(N/A) Power of the medium wave transmitter,$P = 10\; kW = 10^4\; W = 10^4\; J/s$.
Hence,energy emitted by the transmitter per second,$E = 10^4\; J$.
Wavelength of the radio wave,$\lambda = 500\; m$.
The energy of a single photon is given by $E_1 = \frac{hc}{\lambda}$.
Where,$h = 6.6 \times 10^{-34}\; Js$ and $c = 3 \times 10^8\; m/s$.
$E_1 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500} = 3.96 \times 10^{-28}\; J$.
Let $n$ be the number of photons emitted per second.
$n = \frac{E}{E_1} = \frac{10^4}{3.96 \times 10^{-28}} \approx 2.525 \times 10^{31} \approx 3 \times 10^{31}$ photons/s.
Since the number of photons is extremely large,the discrete nature of energy can be ignored,and the wave can be treated as continuous.
$(b)$ Intensity of light,$I = 10^{-10}\; W m^{-2}$.
Area of the pupil,$A = 0.4\; cm^2 = 0.4 \times 10^{-4}\; m^2$.
Frequency of white light,$\nu = 6 \times 10^{14}\; Hz$.
Energy of one photon,$E_p = h\nu = 6.6 \times 10^{-34} \times 6 \times 10^{14} = 3.96 \times 10^{-19}\; J$.
Total energy incident on the pupil per second,$E_{total} = I \times A = 10^{-10} \times 0.4 \times 10^{-4} = 4 \times 10^{-15}\; J/s$.
Number of photons entering the pupil per second,$n = \frac{E_{total}}{E_p} = \frac{4 \times 10^{-15}}{3.96 \times 10^{-19}} \approx 1.01 \times 10^4$ photons/s.
This number is large enough that the human eye cannot perceive individual photons.