Question

The minimum intensity of light to be detected by human eye is $10^{- 10}Wm^{- 2}$ . 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. $400$

Answer 1

Answer: (C)

Intensity of light is given as $I=\frac{P}{A}=\frac{n E}{A}$ , where $n$ is number of photons entering the eye, $E$ is energy of each photon and $A$ is area of pupil.
$n=\frac{A I}{E}=\frac{A I \lambda }{h c}$ , where $\lambda $ is wavelength of photon.
$n=\frac{10^{- 6} \times 10^{- 10} \times 5 \times 10^{- 7}}{6 . 34 \times 10^{- 34} \times 3 \times 10^{8}}=300$
Hence, minimum $300$ photons need to enter the eye of a person to detect light.