SOLUTION: A piece of cardboard is placed midway between two sources of light of equal intensity which are 12 feet apart.Its plane is perpendicular to the line joining the two sources. How fa

Algebra ->  Test -> SOLUTION: A piece of cardboard is placed midway between two sources of light of equal intensity which are 12 feet apart.Its plane is perpendicular to the line joining the two sources. How fa      Log On


   

Question 815399: A piece of cardboard is placed midway between two sources 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?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A piece of cardboard is placed midway between two sources 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?
--------------------
I = intensity of the sources.
At the midpoint, d = 6 for both sides
Energy rcvd at each side = I*k/6^2 at the midpoint
E = 2I*k/36 at midpoint
-----
At any other point:
E = I*k/d^2 + I*k/(12-d)^2
----
For 10x total:
E = I*k/d^2 + I*k/(12-d)^2 = 10*2*I*k/36
solve for d
1/d^2 + 1/(12-d)^2 = 5/9
%28%2812-d%29%5E2+%2B+d%5E2%29%2F%28d%5E2%2812-d%29%5E2%29+=+5%2F9
9%28d%5E2+-+24d+%2B+144+%2B+d%5E2%29+=+5d%5E2%2A%2812-d%29%5E2
18d%5E2+-+216d+%2B+1296+=+5d%5E4+-+120d%5E3+%2B+720d%5E2
5d%5E4+-+120d%5E3+%2B+702d%5E2+%2B+216d+-+1296+=+0
-----------
d = 1.3524199 by graphing
---
It's moved 6 - d
=~ 4.65 feet