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An illuminant shaped as a plane horizont...

An illuminant shaped as a plane horizontal disc of raiud `R = 25 cm` is suspended over a table at a hight `h = 75 cm`. The illuminance of the table below the cnetre of the illuminant is equal to `E_(0) = 70 Ix`. Assuming the source to obey Lambert's law, find its luminosity.

Text Solution

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Consider an angular element of area
`2 pi x dx = 2pi h^(2) tan theta sec^(2) theta d theta`
Light emitted from his ring is
`d Phi = Ld Omega (2pi h^(2) tan theta sec^(2) theta d theta). Cos theta`
Now `d Omega = (dA cos theta)/(h^(2) sec^(2) theta)`
where `dA =` an element of area of the table just below the untre of the illuminant. Then the illuminance at the element `dA` will be
`E_(0) = int_(theta-0)^(theta-alpha) 2pi L sin theta cos theta d theta`
where `sin alpha = (R)/(sqrt(h^(2) + R^(2))`. Finally using luminosity `M = pi L`
`E_(0) = M sin^(2) alpha = M (R^(2))/(h^(2) + R^(2))`
or `M = E_(0) (1+(h^(2))/(R^(2))) = 7001m//m^(2) *(1 1x = 1 (1m)/(m^(2))` dimensionally).
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