The eye can detect 5 times 10^4 photons per s | vedclass

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(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \ \mathring{A}$, we get $E =

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$5$

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(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \ \mathring{A}$, we get $E = \frac{12400}{5000} \ 	ext{eV} = 2.48 \ 	ext{eV}$.Converting this to Joules: $E = 2.48 	imes 1.6 	imes 10^{-19} \ J \approx 3.97 	imes 10^{-19} \ J \approx 4 	imes 10^{-19} \ J$.The minimum intensity detected by the eye $(I_{Eye})$ is the product of photon flux and energy per photon:$I_{Eye} = (5 	imes 10^4 \ 	ext{photons}/m^2s) 	imes (4 	imes 10^{-19} \ J) = 2 	imes 10^{-14} \ W/m^2$.The sensitivity of a detector is inversely proportional to the minimum intensity it can detect.Therefore, the ratio of sensitivity is $\frac{S_{Eye}}{S_{Ear}} = \frac{I_{Ear}}{I_{Eye}} = \frac{10^{-13}}{2 	imes 10^{-14}} = \frac{10}{2} = 5$.Thus, the eye is $5$ times more sensitive than the ear.

The eye can detect $5 \times 10^4$ photons per square metre per second of green light $(\lambda = 5000 \ \mathring{A})$ while the ear can detect $10^{-13} \ W/m^2$. The factor by which the eye is more sensitive as a power detector than the ear is close to:

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