SOLUTION: the intensity I of light varies inversely as the square of the distance D from the source if the intensity of Illumination on a screen 5 ft from a light is 3 foot

SOLUTION: the intensity I of light varies inversely as the square of the distance D from the source if the intensity of Illumination on a screen 5 ft from a light is 3 foot_candles find the

Question 1120409: the intensity I of light varies inversely as the square of the distance D from the source if the intensity of Illumination on a screen 5 ft from a light is 3 foot_candles find the intensity on a screen 15 ft from the light.
A. 1(1/3) Foot-candles
B.1/3 Foot-Candles
c.1/4 Foot- Candles
D. 2 Foot - Candles
Found 2 solutions by rothauserc, greenestamps:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
I = k/D^2, where I is intensity and D is distance from source
:
Note k is the constant of proportionality
:
3 = k/5^2
:
k = 3 * 25 = 75
:
I = 75/15^2 = 75/225 = 1/3 foot_candle
:
answer is B.
:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Solving the problem as the other tutor did, using the formal definition of inverse proportion to find the constant of proportionality, is of course a valid method.

And you should be able to use that method, as it may be required in a more complicated problem.

But for a simple problem like this, I personally find it far easier simply to say that since the distance is multiplied by 3, the intensity is multiplied by 1/(3^2) = 1/9.

Then 1/9 of 3 foot-candles is 1/3 foot-candle, answer B.
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