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(A) The total energy emitted by the lamp in time $t$ is $E_{total} = P 	imes t$.Since the lamp radiates light uniformly in all directions,the intensi

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$\frac{P\lambda d^2 t}{hc l^2}$

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(A) The total energy emitted by the lamp in time $t$ is $E_{total} = P 	imes t$.Since the lamp radiates light uniformly in all directions,the intensity at a distance $l$ is $I = \frac{P}{4\pi l^2}$.The sensor has a circular opening of diameter $4d$,so its radius is $r = 2d$.The area of the circular opening is $A = \pi r^2 = \pi (2d)^2 = 4\pi d^2$.The energy $E$ incident on the sensor is $E = I 	imes A 	imes t = \left( \frac{P}{4\pi l^2} ight) 	imes (4\pi d^2) 	imes t = \frac{P d^2 t}{l^2}$.If $n$ is the number of photons,then $E = n 	imes \frac{hc}{\lambda}$.Equating the two expressions for energy: $n \frac{hc}{\lambda} = \frac{P d^2 t}{l^2}$.Solving for $n$: $n = \frac{P \lambda d^2 t}{hc l^2}$.

$A$ sensor is exposed for time $t$ to a lamp of power $P$ placed at a distance $l$. The sensor has a circular opening that is $4d$ in diameter. Assuming all energy of the lamp is given off as light,the number of photons entering the sensor if the wavelength of light is $\lambda$ is (given $l >> d$):

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