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An isotropic popint source emits light with wavelength 589 nm. The radiation power of the source is P = 10 W. (a) Find the number of photons passing through unit area per second at a distance of 2 m from the source (b) Also calculate the distance between the source and the point at which the mean concentration of the photons is 100//cm^(3) . |
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Answer» Solution :(a) `P =n_(0) (hc)/(lambda)` `n_(0) =(P lambda)/(hc)` , where `n_(0)`: number of photons/sec,  At distance r from POINT source, number of photons / area / TIME `n' =(n_(0))/(4pi r^(2)) = (P lambda)/(hc.4pi r^(2))` `=(10xx589xx10^(-9))/(6.6xx10^(-34)xx3xx10^(8)xx4pixx(2)^(2))` `=5.92xx10^(18)//m^(2)`. sec (b) Consider a SPHERICAL shell of RADIUS r and thickness DR  Volume of this shell `dV =4pi r^(2)dr` If n: number of photons/volume Number of photons crossing this shell per unit time `n_(0) =n(4pi r^(2)(dr)/(dt))` `[where(dr)/(dt) =c]` `n =(n_(0))/(4pi r^(2)c) =(P lambda//hc)/(4pi r^(2)c)` `r=sqrt((P lambda)/(4pi hc^(2)n)` `=sqrt((10xx589xx10^(-9))/(4xx3.14xx6.6xx10^(-34)xx(3xx10^(8))xx100xx10^(6)))` `=8.88m` |
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