Two light sources of $8 C d$ and $12 C d$ are placed on the same side of the photometer screen at a distance of $40 c m$ from it. Where should a $80 C d$ source be placed to balance the illuminance?

40 c m

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60 c m

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20 c m

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80 c m

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Solution

The correct option is D $80 c m$
Intensity of illumination = $\frac{L}{d^{2}}$

Therefore
$I_{1} + I_{2} = I$

$\frac{8}{{0.4}^{2}} + \frac{12}{{0.4}^{2}} = \frac{80}{d^{2}}$

=> $\frac{20}{0.16} = \frac{80}{d^{2}}$

=> d = 0.8 m =80 cm

Answer. D) 80 cm

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Two light sources of 8Cd and 12Cd are placed on the same side of the photometer screen at a distance of 40cm from it. Where should a 80Cd source be placed to balance the illuminance?

The correct option is D 80cmIntensity of illumination = Ld2ThereforeI1+I2=I80.42+120.42=80d2=>200.16=80d2=> d = 0.8 m =80 cmAnswer. D) 80 cm

Two light sources of $8 C d$ and $12 C d$ are placed on the same side of the photometer screen at a distance of $40 c m$ from it. Where should a $80 C d$ source be placed to balance the illuminance?

The correct option is D $80 c m$
Intensity of illumination = $\frac{L}{d^{2}}$

Therefore
$I_{1} + I_{2} = I$

$\frac{8}{{0.4}^{2}} + \frac{12}{{0.4}^{2}} = \frac{80}{d^{2}}$

=> $\frac{20}{0.16} = \frac{80}{d^{2}}$

=> d = 0.8 m =80 cm

Answer. D) 80 cm