A monochromatic beam of light of wavelength 4 | vedclass

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(C) The radiation pressure $P$ exerted by a beam with intensity $I$ is given by $P = \frac{R}{c} I + \frac{A}{c} I$,where $R$ is the reflection coeffi

Vedclass · Accepted Answer

$19.2$

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(C) The radiation pressure $P$ exerted by a beam with intensity $I$ is given by $P = \frac{R}{c} I + \frac{A}{c} I$,where $R$ is the reflection coefficient and $A$ is the absorption coefficient. Here,$R = 0.25$ and $A = 0.75$. Since the reflected light exerts $2I/c$ and absorbed light exerts $I/c$,we have: $P = 0.25 \left(\frac{2I}{c}ight) + 0.75 \left(\frac{I}{c}ight) = \frac{0.5I + 0.75I}{c} = \frac{1.25I}{c}$. Given $P = 5 	imes 10^{-7}\,N/m^2$,we find intensity $I = \frac{P 	imes c}{1.25} = \frac{5 	imes 10^{-7} 	imes 3 	imes 10^8}{1.25} = 120\,W/m^2$. The number of photons incident per second $n_p = \frac{I 	imes A_{area}}{E_{photon}} = \frac{I 	imes A_{area} 	imes \lambda}{hc}$. $n_p = \frac{120 	imes (5 	imes 10^{-4}) 	imes 400 	imes 10^{-9}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 1.206 	imes 10^{17} \approx 1.2 	imes 10^{17}\,s^{-1}$. Given $0.1\%$ of photons produce electrons,$n_e = 10^{-3} 	imes 1.2 	imes 10^{17} = 1.2 	imes 10^{14}\,s^{-1}$. Saturation current $i_s = n_e 	imes e = 1.2 	imes 10^{14} 	imes 1.6 	imes 10^{-19} = 1.92 	imes 10^{-5}\,A = 19.2\,\mu A$.

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