Question

(a) Figure 9.28 shows a cross-section of a light pipe made of a glass fiber of a refractive index of 1.68. The outer covering of the pipe is made of a material with a 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) Refractive index of the glass fiber,\(\mu\)₁= 1.68
The refractive index of the outer covering of the pipe,\(\mu_2\)= 1.44
The angle of incidence= i
The angle of refraction= r
The angle of incidence at the interface= i'
The refractive index(μ)of the inner core-outer core interface is given as:
\(\mu=\frac{\mu_2}{\mu_1}=\frac{1}{sin\,i'}\)
\(sin\,i'=\frac{\mu_1}{\mu_2}\)
=\(\frac{1.44}{1.68}\)= 0.8571
∴i'= 59^º
For the critical angle,total internal reflaction(TIR)takes place only when i>i'.i.e.,i>59º
Maximum angle of reflaction,rmax= 90º-i'= 90º-59º= 31º
Let, imax be the maximum angle of incidence.
The refractive index at the air-glass interface, \(\mu_1\)= 1.68
\(\mu_1\)= \(\frac{sin\,i_{max}}{sin\,r_{max}}\)
sin i_max= μ₁ sin r_max
= 1.68 sin31^º
= 1.68×0.5150
= 0.8652
∴i_max= sin^-1 0.8652≈60º
Thus, all the rays incident at angles lying in the range 0< i< 60º will suffer total internal reflection.
(b) If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ₁= Refractive index of air= 1
For the angle of incidence i=90º, we can write Snell's law at the air-pipe interface as:
\(\frac{sin_i}{sin_r}\)= μ₂= 1.68
sin_r= \(\frac{sin90^{\circ}}{1.68}\)= \(\frac{1}{1.68}\)
r= sin^-1(0.5952)
= 36.5^º
∴i'= 90^º-36.5^º=53.5^º
Since i'>r, all incident rays will suffer total internal reflection.