(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of aglass fibre of refractive index 1.68. The outer covering of thepipe is made of a material of refractive index 1.44. What is therange of the angles of the incident rays with the axis of the pipefor which total reflections inside the pipe take place, as shownin the figure (b) What is the answer if there is no outer covering of the pipe?
a)
Given: The refractive index of the glass pipe is 1.68 , and the refractive index of the outer covering of the pipe is 1.44 .
The refractive index of the inner core-outer core interface can be calculated as,
μ = μ g μ 0
where, the refractive index of glass fiber is μ g , and the refractive index of the outer covering of the pipe is μ o .
By substituting the given values in the above equation, we get,
μ = 1.68 1.44 (1)
The refractive index of the inner core-outer core interface is given as,
μ = 1 sin i ′
where, the angle of incidence at the interface is i ′ .
By substituting the given values in the above equation, we get,
1.68 1.44 = 1 sin i ′ sin i ′ = 1.44 1.68 sin i ′ = 0.857 i ′ = 59 °
The angle of incidence should be greater than the angle of incidence at the interface for total internal reflection to occur.
i > i ′ i > 59 °
The maximum angle of reflection can be calculated as,
r m = 90 − i ′
By substituting the given values in the above equation, we get,
r m = 90 ° − 59 ° = 31 °
The maximum angle of incidence can be calculated as,
μ g = sin i m sin r m sin i m = μ g × sin r m i m = sin − 1 ( μ g × sin r m )
By substituting the given values in the above equation, we get,
i m = sin − 1 ( 1.68 × sin 31 ° ) = sin − 1 ( 0.8652 ) ≈ 60 °
Thus, the range of angle of incident angles of the rays which will suffer total internal reflection is from 0 ° to 60 ° .
b)
As there is no outer covering of pipe; hence, in place of refractive index of outer pipe we will use refractive index of air which is 1 .
The angle of reflection can be calculated as,
sin i sin r = μ o sin r = sin i μ o r = sin − 1 ( sin i μ o )
where, the angle of incidence is i .
By substituting the given values in the above equation, we get,
r = sin − 1 ( sin 90 ° 1.68 ) = sin − 1 ( 0.5932 ) ≈ 36.5 °
The angle of incidence at interface can be calculated as,
i ′ = 90 ° − r
By substituting the given values in the above equation, we get,
i ′ = 90 ° − 36.5 ° = 53.5 °
As i ′ > r thus, all incident rays will suffer total internal reflection.
a)
Given: The refractive index of the glass pipe is
The refractive index of the inner core-outer core interface can be calculated as,
where, the refractive index of glass fiber is
By substituting the given values in the above equation, we get,
The refractive index of the inner core-outer core interface is given as,
where, the angle of incidence at the interface is
By substituting the given values in the above equation, we get,
The angle of incidence should be greater than the angle of incidence at the interface for total internal reflection to occur.
The maximum angle of reflection can be calculated as,
By substituting the given values in the above equation, we get,
The maximum angle of incidence can be calculated as,
By substituting the given values in the above equation, we get,
Thus, the range of angle of incident angles of the rays which will suffer total internal reflection is from
b)
As there is no outer covering of pipe; hence, in place of refractive index of outer pipe we will use refractive index of air which is
The angle of reflection can be calculated as,
where, the angle of incidence is
By substituting the given values in the above equation, we get,
The angle of incidence at interface can be calculated as,
By substituting the given values in the above equation, we get,
As
