An electric bulb is hanging over a table at a height of 1m above it.The illuminance on the table directly below the bulb is 40 lux. the illuminance at a point on the table 1 m away from the first point will be about

10.5 lux

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14.1 lux

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20.8lux

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28.8 lux

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Solution

The correct option is B 14.1 lux
Let $I$ be the intensity of bulb then
case I: $40 = \frac{I}{1 \times 1} \Rightarrow I = 40 c d$
case II: $E = \frac{40}{{(1^{2} + 1^{2})}^{3 / 2}} = \frac{40}{\sqrt{8}} = 14.1 l u x$

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An electric bulb is hanging over a table at a height of 1m above it.The illuminance on the table directly below the bulb is 40 lux. the illuminance at a point on the table 1 m away from the first point will be about

The correct option is B 14.1 luxLet I be the intensity of bulb thencase I: 40=I1×1⇒I=40cdcase II: E=40(12+12)3/2=40√8=14.1lux

The correct option is B 14.1 lux
Let $I$ be the intensity of bulb then
case I: $40 = \frac{I}{1 \times 1} \Rightarrow I = 40 c d$
case II: $E = \frac{40}{{(1^{2} + 1^{2})}^{3 / 2}} = \frac{40}{\sqrt{8}} = 14.1 l u x$