Question

Minimum light intensity that can be perceived by normal human eye is about $ 10^{-10}Wb\,m^{-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} $ .

• A. $ 3.3\times {{10}^{2}} $
• B. $ 3.3\times {{10}^{3}} $
• C. $ 3.3\times {{10}^{4}} $
• D. $ 3.3\times {{10}^{5}} $

Answer 1

Answer: (C)

Intensity, I = $ 10^{ - 10 } \, W bm^{ - 2} \, = 10^{ - 10 } \, Js^{ - 1} \, m^{ - 2} $ .
Let the number of- photons required be $n$.
Then, $ \frac{ nhv }{ 10^{ - 4 } } = 10^{ - 10 } $
Hence, $ n = 10^{ - 10 } \times 10^{ - 4} / hv $
$= 10^{ - 14 } \frac{ \lambda }{ hc } $
$= \frac{ 10^{ - 14 } \times 660 \times 10^{ - 9 }}{ 6.6 \times 10^{ - 34 } \times 3 \times 10^8 } $
$= 3.3 \times 10^4 $