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15 Optics
(c) From (2) it is observed that as d decreases, x also decreases. The smallest
value for x is zero, in which case the distance of the observer from the
tree would be d = 282.8 m when the mirage disappears.
15.5 (a) The purpose of the cladding is to improve the transmission efficiency
of the optical fibre. If cladding is not used then the signal is attenuated
dramatically.
(b) Let a ray be incident at an angle θ , Fig. 15.6, the angle of refraction at P
being θ p . Let C be the critical angle at Q, interface of core and cladding:
sin C =
n 1
n 2
where n 1 and n 2 are the indices of the cladding and core, respectively:
θ p = 90 − θ in
where θ in is the angle of incidence at Q:
n 0 sin θ = n 2 sin θ p = n 2 sin(90 − θ in ) = n 2 cos θ in
For internal reflection θ in > C or cos θ r < cos C
∴
n 0 sin θ ≤ n 2 cos C
But
n 2 cos C = n 2
1 − sin
2 C =
n 2
2 − n 2
1
∴
n 0 sin θ ≤
n 2
2 − n 2
1
This shows that there is a maximum angle of acceptance cone outside of
which entering rays will not be totally reflected within the fibre. For the
largest acceptance cone, it is desirable to choose the index of refraction
of the cladding to be as small as possible. This is achieved if there is no
cladding at all. However, this leads to other problems associated with the
loss of intensity.
(c) The transmission is reduced due to multiple reflections and the absorption
of the fibre core material due to impurities.
15.3.2 Prisms and Lenses
15.6 With reference to Fig. 15.6, using Snell’s law
n 2 sin r 1 = n 1 sin 20
◦
15 Optics
(c) From (2) it is observed that as d decreases, x also decreases. The smallest
value for x is zero, in which case the distance of the observer from the
tree would be d = 282.8 m when the mirage disappears.
15.5 (a) The purpose of the cladding is to improve the transmission efficiency
of the optical fibre. If cladding is not used then the signal is attenuated
dramatically.
(b) Let a ray be incident at an angle θ , Fig. 15.6, the angle of refraction at P
being θ p . Let C be the critical angle at Q, interface of core and cladding:
sin C =
n 1
n 2
where n 1 and n 2 are the indices of the cladding and core, respectively:
θ p = 90 − θ in
where θ in is the angle of incidence at Q:
n 0 sin θ = n 2 sin θ p = n 2 sin(90 − θ in ) = n 2 cos θ in
For internal reflection θ in > C or cos θ r < cos C
∴
n 0 sin θ ≤ n 2 cos C
But
n 2 cos C = n 2
1 − sin
2 C =
n 2
2 − n 2
1
∴
n 0 sin θ ≤
n 2
2 − n 2
1
This shows that there is a maximum angle of acceptance cone outside of
which entering rays will not be totally reflected within the fibre. For the
largest acceptance cone, it is desirable to choose the index of refraction
of the cladding to be as small as possible. This is achieved if there is no
cladding at all. However, this leads to other problems associated with the
loss of intensity.
(c) The transmission is reduced due to multiple reflections and the absorption
of the fibre core material due to impurities.
15.3.2 Prisms and Lenses
15.6 With reference to Fig. 15.6, using Snell’s law
n 2 sin r 1 = n 1 sin 20
◦