈ 90°

Thus all incident rays at first surface 0° to 90° will suffer total internal reflection inside core.