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I am reading the book: Kaushal, H., Jain, V.K. and Kar, S., 2017. Free space optical communication. New Delhi: Springer India.

I have asked before about a reason of choice of wavelength for beacon signal.

I have a doubt about one reason why 1550 nm is the best choice. Let me cite its explanation:

High transmitter power: At 1550 nm a much higher power level (almost 50 times) than at lower wavelengths is available to overcome various losses due to attenuation.

It is confused… I am sure if we have low wavelength we will have more power and if we have high wavelength, we will have low power. Let take as example lasers. GaAlAs-Laser operates in 780-890 nm with power 200 mW, InGaAsP operates in 1300nm, its power is <50 mW.

Did I understand this explanation wrong? If it is correct, could you explain the reason ?

update: Here is the passage itself:

3.1 Optical Transmitter

The optical transmitter including the choice of laser, concept of ATP system, and various types of modulation schemes and coding techniques used in FSO communications are discussed in this section. Further, the details of communication and beacon detectors in FSO receivers are also discussed. The transmitter converts the source information into optical signals which are transmitted to the receiver through the atmosphere. The essential components of the transmitter are (i) modulator, (ii) driver circuit for the optical source to stabilize the optical radiations against temperature fluctuations, and (iii) collimator that collects, collimates, and direct the optical signals toward the receiver via atmospheric channel. The optical sources that are used for FSO transmission lie in the atmospheric transmission window that is ranging from 700 to 10,000 nm wavelength. The wavelength range from 780 to 1064 nm is most widely used as a beacon operating wavelength due to following reasons:

  1. Reduced background noise and Rayleigh scattering: The absorption coefficient for the Rayleigh scattering has a functional dependence with the wavelength λ as λ-4. Consequently there is almost negligible attenuation at higher operating wavelengths as compared to those at the visible range.
  2. High transmit power: At 1550 nm a much higher power level (almost 50 times) than at lower wavelengths is available to overcome various losses due to attenuation.
  3. Eye-safe wavelength: The maximum permissible exposure (MPE) for eye is much higher at 1550 nm wavelength at 850 nm. This difference can be...

original screen shot

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    $\begingroup$ @uhoh I added the screenshot of this quates from book. As you see there is no explanation about, only this line $\endgroup$ Commented Nov 11, 2021 at 13:43
  • $\begingroup$ That's great, thanks! I've typed it in as a block quote. It's possible that somewhere later in the chapter they elaborate. This is going to be a difficult question to get answered here in Stack Exchange as it may simply be an unclear sentence. We usually use "longer" and "shorter" when comparing wavelengths, not "lower". Instead we use "lower" and "higher" for frequency. Certainly for lower frequencies (longer wavelengths) than the 1550 nm band the lasers that do exist are likely to be lower power when feeding a single mode fiber, that might be what they are really trying to say. $\endgroup$
    – uhoh
    Commented Nov 11, 2021 at 23:19

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Photons with a shorter wavelength and a higher frequency have more energy than those with longer wavelength.

I am sure if we have low wavelength we will have more power and if we have high wavelength, we will have low power. Let take as example lasers. GaAlAs-Laser operates in 780-890 nm with power 200 mW, InGaAsP operates in 1300nm, its power is <50 mW.

These power levels are the specifications of the lasers but not the power needed for a communication over a certain distance and data rate. If you have a low wavelength and more beam power you need less photons per mW, for high wavelength and low beam power there are more photons per mW.

But the energy of a photon measured in Joule or Wattseconds and the power of a beam measured in Watt are two very different things.

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    $\begingroup$ Photon energy is inversely proportional to wavelength. If you have a low wavelength, you will have high photon energy. If you have a high wavelength, you will have low photon energy, right? $\endgroup$ Commented Nov 18, 2021 at 10:16

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