Photon energy and transmit power of an optical transmitter

I have a big doubt.

I have found an answer of other community a question about a relation between wavelength and transmit power.

My doubt is according to the reply

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.

As I understand, Photon energy is inversely proportional to wavelength. If we have a low wavelength, we will have high photon energy. If we have a high wavelength, wewill have low photon energy, right?

• so, you can answer this based on en.wikipedia.org/wiki/Photon#Physical_properties . The formula for $E$ is literally there. Could you maybe specify where your doubt arises from? We don't just want to repeat what has already been written in the reply you quote or the wikipedia article that I bet you've already read! Nov 18 '21 at 10:57
• Don't cross-post questions. Nov 18 '21 at 11:28
• Crossposted from electronics.stackexchange.com/q/595593/52589 Nov 18 '21 at 11:32
• @MarcusMüller here space.stackexchange.com/questions/55740/… , it is given , at 1550 nm a much higher transmit power level we need than at lower wavelengths, I didnt undersand this point. This post has the answer, which the reason why is correct was explained. If we have a look at gain equation....we wil get opposite result. Am i wrong? Nov 18 '21 at 11:43