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- Thread starter MarkFL
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- Jan 26, 2012

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[JUSTIFY]I have to say Mark that the challenge is severely underdetermined. Assuming the most likely interpretation, my solution is below:

[JUSTIFY]Let the first light be located at position $x = 0$, and the second light at position $x = a$. Then, at any position $x$, the amount of light received from the first light is $I_1 x^{-2}$ and the amount of light received from the second light is $I_2 (a - x)^{-2}$. So the total light received is:

$$I(x) = I_1 x^{-2} + I_2 (a - x)^{-2}$$

Differentiating with respect to $x$ to find the minimum:

$$\frac{d}{dx} I(x) = -2 I_1 x^{-3} + 2 I_2 (a - x)^{-3}$$

And setting this to zero:

$$-2 I_1 x^{-3} + 2 I_2 (a - x)^{-3} = 0$$

$$I_2 (a - x)^{-3} - I_1 x^{-3} = 0$$

$$I_2 (a - x)^{-3} = I_1 x^{-3}$$

$$\left (\frac{a - x}{x} \right )^{-3} = \frac{I_1}{I_2}$$

$$\left (\frac{a - x}{x} \right )^3 = \frac{I_2}{I_1}$$

$$\frac{a - x}{x} = \sqrt[3]{ \frac{I_2}{I_1}}$$

$$\frac{a}{x} - 1 = \sqrt[3]{ \frac{I_2}{I_1}}$$

$$\frac{a}{x} = \sqrt[3]{ \frac{I_2}{I_1}} + 1$$

$$\frac{x}{a} = \frac{1}{\sqrt[3]{ \frac{I_2}{I_1}} + 1}$$

$$x = \frac{a}{\sqrt[3]{ \frac{I_2}{I_1}} + 1}$$

So this would be the position where the light received is minimized. Notice the solution is linear in $a$. Indeed, the problem is scale-invariant, since increasing the distance between the two lights does not change their intensity ratio.

Note the singularity at $I_1 = 0$. Let us depart from the abstract realm of mathematics for a moment and consider the physical implications of turning off the first light. Take a few minutes to convince yourself that the solution is going to be $x = 0$ ("as far as possible from the second light") and the issue is resolved.

Here is a plot of where you should be (in terms of $\frac{x}{a}$), against $\frac{I_2}{I_1}$ ratios (higher $x$-values mean the second light is brighter than the first):

Notice that $\frac{x}{a} = \frac{1}{2}$ when $I_1 = I_2$, as can be reasonably expected.

[/JUSTIFY]

Of course this analysis is not very rigorous, but then again, radiometry is not well-defined in one dimension (though I think it could easily be made to be) so I think this is what you meant.

[/JUSTIFY]

$$I(x) = I_1 x^{-2} + I_2 (a - x)^{-2}$$

Differentiating with respect to $x$ to find the minimum:

$$\frac{d}{dx} I(x) = -2 I_1 x^{-3} + 2 I_2 (a - x)^{-3}$$

And setting this to zero:

$$-2 I_1 x^{-3} + 2 I_2 (a - x)^{-3} = 0$$

$$I_2 (a - x)^{-3} - I_1 x^{-3} = 0$$

$$I_2 (a - x)^{-3} = I_1 x^{-3}$$

$$\left (\frac{a - x}{x} \right )^{-3} = \frac{I_1}{I_2}$$

$$\left (\frac{a - x}{x} \right )^3 = \frac{I_2}{I_1}$$

$$\frac{a - x}{x} = \sqrt[3]{ \frac{I_2}{I_1}}$$

$$\frac{a}{x} - 1 = \sqrt[3]{ \frac{I_2}{I_1}}$$

$$\frac{a}{x} = \sqrt[3]{ \frac{I_2}{I_1}} + 1$$

$$\frac{x}{a} = \frac{1}{\sqrt[3]{ \frac{I_2}{I_1}} + 1}$$

$$x = \frac{a}{\sqrt[3]{ \frac{I_2}{I_1}} + 1}$$

So this would be the position where the light received is minimized. Notice the solution is linear in $a$. Indeed, the problem is scale-invariant, since increasing the distance between the two lights does not change their intensity ratio.

Note the singularity at $I_1 = 0$. Let us depart from the abstract realm of mathematics for a moment and consider the physical implications of turning off the first light. Take a few minutes to convince yourself that the solution is going to be $x = 0$ ("as far as possible from the second light") and the issue is resolved.

Here is a plot of where you should be (in terms of $\frac{x}{a}$), against $\frac{I_2}{I_1}$ ratios (higher $x$-values mean the second light is brighter than the first):

Notice that $\frac{x}{a} = \frac{1}{2}$ when $I_1 = I_2$, as can be reasonably expected.

[/JUSTIFY]

Of course this analysis is not very rigorous, but then again, radiometry is not well-defined in one dimension (though I think it could easily be made to be) so I think this is what you meant.

[/JUSTIFY]

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