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how far must it be moved

a piece of cardboard is placed midway between two surfaces of light of equal intensity which are 12 feet apart. its plane is perpendicular to the line joining the two sources. how far must it be moved, being kept always in this line and perpendicular to it, so that the total illumination which it receives (i.e., on both sides) will be ten times as great?
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2 Answers

Hi Emilyn,
 
If we let D= distance between the light sources (12ft)
and d= distance from the cardboard sheet to the nearest light source and assuming the intensity of light varies inversely as the square of the distance from the source then we can make the following calculations
 
The total intensity initially when the sheet is midway between is I0=I/(D/2)2 + I/(D/2)= 8I/Dso we have to now find out where to place the sheet so that the total intensity is 10*I0
The total intensity when the sheet is a distance d form one of the light sources is
 
It=I/d+ I/(D-d)2 we then set this equal to 10*I0 and we get
 
I/d2 + I/(D-d)2 =80I/D2 the I's cancel and we are left with an equation with d being the only unknown which has to be >0, D=12ft.
I got d=1.35 and 10.65ft which are the positions that the sheet can be placed so the total intensity is ten time that of the mid position.
 
Hope this helps
Jim
I = k/(d^2)

10*2k/(6^2) = k/((6–x)^2) + k/((6+x)^2)

20/36 = 5/9 = ( (6+x)^2 + (6–x)^2 ) / ( ((6–x)^2)((6+x)^2) )

5((6–x)^2)((6+x)^2) = 9 ( (6+x)^2 + (6–x)^2 )

5 (36–x^2)^2 = 9 ( 36 + 12x + x^2 + 36 - 12x +x^2 )

5 (36^2–72x^2+x^4) = 9 ( 72 + 2x^2 )

5*36^2 – 5*72x^2 + 5x^4 = 9*72 + 18x^2 )

5x^4 – (5*72 + 18)x^2 + 5*36^2 – 9*72 = 0

5x^4 – (5*8*9 + 2*9)x^2 + 5*4*9*4*9 – 9*9*8 = 0

5x^4 – 2*9(5*4 + 1)x^2 + 9*9*8(5*2 – 1) = 0

5x^4 – 2*3*7*9x^2 + 8*9*9*9 = 0

h = – – 2*3*7*9/10 = 3*3*3*7/5

k = 8*9*9*9 – 5*(3*3*3*7/5)^2

k = 5*8*9*9*9/5 – 9*9*9*49/5

k = (5*8 – 49)*9*9*9/5 = – 9*9*9*9/5

5(x^2 – 3*3*3*7/5)^2 – 9*9*9*9/5 = 0

(x^2 – 3*3*3*7/5)^2 = 9*9*9*9/25

x^2 = 3*3*3*7/5 ± 9*9/5 = (7 ± 3)(3*9/5)

x^2 = 2*3*3*3 or 2*2*3*3*3*5/25

x = 3√(6) or 6√(15)/5 feet from center.

x ≈ 7.34846922834953 or 4.6475800154489 feet from center.

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