Correct Answer - A
`E = (12375)/(5000) = 2.475 eV ~~ 4 xx 10^(-19) J`
So the minimum intensity to which the eye can respond
`I_(Eye) = ("Photon flux")` xx `(" Energy of a photon")`
`rArr I_(Eye) =( 5 xx 10^(4)) xx ( 4 xx 10^(-19)) = 2 xx 10^(-14) (W//m^(2))`
Now as lesser the intensity required by a detector for detection , more sensitive it will be
`(S_(Eye))/(S_(Ear)) = (I_(Ear))/(I_(Eye)) = (10^(-13))/(2 xx 10^(-14)) = 5` i.e., as intensity `("power")`
detector , the eye is five times more sensitive than ear.