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.
$E=I$
$ \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 : $n_A = n \times A$
$= 2.52 \times 10^8 \times 0.4 \times 10^{-4}$
$= 1.008 \times 10^4 s^{-1}$