InterviewSolution
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InterviewSolution
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An isotropic point source emits light with wavelength `lambda = 589 nm`. The radiation power of the source is `P = 10W`. Find: (a) the mean density of the flow of photons at a distance `r = 2.0 m` from the osurce, (b) the distance between the source and the point at which the mean concentration of photons is equal to `n = 100 cm^(-3)`. |
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Answer» (a) The means density of the flow of photons at a disatnce `r` is `lt j gt = (P)/(4pir^(2))//(2picancelh c)/(lambda) =(P lambda)/(8pi^(2)cancelh cr^(2))m^(-2)s^(-2)` `= (10xx589xx10^(-6))/(8pi^(2)xx1.054xx10^(-34)xx10^(8)xx4)m^(-2)s^(-1)` `= (10xx.589)/(8pi^(2)xx1.054xx12)xx10^(16)cm^(-2)s^(-1)` `= 5.9xx10^(13)cm^(-2)s^(-1)` (b) If `n(r)` is the mean concentration (number per unit volume) of photons at a distacne `r` form the source, then since all photons are moving outwards with a velocity `c`, there is an outward flux of `cn` which is balanced by the flux form the source. In steady state, the two are equal and so `n(r) = (lt j gt)/(c ) = (P lambda)/(8pi^(2)cancelh c^(2)r^(2)) = n` or `r = (1)/(2pic) sqrt((P lambda)/(2cancelh n))` `= (1)/(6pixx10^(8)) sqrt((10xx589xx10^(-6))/(2xx1.054xx10^(-34)xx10^(2)xx10^(6)))` `= (10^(2))/(6pi) sqrt((5.89)/(2.108)) = 8.87` metre |
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