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 $300$

On using $l=\frac{P}{A}$, where $P=$ radiation power
$\Rightarrow P=l\times A=\frac{nhc}{t\lambda }=lA\Rightarrow \frac{n}{t}=\frac{lA\lambda }{hc}$
Hence the number of photons entering per second
eye$\left(\frac{n}{t}\right)=\frac{{10}^{-10}\times {10}^{-6}\times 5.6\times {10}^{-7}}{6.6\times {10}^{-34}\times 3\times {10}^8}=300$