Minimum light intensity that can be perceived by normal human eye is about $10^{-10}Wbm^{-2}$. What is the minimum number of photons of wavelength 660 nm that must enter the pupil in 1 s for one to see the object? Area of cross-section of the pupil is $10^{-4}m^2$.

$3.3×10^4$

Intensity, $I=10^{-10}Wbm^{-2}=10^{-10}Js^{-1}m^{-2}$. Let the number of photons required be n.

Then, $I=\frac{nhv}{A}=\frac{nhv}{10^{-4}}=10^{-10}$

Hence, $n=10^{-10}×10^{-4}/hv=10^{-14}\frac{λ}{hc}$

$=\frac{10^{-14}×660×10^{-9}}{6.6×10^{-34}×3×10^8}=3.3×10^4$