Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by ____________ .

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KollegeApply · Accepted Answer

2% ​Illuminance is given by E = I_o \cos$\theta$/r^2 $\theta$ = 0^0 \Delta r/r = 1 \% E = I_o/r^2 Differentiating, dE = - 2I_o/r^3dr As approximation differentials are replaced by \Delta, \Delta E = - 2I_o/r^3\Delta r \Rightarrow \Delta E = - 2I_o/r^2\left( \Delta r/r \right) \Rightarrow \Delta E = - 2E\left( \Delta r/r \right) \Rightarrow \Delta E/E = - 2\left( \Delta r/r \right) \Rightarrow \Delta E/E = - 2 $\times$ 1 \% = - 2 \% Since, negative sign implies decrease; hence, illuminance decreases by 2% .

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