Welcome to our community

Be a part of something great, join today!

The light hurts my eyes

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose two bright lights are separated by a distance $a$, with one light having intensity $I_1$ and the other having intensity $I_2$. I am situated between the lights, trapped in a narrow passageway, and the light is blinding me. (Sweating) Please find the point at which the light I receive is a minimum. (Whew)
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
[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]
 
Last edited:
  • Thread starter
  • Admin
  • #3

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, it is intended to be in one dimension, where the inverse square law of radiation is used to determine the sum of the intensities. Your answer is correct! (Sun)