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Problem 82 Hard Difficulty

Two light sources of identical strength are placed $ 10 $ m apart. An object is to be placed at a point $ P $ on a line $ \ell $, parallel to the line joining the light sources and at a distance $ d $ meters from it (see the figure). We want to locate $ P $ on $ \ell $ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source.
(a) Find an expression for the intensity $ I(x) $ at the point $ P $.
(b) If $ d = 5 $ m, use graphs of $ I(x) $ and $ I'(x) $ to show that the intensity is minimized when $ x = 5 $ m, that is, when $ P $ is at the midpoint of $ \ell $.
(c) If $ d = 10 $ m, show that the intensity (perhaps surprisingly) is not minimized at the midpoint.
(d) Somewhere between $ d = 5 $ m and $ d = 10 $ m there is a transitional value of $ d $ at which the point of minimal illumination abruptly changes. Estimate this value of $ d $ by graphical methods. Then find the exact value of $ d $.

Answer

a) $I(x)=\frac{k}{x^{2}+d^{2}}+\frac{k}{(10-x)^{2}+d^{2}}=\frac{k}{x^{2}+d^{2}}+\frac{k}{x^{2}-20 x+100+d^{2}}$
b) $I(x)$ has a minimum at $x=5 \mathrm{m}$
c) $I(x)$ has is minimized at $x=0$ and $x=10 \mathrm{m} .$ And the midpoint is the most brightly lit point.
d) 5$\sqrt{2}$ which is approximately 7.07

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Video Transcript

to begin this problem, as you can see in the corner, I kind of drew what this problem looks like. Um Just so you can visually understand it better and it makes more sense to follow. So right here are the two bulbs that are mentioned. I label them both one and bulb to which you'll see in a second why I did that. Um and then I should right here, the distance between them was told to be 10 m. We also were told there was a line that was parallel to this 10 m line. And that is called line L. And Lionel has a midpoint that's marked as pii. Then X. Is going to be representing that um P. Point all the way to the left side of line L. And this represents the distance in between the two lines. So now that's kind of easier to understand and see. We're gonna start using that information. So we're gonna start with a. So A is asking for an expression for the intensity of I. In terms of X. At point P. So this is something where we were basically gonna work With Bulb one and build two separately. Find an expression for each and then combine them. So we're gonna first start what the illumination do. That's right illumination Due to Bulb one. Okay, so what this is gonna look like um is we need to remember that the intensity of illumination for a single source is directly proportionate to the strength of the source and inversely inversely proportional to the square of the distance that source. So it's kind of wordy. But if you keep that in mind, you'll understand um what this is going to me is gonna put K over X squared plus D squared. That's exactly were just talking about um and we're told to assume that K. Is one, so everything else is the same. And that's what it looks like. Then we need the illumination due to bulb two and that's gonna look also very similar to me okay, over the X squared plus B squared. And that's actually going to look a little bit different. So the case still assumed to be one. But now what's different about this is that we need to um express a distance to be 10 m away from Bulb one. So we're gonna do 10 -X. So this is going to show the difference in the distances but they're not just right next to each other. So that's very key to remember. So this is going to be very crucial and necessary in solving this. And so is this one. So like I said before, we're still not there yet to find the expression because what we have to do is we have to combine these, we have to add these two up and what that's gonna look like we're gonna, right I have X. That's the expression we're looking for Is going to equal one over um X squared plus D squared plus one. Over D squared plus 10 minus x squared and right there is your expression for part A. And now we can start putting this to use so that's over here is gonna be B. So B. Is asking um if D equals five m we're gonna use a graph to to show the intensity of I. In terms of X. And also the inverse of that. Um So we put this in another colour that some yellow. So we're gonna show the graph right here. Mhm. Um label this we can put 0.6, 0.4 and 0.2 and 123456 A little bit longer Which is Marcus is five and 1. So in order to to graph this we first need to make sure we know so D. We know is five. Right? So D. Equals five. So when you grasp this where first can be graphing by of X. And the universe which looks like that little mark. So you can you take your time to kind of graphic. Mine is gonna be a little bit more of a sloppy version but it should look something like this where it goes ah And a little down a little bit of curvature to it and same thing here. Um This would be I. Of X. Which is this equation over to the right and then the inverse of I have extra looked like this. Sure, just to go a little bit down and maybe do that. Um Try second. So it's not like this, it's gonna go down and then it's gonna come right back up and you're going to actually see it Make contact on the axis, right by five x equals five. That makes contact. So what that's called is a critical point. Right here, this is a critical point. So we're also right here, this is the inverse of Iraq's and that's written, I'm not sure if it's really necessary in this problem, but I'll write it just in case and B D over dx times I of X at the inverse equation. So because the critical point is that X equals five, that's how we know that. The answer for this is X equals five. That is where um if d goes five m, that it's minimized when X equals five m. Um because this is a minimum point. Um and the X value to the minimum intensity would be somewhere between them. So keep that in mind and then we can go on to see so let's see is going to be doing Is doing the same thing but at D equals 10. So we're gonna draw another graph for that. And this graph We can put .015 and this is only a rough draft kind of graph. So Gonna be a little bit sloppy .005 And we need to mark the 10 point because we're looking at vehicles 10 and what this graph is gonna look like is something like this, it's going to go like this, this is going to be a nice curve And this minimum point is going to be at x equals 10 And this minimum points to be an X equals zero. And that is key to realize because that answers the question right there. It's asking us um to show the intensity that is it's not minimized at the midpoint. And that's true because it's not the midpoint would be here and here. Is that the two ends which is X equals 10 and X equals zero. So this proves that it's not intensified at the midpoint. Your maximum the minimum um Is actually at the two ends. So That is two x. value to the minimum intensities X equals zero and X equals 10. The minimum is not at the midpoint. So that is what see what's really just looking for. And lastly, d um do you asking about The point between the Eagles five m and Eagles 10 m and at what point the minimum elimination abruptly changes. So how we're gonna do this is use those two equations that we first got in the beginning. So we're gonna plug in all of our exes that we found, I'm gonna plug in five for X. Yeah. And then we want to, you want to show the yeah, let's see it's asking for between Diego's five and 10. So this one this part's gonna be five. So the equation Up here tells us when you do 10 -X. In this case we know the X on the left side is gonna represent five and that's gonna be squared equals. Now we want to show it for 10. So the X. Is now 10 on this side, so 10 squared plus D squared before I continue this equation. Let me just Right that here that this side represents x equals five. The left side and the right side represents X Equal 10. Does that make sense when you look back at it? And then here one will be over D squared plus 10 minus 10 squared. I'm gonna run out of room before I can do this. But once you find the equation um and you solve for D. We should get this, it should be D equals mhm five and the Square Root of two. And this is your answer attribute meters and that is problems a through deep