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

Options

A. `3xx10^2`
B. `3 xx 10 ^3`
C. `3 xx 10 ^4`
D. `3 xx 10 ^5`

Answer

A

Explanation

Here , `I = 10^(-10) Wm^(-2) = 10^(-10) Js^(-1) m^(-2)`
Let the number of photons that enter the pupil second be
N. Then `(Nhupsilon)/(10^(-6)) = 10^(-10)`
or `N=(10 ^(-6) xx 10^(-10))/(hupsilon) = (10^(-16)lamda)/ (hc) " "( :. c = upsilonlamda`)
= `(10^(-6)xx(5.6 xx10^(-7)))/((6.6 xx10^(-34))(3xx10^(8))) = 0.28 xx10^(3) ~~ 3 xx 10^(2) ` photons/s