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adelina 88 [10]
3 years ago
7

The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportiona

l to the square of the distance from the source. if two light sources, one four times as strong as the other, are placed 20 ft apart, how far away from the stronger light source should an object be placed on the line between the two sources so as to receive the least illumination
Chemistry
1 answer:
ElenaW [278]3 years ago
5 0
<span>Exactly 4(4 - 2*2^(1/3) + 2^(2/3)) feet, or approximately 12.27023581 feet. Let's first create an equation to calculate the relative intensity of the light based upon the distance D from the brighter light source. The distance from the dimmer light source will of course be (20-D). So the equation will be: B = 4/D^2 + 1/(20-D)^2 The minimum and maximum can only occur at those points where the slope of the equation is 0. And you can determine the slope by using the first derivative. So let's calculate the first derivative. B = 4/D^2 + 1/(20-D)^2 B' = d/dD [ 4/D^2 + 1/(20-D)^2 ] B' = 4 * d/dD [ 1/D^2 ] + d/dD [ 1/(20-D)^2 ] B' = 4(-2)D^(-3) + (-2)(20 - D)^(-3) * d/dD [ 20-D ] B' = -8/D^3 - 2( d/dD [ 20 ] - d/dD [ D ] )/(20 - D)^3 B' = -8/D^3 - 2(0 - 1)/(20 - D)^3 B' = 2/(20 - D)^3 - 8/D^3 Now let's find a zero. B' = 2/(20 - D)^3 - 8/D^3 0 = 2/(20 - D)^3 - 8/D^3 0 = 2D^3/(D^3(20 - D)^3) - 8(20 - D)^3/(D^3(20 - D)^3) 0 = (2D^3 - 8(20 - D)^3)/(D^3(20 - D)^3) 0 = 2D^3 - 8(20 - D)^3 8(20 - D)^3 = 2D^3 4(20 - D)^3 = D^3 4(8000 - 1200D + 60D^2 - D^3) = D^3 32000 - 4800D + 240D^2 - 4D^3 = D^3 32000 - 4800D + 240D^2 - 5D^3 = 0 6400 - 960D + 48D^2 - D^3 = 0 -6400 + 960D - 48D^2 + D^3 = 0 D^3 - 48D^2 + 960D - 6400 = 0 We now have a simple cubic equation that we can use the cubic formulas to solve. Q = (3*960 - (-48)^2)/9 = 64 R = (9*(-48)*960 - 27*(-6400) - 2*(-48)^3)/54 = -384 D = Q^3 + R^2 = 64^3 + (-384)^2 = 409600 Since the value D is positive, there are 2 imaginary and 1 real root. We're only interested in the real root. S = cbrt(-384 + sqrt(409600)) S = cbrt(-384 + 640) S = cbrt(256) S = 4cbrt(4) T = cbrt(-384 - sqrt(409600)) T = cbrt(-384 - 640) T = cbrt(-1024) T = -8cbrt(2) The root will be 4cbrt(4) - 8cbrt(2) + 48/3 So simplify 4cbrt(4) - 8cbrt(2) + 48/3 =4cbrt(4) - 8cbrt(2) + 16 =4(cbrt(4) - 2cbrt(2) + 4) = 4(4 - 2*2^(1/3) + 2^(2/3)) Which is approximately 12.27023581 This result surprises me. I would expect the minimum to happen where the intensity of both light sources match which would be at a distance of 2/3 * 20 = 13.3333 from the brighter light source. Let's verify the calculated value. Using the brightness equation at the top we have: B = 4/D^2 + 1/(20-D)^2 Using the calculated value of 12.27023581, we get B = 4/D^2 + 1/(20-D)^2 B = 4/12.27023581^2 + 1/(20-12.27023581)^2 B = 4/12.27023581^2 + 1/7.72976419^2 B = 4/150.5586868 + 1/59.74925443 B = 0.026567713 + 0.016736611 B = 0.043304324 And the intuition value of 13.33333333 B = 4/D^2 + 1/(20-D)^2 B = 4/13.33333333^2 + 1/(20-13.33333333)^2 B = 4/13.33333333^2 + 1/6.666666667^2 B = 4/177.7777778 + 1/44.44444444 B = 0.0225 +0.0225 B = 0.045 And the calculated value is dimmer. So intuition wasn't correct. So the object should be placed 4(4 - 2*2^(1/3) + 2^(2/3)) feet from the stronger light source, or approximately 12.27023581 feet.</span>
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