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Find the number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that humans can perceive $(10^{-10}W/m^2$ ).( Assume area of pupil to be $0.4 cm^2$ , and average frequency of white light to be $6 \times 10^{14}Hz)$

$\begin {array} {1 1} (a)\;1.008 \times 10^4s^{-1} & \quad (b)\;10.08 \times 10^3s^{-1} \\ (c)\;1.008 \times 10^3s^{-1} & \quad (d)\;10.08 \times 10^4s^{-1} \end {array}$


1 Answer

Ans : (a)
$I=10^{-10} \large\frac{W}{m^2}$
$A = 0.4 cm^2 = 0.4 \times 10^{-4} m^2 \: \: \: \: v=6 \times 10^{14}Hz$
So, $E=hv = 6.6 \times 10^{-34} \times 6 \times 10^{14} Js$
$= 3.96 \times 10^{-19} J$
Let n be total number of photons falling per second, per unit area of the pupil.
So, total energy per unit for n falling photons : $E = n \times 3.96 \times 10^{-19} J/sm^2$
Energy per unit area per second is intensity of light.
$ \Rightarrow n \times 3.96 \times 10^{-19} = 10^{-10}$
$ \Rightarrow n= \large\frac{10^{-10}}{3.96 \times 10^{-19} } = 2.52 \times 10^8 m^2 s^{-1}$
So, total no. of photons entering per second is : $n_A = n \times A$
$= 2.52 \times 10^8 \times 0.4 \times 10^{-4}$
$= 1.008 \times 10^4 s^{-1}$


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