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 four times as strong as the other, are placed 11 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?
I did a similar question last night
just change the numbers
http://www.jiskha.com/display.cgi?id=1417144323
Well, if you're looking for the least illumination, I would suggest placing the object right smack in the middle of the two light sources. That way, you can bask in the glorious mediocrity of equal illumination from both sources! Plus, it saves you the trouble of doing any pesky calculations. You're welcome for the enlightening answer!
To find the distance from the stronger light source to place the object for the least illumination, we need to find the point where the total illumination from the two light sources is minimized.
Let's assume that the distance from the object to the stronger light source is x ft. Then the distance from the object to the weaker light source is (11 - x) ft.
According to the given conditions, the illumination at the object from the stronger light source is directly proportional to the strength of the source. Thus, if the strength of the stronger light source is S, the illumination at the object from the stronger light source is S.
The illumination at the object from the weaker light source is inversely proportional to the square of the distance from the source. Thus, if the strength of the weaker light source is 1/4S, the illumination at the object from the weaker light source is (1/4S)/(11 - x)^2.
To find the total illumination at the object, we add the illumination from the stronger light source and the weaker light source: S + (1/4S)/(11 - x)^2.
Now, to minimize the total illumination, we can differentiate it with respect to x and equate it to zero:
d/dx (S + (1/4S)/(11 - x)^2) = 0
Differentiating, we get:
0 - 2*(1/4S)/(11 - x)^3 = 0
Simplifying, we have:
-(1/4S)/(11 - x)^3 = 0
Multiplying both sides by -4S and (11 - x)^3, we get:
1/(11 - x)^3 = 0
Taking the cube root of both sides, we have:
1/(11 - x) = 0
But this is not possible as the denominator cannot be zero.
Therefore, there is no distance on the line between the two sources where the object can be placed to receive the least illumination.
To find the distance from the stronger light source where the object should be placed to receive the least illumination, we need to understand the relationship between illumination, strength of the source, and distance.
According to the given problem, the illumination of an object is directly proportional to the strength of the light source and inversely proportional to the square of the distance from the source. Mathematically, we can express this relationship as:
I ∝ S / d²
where I represents the illumination, S represents the strength of the light source, and d represents the distance from the source.
Given that one light source is four times as strong as the other, we can express the relative strengths as:
S1 = 4S2
Next, we're given that the two light sources are placed 11 ft apart. Let's denote the distance from the stronger light source to the object as x. So, the distance from the weaker light source to the object would be (11 - x).
Now, we can write the equation for the illumination of the object when considering both light sources:
I1 = (S1 / x²) (illumination from the stronger light source)
I2 = (S2 / (11-x)²) (illumination from the weaker light source)
To find the total illumination (I_total), we add the illuminations from both sources:
I_total = I1 + I2
Since we want to find the distance from the stronger light source with the least illumination, we need to minimize the value of I_total. To do this, we can take the derivative of I_total with respect to x and set it equal to zero to solve for x.
I'll calculate it:
I=kS/d²
where
I=illumination,
k=proportional constant >0
S=source strength >0
d=distance
One light source is S, and the other is 4S.
Let d=distance from weaker source, then
11-d =distance from stronger source.
We have illumination as a function of d
I(d) = kS/d²+k(4S)/(11-d)²
To find the minimum illumination, we equate first derivative to zero and check that it is a minimum.
I'(d)=-2kS/d³+8kS/(11-d)²
Equating to zero:
2kS/d³=8kS/(11-d)²
and solve for d:
(11-d)³=4d³
11-d=4^(1/3)d
d=11/(1+4^(1/3))=4.25 feet (approx.)
Check that it is a minimum.
I"(d)=(6kS)/d^4+(24kS)/(11-d)^4
I"(4.25)=0.02994>0 therefore minimum