Two lamps of luminous intensity of 8 , Cd and | vedclass

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(C) The illuminance $I$ at a distance $r$ from a source of luminous intensity $L$ is given by $I = \frac{L}{r^2}$.For the two faces of the screen to b

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$40 \, cm$ from $8 \, Cd$ lamp

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(C) The illuminance $I$ at a distance $r$ from a source of luminous intensity $L$ is given by $I = \frac{L}{r^2}$.For the two faces of the screen to be equally illuminated,the illuminance from both sources must be equal:$\frac{L_1}{r_1^2} = \frac{L_2}{r_2^2}$Given $L_1 = 8 \, Cd$,$L_2 = 32 \, Cd$,and the total distance $d = 1.2 \, m = 120 \, cm$.Let the screen be placed at a distance $x$ from the $8 \, Cd$ lamp. Then the distance from the $32 \, Cd$ lamp is $(120 - x) \, cm$.Substituting the values:$\frac{8}{x^2} = \frac{32}{(120 - x)^2}$Taking the square root on both sides:$\frac{\sqrt{8}}{x} = \frac{\sqrt{32}}{120 - x}$$\frac{2\sqrt{2}}{x} = \frac{4\sqrt{2}}{120 - x}$$\frac{1}{x} = \frac{2}{120 - x}$$120 - x = 2x$$3x = 120$$x = 40 \, cm$.Thus,the screen should be placed $40 \, cm$ from the $8 \, Cd$ lamp.

Two lamps of luminous intensity of $8 \, Cd$ and $32 \, Cd$ respectively are lying at a distance of $1.2 \, m$ from each other. Where should a screen be placed between the two lamps such that its two faces are equally illuminated due to the two sources?

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