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5 Optical Power Coupling
In this case the power loss in decibels is
L power (dB) = −10 log(1 − R) − 10 log 0.662 = 1.791 dB
5.1.3 Optical Coupling Versus Wavelength
It is of interest to note that the optical power launched into a fiber does not depend on
the wavelength of the source but only on its radiance. To explore this concept a little
further, Eqs. (2.30) and (2.42) show that the number of modes that can propagate in
a multimode fiber (either step-index or graded-index) is proportional to the inverse
wavelength squared:
M ∝ λ
−2
(5.11)
Thus, for example, twice as many modes propagate in a given fiber at 900 nm
than at 1300 nm.
The radiated power per mode, P s /M, from a source at a particular wavelength is
given by the radiance multiplied by the square of the nominal source wavelength [4],
P s
M
= L 0 λ
2
(5.12)
Thus twice as much power is launched into a given mode at 1300 nm than at
900 nm. Hence, two identically sized sources operating at different wavelengths but
having identical radiances will launch equal amounts of optical power into the same
fiber.
Drill Problem 5.5 Consider two identically sized optical sources, one emitting
at 895 nm and the other at 1550 nm. (a) Verify that about three times as many
modes propagate in the same multimode fiber at 895 nm than at 1550 nm. (b)
Show that if the two sources have equal radiances then about three times as
much power is launched into a given mode at 1550 nm compared to 895 nm.
(c) Using Eqs. (5.11) and (5.12) show that these two sources will launch equal
amounts of power into the same fiber.
5 Optical Power Coupling
In this case the power loss in decibels is
L power (dB) = −10 log(1 − R) − 10 log 0.662 = 1.791 dB
5.1.3 Optical Coupling Versus Wavelength
It is of interest to note that the optical power launched into a fiber does not depend on
the wavelength of the source but only on its radiance. To explore this concept a little
further, Eqs. (2.30) and (2.42) show that the number of modes that can propagate in
a multimode fiber (either step-index or graded-index) is proportional to the inverse
wavelength squared:
M ∝ λ
−2
(5.11)
Thus, for example, twice as many modes propagate in a given fiber at 900 nm
than at 1300 nm.
The radiated power per mode, P s /M, from a source at a particular wavelength is
given by the radiance multiplied by the square of the nominal source wavelength [4],
P s
M
= L 0 λ
2
(5.12)
Thus twice as much power is launched into a given mode at 1300 nm than at
900 nm. Hence, two identically sized sources operating at different wavelengths but
having identical radiances will launch equal amounts of optical power into the same
fiber.
Drill Problem 5.5 Consider two identically sized optical sources, one emitting
at 895 nm and the other at 1550 nm. (a) Verify that about three times as many
modes propagate in the same multimode fiber at 895 nm than at 1550 nm. (b)
Show that if the two sources have equal radiances then about three times as
much power is launched into a given mode at 1550 nm compared to 895 nm.
(c) Using Eqs. (5.11) and (5.12) show that these two sources will launch equal
amounts of power into the same fiber.