The minimum intensity of light to be detected | vedclass

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(C) The intensity $I$ is defined as power per unit area: $I = \frac{P}{A}$.Given $I = 10^{-10} \ W/m^2$ and $A = 10^{-6} \ m^2$,the power $P$ entering

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$300$

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(C) The intensity $I$ is defined as power per unit area: $I = \frac{P}{A}$.Given $I = 10^{-10} \ W/m^2$ and $A = 10^{-6} \ m^2$,the power $P$ entering the eye is $P = I 	imes A = 10^{-10} 	imes 10^{-6} = 10^{-16} \ W$.The power of a beam of $n$ photons per second is given by $P = \frac{n \cdot h \cdot c}{\lambda}$.Rearranging for the number of photons per second $\frac{n}{t} = \frac{P \cdot \lambda}{h \cdot c}$.Substituting the values: $\frac{n}{t} = \frac{10^{-16} 	imes 5.6 	imes 10^{-7}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.$\frac{n}{t} = \frac{5.6 	imes 10^{-23}}{19.89 	imes 10^{-26}} \approx 281.5 \approx 300$ photons per second.

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

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