The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one 10 times as strong as the other, are placed 9 ft apart, how many feet from the brighter source should an object be placed on the line between the sources so as to receive the least illumination?
2 answers
10 times 9 = 90ft apart
let the light be situated x ft from one light and
9-x from the other.
Let the power be P, where P is a constant
let I be the total illumination.
I = P/x^2 + 10P/(9-x)^2
= Px^-2 + 10P(9-x)^-2
dI/dx = -2Px^-3 - 10P(9-x)^-3 (-1)
= 0 for a min of I
10P/(9-x)^3 = P/x^2
divide by P , which is a constant
10/(9-x)^3 = 1/x^3
10x^3 = (9-x)^3
10^(1/3) x = 9-x
10^(1/3) x + x = 9
x(10^(1/3) - 1 = 9
x = 7.796 ft
It should be placed appr 7.8 ft away from the stronger light
or 1.2 ft from the weaker light
9-x from the other.
Let the power be P, where P is a constant
let I be the total illumination.
I = P/x^2 + 10P/(9-x)^2
= Px^-2 + 10P(9-x)^-2
dI/dx = -2Px^-3 - 10P(9-x)^-3 (-1)
= 0 for a min of I
10P/(9-x)^3 = P/x^2
divide by P , which is a constant
10/(9-x)^3 = 1/x^3
10x^3 = (9-x)^3
10^(1/3) x = 9-x
10^(1/3) x + x = 9
x(10^(1/3) - 1 = 9
x = 7.796 ft
It should be placed appr 7.8 ft away from the stronger light
or 1.2 ft from the weaker light