The intensity of illumination at any point from a light source...

Question

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least?

Accepted Answer

Welcome back, everyone. In a photography studio, 2 flash units are used to illuminate a subject between them. One flash unit is 8 times as powerful as the other, and they are positioned 3 m apart. The intensity of illumination is inversely proportional to the square of the distance between a light source and a point. To achieve the best lighting and to avoid overexposure, the photographer wants to minimize the total illumination. How far from the more powerful flash should the subject be positioned? A 1.75 m, B 2 m, C 2.25 m, and D 2.75 m. Now, let's try to draw this scenario to better understand what's going on here. So we're saying that we have 2 flash units in a photography studio, OK? Let's say this is the 1st 1 and this is the 2nd 1. I know, we also know that these, that both of these units are positioned 3 m apart, OK? So the distance between them is 3 m. Next, we know that there is a, we, we have an, a, a subject that we want to put in between both of these flash units. OK. So, let me just make some space here and put our subject. And we can say then that from our subject to one of the lights, in this case from the more powerful flash unit to the subject, let's say the distance is X. Which means then that the distance from the unit to the less powerful flash is going to be 3 minus X, OK? I know we can score I want the intensity of the less powerful flash unit, OK. And I too, as the more powerful flash unit where. I It is equal to 811. It's 8 times as powerful as the less unit. What else do we know? Well, the problem tells us that the intensity of illumination is inversely proportional to the square of the distance between a light source and a point. What does that mean? Well, that means then that I1 will be equal to K divided by its distance from the from the object 3 minus X squarek where K is a proportionality constant and I2 will be equal to 8 multiplied by that. So that's gonna be 8K divided by X quad. The distance from the more powerful flash to our object, OK. And thus this tells us then that the total illumination I total is going to be equal to 11 plus 12, so that's K divided by 3 minus X2 + 8k divided by X2. OK. And now for simplicity, let's let the proportionality constant be 1. So if K equals 1, then we can rewrite our equation. As 1 divided by 3 minus X squared. Plus 8 divided by X2. OK. That is if K equals 1. So now, we have an equation for our total intensity. And using this information, we want to figure out how far from the more powerful flash should the subject be positioned. Now to find that, essentially we need to find the value of X that minimizes our intensity. OK? We wanna, we want like our problem says, we would like to minimize the total illumination under this condition. So now, to do this, to find the value of X that minimizes the total intensity, we can find the derivative of our total intensity with respect to 0, with respect to X sorry, set it to 0, solve for X, and then confirm if this X value is a minimum or a maximum. So let's go ahead and do that, OK? So let's go ahead and basically minimize our total intensity. Now, first, to do that, let me put that in red actually. To find the derivative of our total intensity with respect to X, it means that we are going to differentiate with respect to X 1 divided by 3 minus X 2 or 3 minus X raises to the power of -2, OK. And We're going to differentiate with respect to X8 divided by X2, or 8 X to the negative 2, OK? Now by applying the chain and power rule respectively, then we are going to get the derivative of the first term to be negative 2 multiplied by 3 minus X-rays to the part of -3 multiplied by -1 plus the derivative of for a second term 8 multiplied by -2 multiplied by X to the power of -3. And we could rewrite this then as 2 or 2 multiplied by -1 is 2, and all of that will be divided by 3 minus X cube. Minus, OK, it multiplied by -2, gives us -16, so that's m16 divided by XQ. And now remember. To find our critical point and figure out if it's a minimum, we are going to set this to 0, OK? So now at our first derivative for intensity equal to 0. OK, then we're gonna get this expression. I'll just copy and paste it here. We're gonna get this expression to be equal to 0. Now note that the derivative will be undefined if X equals 0, but we need to discard this as the critical point because the subject and light source should not coincide because remember X helps us to figure out our distance and it would make sense for the both of them to be at the same place. We know that they are at some distance apart. So now, ignoring when X equals 0, then since we've set this to 0. Or derivative to 0, that means we could add 16 divided by X cubed to both sides and cross multiply to get 2 X cubed equal to 16 multiplied by 3 minus X cubed. Now if we divide both sides by 2. Then we'll get executed. To be equal to 8 multiplied by 3 minus X cubed, which means then. That we could rewrite this as X cubed equals 2 multiplied by 3 minus X. OK. I cured And if we find a cube root of both sides, then we're going to get X to be equal to 2 multiplied by 3 minus X. That equals 6 minus 2 X. And if X equals 6 minus 2 X, then 3 X must be 6, which means that X equals 6 divided by 3 or 2. So, X equals 2. Is going to be a critical point of our total illumination. But now we need to figure out if this value of X would give the least illumination, and we can do that by performing the first derivative test, OK? Now how can we do that? Well, let's let's make some space here, OK. To perform our test, right? And now, let me just bring this over a little bit. No, I put this in black here. By the first derivative test, basically what we want to see is that if we look at either sides of the first derivative for that critical point, if. It changes from a negative to positive as it increases to the point. The critical point would be a minimum, but if not, then it would be a maximum. So now, let's take by the first derivative test 2 points on either side of X equals 2 and look at what their values, their signs would be. To the left, OK, let's take X to be equal to 1.9. In that case, then our first derivative is going to be equal to 2 divided by 3 minus 1.9 cubed. OK, minus 16 divided by 1.9 cubed, and that would be approximately equal to -0.83. On the other hand, if we take a value on the right, say when X equals 2.1, then we'll get our total illumination or derivative, sorry, of our total illumination with respect to X to be approximately equal to 1.02. So now, since the first derivative changes from negative to positive as X increases through X equals 2, this critical point would indeed be a minimum. So we can say thus X equals 2 is a minimum. It minimizes our total illumination and in that case. If we put it back in context, that means to minimize our illumination, the subject should be positioned 2 m from the more powerful flash unit, OK? So we can say well. Let me highlight it here. It should be 2 m from the more powerful flash unit because when we go back to our answer choices we can see that X represented the distance from the more powerful unit to the object 2 m from the more powerful unit. Thanks for watching everyone. I hope this video helped.

Key Concepts

Inverse Square Law

Optimization

Superposition of Light Intensities

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