The minimum intensity of light to be detected by human eye is `10^(-10)W//m^(2)`. The number of photons of wavelength `5.6xx10^(-7)`m entering they eye, with pupil area `10^(-6)` per second for vision will be nearly

Options

A. 100
B. 200
C. 300
D. 400

Answer

C

Explanation

Intensity (I) `=("Power")/("area")`
`:. P=IxxA = (nhc)/(tlamda)=IA`
`implies n/t = (10^(-10) xx 10^(-6) xx5.6 xx10^(-7) )/(6.6 xx10^(-34)xx3 xx10^8) = 282.8 ~~300`