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

Answer

C

Explanation

By using `I = (P)/(A)`, where `P = radiation power`
`rArr P = I xx A rArr (nhc)/(t lambda) = I A rArr (n)/(t) = (I A lambda)/(hc)`
Hence number of photons entering per sec the eye
`((n)/(t)) = (10^(-10) xx 10^(-6) xx 5.6 xx 10^(-7))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 300`.