Question

The minimum intensity of light to be detected by human eye is ${10}^{-10}W/m^2$. The number of photons of wavelength $5.6\times {10}^{-7}m$ entering the eye, with pupil area ${10}^{-6}{m}^2$, per second for vision will be nearly

• A. 100
• B. 200
• C. 300
• D. 400

Answer 1

The correct option is C 300

By using $I=\frac{P}{A};$ where P = radiation power
$\Rightarrow P=I\times A\Rightarrow \frac{nhc}{t\lambda }=IA\Rightarrow \frac{n}{t}=\frac{IA\lambda }{hc}$ Hence number of photons entering per sec the eye $\left(\frac{n}{t}\right)=\frac{{10}^{-10}\times {10}^{-6}\times 5.6\times {10}^{-7}}{6.6\times {10}^{-34}\times 3\times {10}^8}=300$.