(III) A bright object is placed on one side of a converging le...

Question

(III) A bright object is placed on one side of a converging lens of focal length f, and a white screen for viewing the image is on the opposite side. The distance d_T = dᵢ + dₒ between the object and the screen is kept fixed, but the lens can be moved.

(a) Show that if d_T > 4ƒ, there will be two positions where the lens can be placed and a sharp image will be produced on the screen.

(b) If d_T < 4ƒ, show that there will be no lens position where a sharp image is formed.

(c) Determine a formula for the distance between the two lens positions in part (a), and the ratio of the image sizes.

Accepted Answer

Welcome back. Everyone. In this problem. A photographer uses a convex lens with a focal length. If the focus light from a distant object onto a camera sensor, the total distance between the object and the sensor is DT and the lens can be moved along the axis to adjust the focus. The distance DT remains fixed first. If DT equals six F, at what two distances from the object should the lens be placed to achieve a sharp focus on the sensor? And second, if DT equals three F explain why it is impossible to achieve a sharp focus on the sensor at any lens position. For our answer choices. A says the two distances should be three F plus or minus the square root of three multiplied by F and the total distance is too short for real image formation. If DT equals three FB says that it's three F plus R minus the square root of three multiplied by F. And the total distance is too long for a real image formation. C says it's five F plus or minus the square root of two multiplied by F. And the total distance is too short for real image formation. And the D says it's five F plus R minus the square root of two multiplied by F and the total distance is too long for real image formation. Now, what do we know based on the information that we're given? Well, the main thing that we can surmise from our problem statement is that the total distance between the object and the sensor is DT OK. Now this tells us then OK, that the total distance must be equal to the object distance plus the image distance. OK. So we know that the image distance is going to be equal to the total distance divided by the object distance. Now, why is this useful? Well, if we substitute this expression into our lens equation, then we should be able to use it to solve for our object distance and figure out the first part of our problem that is the distances at which the object sorry at what two distances from the object the lens should be placed to achieve a sharp focus. So let's use it in the lens equation. Recall that by the lens equation, the reciprocal of the focal length is going to be equal to the reciprocal of the object distance plus the reciprocal of the image distance. Using what we know this must mean that the reciprocal of a focal length equals one divided by D plus one divided by DT minus D. So we've expressed the image distance in terms of the object distance. Now, if we combine, if we simplify our fraction on the right hand side of the equation, OK. That must mean one divided by F is going to be equal to DT minus D plus D, OK. Divided by D multiplied by DT minus. Do. Now, we can cancel these for our object distances and numerator, they get DT divided by D multiplied by DT minus D. Remember we're solving for do so let's multiply, let's cross multiply sorry for our equation. So now this tells us when we do that, that F DT equals DDT. We expand our bracket here minus D sorry D squared. And now we're solving for do it's a squared term that means we'll need to solve quadratic. So making zero the subject of our equation, then we'll get D squared minus DTD plus FDT to be equal to zero. Now, this resembles a quadratic equation. OK. Where the coefficient of our squared term is one, the coefficient of our linear term is DT, OK. B equals DT or negative DT rather. OK. And the constant C is F DT. So we can solve quadratic, OK? Because when we do that solving quadratic, because we know the quadratic formula tells us that our, our unknown variable is going to be equal to the negative value of B plus or minus the square root of B squared minus four AC divided by two A. OK. And now when we substitute those known values of A B and C, this tells us that our object distance is going to be equal to the negative value of negative DT or positive DT plus R minus the square root of DT squared minus four FDT. OK. And all of that is going to be divided by two multiplied by one which equals two. So now we have an expression for our object distance. Now, in the first part of our problem, we wanted to find the two distances when DT equals six F six multiplied by the focal length. OK. So if we do that, let me move this up here just to make some space, right? I think I can carry it here. No, when we do that, let me just get rid of this as well. When let me put this in blue, when DT our total distance is six FK, then that means our object distance. OK. If we substitute that entire formula is going to be six F plus or minus the square root. OK? Of six F squared minus four F multiplied by six F and all of that is being divided by two. Now, if we simplify what's going on under our square root here, OK. Uh That's 36 F squared minus 24 F squared or 12 F squared. OK. So this is gonna be six F plus or minus the square root of 12 F squared divided by two. And now we know that the square root of F squared is F and we can rewrite root 12 as two root three. So that means our object distance is going to be six F plus or minus two root three multiplied by F all divided by two. If we divide through by +26 F divided by two is three F and two root three divided by two equals root three. Thus, when our total distance is six F, then the object distance is three F plus or minus the square root of three multiplied by F. In other words, the two possible values are three F plus root three F and three F minus root three F. Now let's move on to our second part of the problem that tells us to explain why we can't achieve a sharp focus on the sensor if our total distance equals three F, OK. Now, when the total distance is three F, if we're going to show it's impossible, then mathematically, it will also be impossible to solve our quadratic equation. In other words, we're going to get an error. So let's see if we can eventually get to that. If we substitute our total distance to be three F. When we do that, the object distance will be six, sorry, three F, OK, plus R minus the square root of three F, all squared minus four F multiplied by three F. OK? All divided by two. Now, under our square root here, um three F squared is going to be nine F squared uh four F multiplied by three F equals 12 F squared. OK. And now we're gonna have nine F squared minus 12 F squared, which under the square root is going to leave us with negative three F squared. OK. That's all divided by two. Now, we don't have to go any further notice that the expression inside the square root is negative. So there are no real solutions for our object distance that tells us that no position of the lens will produce a sharp image when the total distance is three F. Because the total distance is too short notice here, this was nine F squared. And if it was greater, the greater it gets like when it was at six F, then we were able to find a, a total distance here. That was long enough for an image formation. Those this tells us that if DT equals six F, then the two distances from the object should be three F plus root three F and three F minus root three F. And when the total distance is three F, then it is too short for a real image formation. If we look back at our answer choices that tells us that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Converging Lens and Focal Length

Lens Formula

Conditions for Image Formation

Watch next