Q. The minimum intensity of light to be detected by human eye is $10^{- 10}W/m^{2}.$ The number of photons of wavelength $5.6 \times 10^{- 7}m$ entering the eye, with pupil area $10^{- 6}m^{2},$ per second for vision will be nearly
A) 100
B) 200
C) 300
D) 400
Explanation
(c) By using $I = \frac{P}{A};$ where $P =$ radiation power
$\Rightarrow P = I \times A \Rightarrow \frac{nhc}{t\lambda} = IA \Rightarrow \frac{n}{t} = \frac{IA\lambda}{hc}$
Hence number of photons entering per sec the eye
$\left( \frac{n}{t} \right) = \frac{10^{- 10} \times 10^{- 6} \times 5.6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 300$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.