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Standard VIII

Physics

Luminous Intensity

Two light sou...

Question

Two light sources with equal luminous intensity are lying at a distance of 1.2 m from each other. Where should a screen be placed between them such that illuminance on one of its faces is four times that on another face?

0.2 m

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0.4 m

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0.8 m

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1.6 m

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Solution

The correct option is B 0.8 m
$E_{2} = 4 E_{1}$ . If x is distance from 1st source,
then, $\frac{I}{(1.2 - x)^{2}} = \frac{4 I}{x^{2}}$ or $\frac{1}{1.2 - x} = \frac{2}{x}$
$3 x = 2.4, x = 0.8 m$

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Two light sources with equal luminous intensity are lying at a distance of 1.2 m from each other. Where should a screen be placed between them such that illuminance on one of its faces is four times that on another face?

The correct option is B 0.8 mE2=4E1. If x is distance from 1st source,then, I(1.2−x)2=4Ix2 or 11.2−x=2x3x=2.4,x=0.8m

The correct option is B 0.8 m
$E_{2} = 4 E_{1}$ . If x is distance from 1st source,
then, $\frac{I}{(1.2 - x)^{2}} = \frac{4 I}{x^{2}}$ or $\frac{1}{1.2 - x} = \frac{2}{x}$
$3 x = 2.4, x = 0.8 m$