InterviewSolution
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InterviewSolution
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A point source of monochromatic light emitting a lumninous flux `Phi` is positioned at the cnetre of a spherical layer of substance. The inside radius of the layer is `a`, the outside one is `b`. The coefficient of linear absorption of the substance is equal to `x`, the reflection coefficient of the surfcaes is equal to `rho`. Neglecting the secondary reflections, find the intensity of light taht passes through that layer. |
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Answer» We have to derive the law of decrease of intensity in an absorbing medium taking in to account the natural geometrical fall-off (inverse sequare law) as well as absorption. Consider a thin sperical sheel of thickness `dx` and internal radius `x`. Let `I(x)` and `I(x+dx)` be the intersities at the inner and outer surfcaes of this shell. Then `4pix^(2)I(x)e^(-chi dx) = 4pi(x+dx)^(2) I(x+dx)` Except for the factor `e^(-chi dx)` this is the usual equation. we rewrite this as `x^(2)I(x) = I(x+dx) (x+dx)^(2) (1+chi dx)` `=(1+(dI)/(dx)dx) (x^(2)+2xdx)(1+chi dx)` or `x^(2)(dI)/(dx) + chi x^(2) I + 2xI = 0` Hence `(d)/(dx)(x^(2)I) + chi (x^(2)I) = 0` so `x^(2) I = Ce^(-chi x)` where `C` is a constant of intergration. In our case we apply this equation for `ale xle b` For `c le a` the usual inverse square law gives `I(a) = (Phi)/(4pia^(2))` Hence `C = (Phi)/(4pi) e^(chia)` and `I(b) = (Phi)/(4pib^(2)) e^(-chi(b -a))` This does not take into account reflection. When we do that we get `I(b) = (Phi)/(4pib^(2)) (1-rho)^(2) e^(-chi(b-a))` |
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