You place a pencil in front of a concave lens with a focal length of − 20.0 c m -20.0\operatorname{\mathrm{cm}} . The magnification of the image is m = 1 2 m=\frac12 . If you wanted to make the image even smaller, should you move the pencil closer to the lens or farther away?

Welcome back everyone. So this is a little bit more of a conceptual question, but let's take a look. So we have a pencil that we're going to place in front of a concave lens, and it has a focal length of -20 centimeters. Remember, focal lengths are always negative for concave. Now the magnification of this image is M equals 1/2, and it's positive. Some of the words, the image is half as tall as the object and they're both the same orientation. So, if we wanted to make the image even smaller, so we wanted to shrink the image, should you move the pencil closer? Or farther away or neither. Those are our three answer choices. All right. So, it seems kind of conceptual, but we can't actually just draw a quick sketch of what's going on here. So, I've got my optical axis like this, and I've got my concave lens. So I'm gonna draw this in, uh, yeah, let's do red. I've got my concave lens, it's gonna be like this. So remember concave means it sort of caves in instead of bulges out, and I've got my object that's going to be right over here somewhere. It's gonna be my pencil, OK? So this is my pencil. So, if you draw your sort of sketch out your quick, your uh quick uh principal ray diagram, you're gonna have the Pray that goes like this, and then it sort of goes out, um, and it diverges, right? And let's just say that the focal length is uh out here somewhere, all right? What's gonna happen is it's gonna sort of go away from the nearest focal points, it's gonna go like that, uh, and then this far F is gonna be over here somewhere, and we have the F-ray that's gonna go towards the far, and then it doesn't quite make it because it straightens out, right? And again, if you back. trace those because these things diverge. If you trace those things backwards, you should get an image form forming where those two lines sort of meet like this, all right? In other words, we get an image that forms right about over here somewhere. The rays will intersect before actually it crosses through the lens, so they'll intersect on the same side, and it's above the axis. So in other words, we know that this image is going to be upright. And it seems. As if it's about, about, you know, like half as tall as the image itself. So, we, we draw our diagram, you know, roughly correct. All right? So, what the question says is, if we wanted to make this image now even smaller, do we move the pencil? Do we move it to the left so that it's further away from the lens, or do we keep moving it to the right? All right. Well, so let's, we could just try to draw another diagram for this, but let's just sort of look. At the equation for the magnification. We know that magnification is HI over HO, but it's also related to the object and image distances. So let's just not worry about the HI over HO because we don't know what the heights of these things are, but we do know that the magnification is 15, all right? So if we wanted to make this magnification even smaller, it means that we want that number to be smaller. So what that means here is that we want M to go down. There's no Two ways to make this happen. It's either the numerator of this equation goes down, the the DI, or the object distance, the denominator gets bigger, right? So those are the two ways to make this a smaller number, all right? What happens here is the DI basically, if I get the image to be closer, uh, to, to be a smaller number to the lens, then that is going to be, uh, resulting in a smaller magnification. But also, if the object distance, so this DO over here, gets bigger, then that's gonna really, really push the magnification down if the denominator gets a lot a lot a lot larger. So, it seems as if we want the object to go back this way. And so it seems like our answer choice is that we want it to be farther. But what we could do is we could just try to draw another sketch of the object really, really far away to try to sort of confirm this. So, I'm gonna try to draw another sort of sketch like this. I'm gonna try to keep most of the proportions the same. So, that's one way you could have done this. So, I'm gonna draw another concave lens. And I'm gonna draw the focus points roughly at the same exact position, right? Just so I can kind of compare these two things, just so I can kind of see where the rays end up, right? So now I want to have the pencil that's like all the way out here somewhere, and I'm gonna try, try to draw it roughly the same size. So, again, same exact thing. Draw the Pray, it's gonna hit the lens and it's gonna go out in this direction away from F. And then we have the F, uh, the F-ray is gonna go towards this F, but then it's gonna straighten out and it's gonna go parallel like that, all right? So now what happens is if you trace these things backwards. Race these things back, we should see that the image will intersect right over here, all right? So now it's gonna intersect right over here and it's gonna be really, really, really small. Uh, it looks like it's gonna be a slightly smaller and also closer to the focal length, uh, than it was up here, right? So, in other words, the farther you push away this object, the closer this image gets towards the focal point, and it's also gonna get even smaller and smaller, all right? So what happens here is that the DI is getting smaller, but the DO is getting much larger, much faster, and the magnification gets smaller. All right, so again, so we can kind of just confirm this by using our magnification equation, uh, but it is a good idea to kind of sketch it out as best as possible to try to get an idea of what's going on here. But the answer choice is gonna be that you would want to move the pencil farther. Away. That's going to push the magnification way, way down. All right. Eventually what happens is that when this object gets out to infinity, the, uh, or like infinitely super far away, then the image of the pencil will be tiny, tiny, tiny. It would also be basically right at the focal points of that concave lens, all right, at the near the closed F. All right, that's it for this one. Let me know if you have any questions and I'll see you in the next one.

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Physics with Calculus

33. Geometric Optics

Thin Lenses

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You place a pencil in front of a concave lens with a focal length of $negative 20.0$ . The magnification of the image is $m = \frac{1}{2}$ . If you wanted to make the image even smaller, should you move the pencil closer to the lens or farther away?

Closer

Farther

Neither

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Recall the lens formula relating object distance (d_o), image distance (d_i), and focal length (f): $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$.

Recall the magnification formula for lenses: $m = \frac{d_i}{d_o}$, where magnification (m) is the ratio of image height to object height.

Since the lens is concave, the focal length $f$ is negative, and the image formed is virtual, upright, and reduced in size, so $m$ is positive but less than 1.

To make the image smaller (reduce magnification), analyze how changing the object distance $d_o$ affects $m = \frac{d_i}{d_o}$ using the lens formula to express $d_i$ in terms of $d_o$ and $f$.

By increasing the object distance (moving the pencil farther from the lens), the image distance $d_i$ changes such that the magnification $m$ decreases, making the image smaller.

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Padroneggia Ray Diagrams for Convex Lenses con una breve spiegazione video di Patrick Ford

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An object is placed $36 c m$ to the left of a convex lens, an image is formed $20 c m$ to the right of the lens. What is the focal length of this lens?