(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.

Solution:

(a) Refractive index of the glass fibre, $\mu_{1}=1.68$

Refractive index of the outer covering of the pipe, $\mu_{2}=1.44$

Angle of incidence = i

Angle of refraction = r

Angle of incidence at the interface = i’

The refractive index (μ) of the inner core − outer core interface is given as:

$\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\sin i^{\prime}}$

$\sin i^{\prime}=\frac{\mu_{1}}{\mu_{2}}$

$=\frac{1.44}{1.68}=0.8571$

$\therefore i^{\prime}=59^{\circ}$

For the critical angle, total internal reflection (TIR) takes place only when $i>i^{\prime}$, i.e., $i>59^{\circ}$

Maximum angle of reflection, $r_{\max }=90^{\circ}-i^{\prime}=90^{\circ}-59^{\circ}=31^{\circ}$

Let, $i_{\max }$ be the maximum angle of incidence.

The refractive index at the air - glass interface, $\mu_{1}=1.68$

 

We have the relation for the maximum angles of incidence and reflection as:

$\mu_{1}=\frac{\sin i_{\max }}{\sin r_{\max }}$

$\sin i_{\max }=\mu_{1} \sin r_{\max }$

$=1.68 \sin 31^{\circ}$

$=1.68 \times 0.5150$

$=0.8652$

$\therefore i_{\max }=\sin ^{-1} 0.8652 \approx 60^{\circ}$

Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then:

Refractive index of the outer pipe, $\mu_{1}=$ Refractive index of air $=1$

For the angle of incidence i = 90°, we can write Snell’s law at the air − pipe interface as:

$\frac{\sin i}{\sin r}=\mu_{2}=1.68$

$\sin r=\frac{\sin 90^{\circ}}{1.68}=\frac{1}{1.68}$

$\sin r=\frac{\sin 90^{\circ}}{1.68}=\frac{1}{1.68}$

$r=\sin ^{-1}(0.5952)$

$=36.5^{\circ}$

$\therefore i^{\prime}=90^{\circ}-36.5^{\circ}=53.5^{\circ}$

Since $i^{\prime}>r$, all incident rays will suffer total internal reflection.