Physics - Wave Optics Question with Solution | TestHub

The intensity of light at the location of your eye is

$\text{I}=\frac{\text{P}}{4\pi {\text{r}}^2}=\frac{60}{4\pi \times 4^2}W{\text{m}}^{-2}$

The energy entering into your eye per second is

${\text{P}}_1=\text{I}\times \frac{\pi {\text{d}}^2}{4}$

where d is the diameter of pupil

${\text{P}}_1=\frac{60}{4\pi \times 4^2}\times \frac{\pi \times {\left(2\times 10^{-3}\right)}^2}{4}$

$=\text{9.375}\times 10^{-7}{Js}^{-1}$

Let n be the number of photons entering into the eye per second, then

${\text{P}}_1=\text{n}\times \frac{\text{hc}}{\lambda }$

$\text{9.375}\times 10^{-7}=\text{n}\times \frac{1240\times \text{1.6}\times 10^{-19}}{600}$

$\text{n}=\text{2.84}\times 10^{12}photon{\text{s}}^{-1}$

So, the number of photons entering the eye in 0.1 s = 0.1 n = 2.84 x 10¹¹