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Question

(A) The intensity $I$ is given as $10^{-10} \ W/m^2$ and the area $A$ is $10^{-6} \ m^2$.The power $P$ entering the eye is $P = I 	imes A = 10^{-10}

Vedclass · Accepted Answer

$3 	imes 10^2$ photons

Vedclass · Answer

(A) The intensity $I$ is given as $10^{-10} \ W/m^2$ and the area $A$ is $10^{-6} \ m^2$.The power $P$ entering the eye is $P = I 	imes A = 10^{-10} 	imes 10^{-6} = 10^{-16} \ W$.The energy of a single photon is $E = \frac{hc}{\lambda}$.Given $\lambda = 560 \ nm = 560 	imes 10^{-9} \ m$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$:$E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{560 	imes 10^{-9}} = \frac{19.8 	imes 10^{-26}}{560 	imes 10^{-9}} \approx 3.54 	imes 10^{-19} \ J$.The number of photons per second $n$ is given by $n = \frac{P}{E}$.$n = \frac{10^{-16}}{3.54 	imes 10^{-19}} = \frac{1000}{3.54} \approx 282 \approx 3 	imes 10^2$ photons.

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