The minimum intensity of light to be detected by human eye is `10^(-10) W//m^(2)`. The number of photons of wavelength `5.6 xx 10^(-7) m` entering the eye , with pupil area `10^(-6) m^(2)` , per second for vision will be nearly

Answer

`3xx10^(4)`photons

Explanation

`n(hc)/lambda=I.A. or n=(Ialambda)/(hc)`
`n=(10^(-10)xx10^(-4)xx5.84xx10^(-7))/(6.63xx10^(-34)xx3xx10^(8))=3xx10^(4)`