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PHYSICS 20: Light... Illumination problem?

An electric bulb hanging from a ceiling 1.2 m away from the floor. The bulb is risen 3.0 m away from the floor. What is the change in illumination of the floor?

Please explain how you came to your answer

Illumination intensity follows the 1/R^2 law, which says that, all else being equal, intensity is proportional to 1/R^2 where R is the distance from the illumination source. So if "a" is the illumination level at 1.2m, then a(1.2/3)^2 is the new illumination level, or 0.16a.

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The illumination I from the light source is given by the formula I= k/d^2 where k is a const. and d is dist.?

As the distance from the light source doubles, how does the illumination change.

If the distance is doubled, then

I = k/(2d)^2 = k/4d^2

The illumination is 4 times less when the distance from the source is doubled.

Hope this helps.

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