power received will always remain same and does not depends on the distance
it will be same as the power of source
[tex]P = 3 mW[/tex]
now the energy released by one photon is given as
[tex]E = \frac{hc}{\lambda}[/tex]
[tex]E = \frac{6.6* 10^{-34}* 3 * 10^8}{760 * 10^{-9}}[/tex]
[tex]E = 2.6 * 10^{-19} J[/tex]
now let say N photons per second released
[tex]N * 2.6 * 10^{-19} = 3 * 10^{-3}[/tex]
[tex]N = 1.15* 10^{16} per\: second[/tex]
