Two sources of light emit with a power of $200 \ W$. The ratio of the number of photons of visible light emitted per second by each source having wavelengths $300 \ nm$ and $500 \ nm$ respectively,will be:

$A$ transmitter of $100 \ W$ power emits electromagnetic waves of wavelength $540 \ nm$. How many photons are emitted per second? (Given: $h = 6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$)

How is the intensity of radiation determined?

The wavelength of a photon is $5000 \mathring A$. Its energy will be:

If the energy of the photon is increased by a factor of $4$,then its momentum

The minimum intensity of light that can be perceived by the human eye is $10^{-10} \ W/m^2$. If the area of the pupil is $10^{-6} \ m^2$ and the wavelength of light is $560 \ nm$,approximately how many photons per second enter the eye? (Use $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$).