- Convex lenses
- Convex lens examples
- Concave lenses
- Object image and focal distance relationship (proof of formula)
- Object image height and distance relationship
- Thin lens equation and problem solving
- Multiple lens systems
- Diopters, Aberration, and the Human Eye
Multiple lens systems
Some examples of using the thin lens equation with multiple lenses. Created by David SantoPietro.
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- can someone explain why (1/12) - (1/36) = (1/18), please? thanks.(0 votes)
- 1/12-1/36=(3-1)/36=2/36=1/18(0 votes)
- At :40sec, why do you decide to use the lens on the left as the "first lens" and not the second one on the right?(2 votes)
- Because a photon coming from the original image to eye will pass the lens on the left "first."(16 votes)
- Object distance has to be negative right it should be - 1/36 as per sign conventions(5 votes)
- No that's for the mirror formula... Since lens is made up of two spherical mirrors , both of its sides are positive..(2 votes)
- I understand that the image created is virtual, but can you explain why?
Is it because the image went to a concave lens and they always have a virtual image(3 votes)
- did you see a diagram of the light rays?
If so, you will see that the rays look as though they 'come from a point'. so the image LOOKS as though it should be somewhere, but really it is not. you can not shine the rays of light onto the wall and see an image on the wall. You only see an image when you eye focusses the light again to make it real (on the back of your eye)
Hope that helps(5 votes)
- So Im confused, In previous videos the other instructor (Sal I think) always drew the real image of a convex lens LARGER than the object itself. Sal used 2 light rays defracting through the lens that both went through the focal points on either side of the lens to accomplish this. He also drew a diagram in his video "object image height and distance relationship", where the image was larger. SO WHY IS THE IMAGE IN THIS VIDEO SMALLER THAN THE OBJECT FOR THE CONVEX LENS?!(2 votes)
- In Sal's video, the image of an object seen through a convex lens was larger when the object was placed a distance between f and 2f from the lens. When the object is placed at a point past 2f (i.e. 2f or greater), the inverted real image is smaller.
In the example used in this video, we see that f = 12cm, and the object is placed at 36cm (3f). So, the inverted real image is smaller.(5 votes)
- After getting -6 cm as answer, shouldn't he take 6 cm from right hand side of lens as he took +15 from left hand side of lens as object.(3 votes)
- No because for an image a negative means it is on the opposite side of the eye(2 votes)
- So to know how high the image is, the magnification isn't all that precise. So why dont you do the M = hi/ho which is equal to -di/do? Like wouldnt that be simpler to know the height of the object?(3 votes)
- I think in this video the magnification factor was used and not the height because we don't know the original height of the object (h0). If we knew that then yes, using that formula we'd get h1= h0 x (-d1/d0), where (-d1/d0) is the magnification factor.(2 votes)
- How did he get -6 as the final answer? I didn't quite catch what he meant by "flip it over"..(3 votes)
- The multiplicative inverse of (1/-6) is -6. So 1/(1/-6) = (-6). The "one over" corresponds to the narrators "flipping".(2 votes)
- If, say eye would have been on left side, object distance would be negative, right?(3 votes)
- So did the eye see the 2/5 image? If so, is it because it's closer?(2 votes)
- The eye sees an image -1/5 of the original object because the individual magnifications of the two lenses should be multiplied to get the total magnification. The -ve sign indicates that the image is inverted.(3 votes)
- [Voiceover] If I had been handed this problem on a physics test I probably would have freaked out. This looks really intimidating but it's actually not that bad. It's a classic example of a two lens system and overall, before we get lost in details here. The overall idea of how we're going to approach is this. We've got two lenses. This 1st lens is going to create an image of this object over here. It creates an image of this object. Then the 2nd lens is going to create an image of that image and so what we're basically going to do is we're going to use the thin lens formula. We're first going to pretend like this 2nd lens doesn't exist. We're going to figure out, what image does this 1st lens create? Then, we'll pretend like the 1st lens doesn't exist and we'll treat that image that the 1st lens creates as if it's the object for the 2nd lens. We'll do another calculation, figure out where this 2nd lens creates an image of the thing it thinks is the object but it's actually the image from the 1st one and so we're going to figure out first the object gets turned into an image from the 1st lens, then the 2nd lens turns that image into another image and that's what our eye is going to see. So that's the question we want to answer is what image would our eye see in this case if we had it over here looking through these two lenses? So let's do this. First thing we got to do is I'm going to pretend like the 2nd lens doesn't exist so it doesn't confuse me. We use the thin lens equation. Alright here we go. 1 over f, so 1 over, okay here we go, 12 centimeters, do I make it positive or negative? It's on the left. Maybe I make it negative, nope. The only thing you look at is what kind of lens this is. This is a convex lens. I can tell by the way it's drawn. Since this is a convex lens my focal length is going to be positive 12 centimeters. So that's what I plug in, Positive 12 centimeters equals, alright, object distance so 1 over the object distance that's 24 right? No, all distances are measured from the center of the lens to the thing you're looking at and so for an object distance I'd have to go from the center of the lens all the way to the object. Don't just plug in any number they're giving you. You've got to plug in that distance and that's going to be 24 plus 12. It's actually 36. So I'd have 1 over and this is going to be 36 centimeters. Positive or negative? Well that's going to be positive here. We've got two lenses but for this first case it's on the left hand side. Everything's cool, it's on the opposite side as my eye. It's going to be a positive 36. I'll show you in a minute how this could possibly be negative in a second. Plus 1 over the image distance. I'll just label that with a di. Alright so now we just got to solve. If you solve this you going to go 1 over 12 centimeters minus 1 over 36 centimeters equals 1 over di what you're going to end up getting is 1 over 18 centimeters on the left hand side equals 1 over di but that's 1 over di, if you solve that for di you'll get that di equals 18 centimeters positive 18 centimeters. And so what does that mean? Where is this? Well, alright, di, positive image distance. Remember, positive means on the same side as your eye for a lens. So I'm going to have an image that's formed over on this side of the lens and 18 means 18 centimeters from where? From the center of the lens. So if I draw this. I'm going to go 18 from here. That's going to be past the focal point because I know the focal point is only 12 so I'm going to be somewhere around here. I'm going to label that and that's going to be 18 centimeters and that's where the first images, so that's where the image is going to be that this 1st lens creates. It's going to be right there. Okay so now what do I do? I told you, here's what we do. We pretend, we'll bring the 2nd lens back. We know the image that the 1st lens created is right here. This 2nd lens is going to think that that's the object. It's going to create an image of that image as if that were an object. So we do another thin lens formula but this time we treat this positive 18 not at if it's the image, we treat it like it's the object. So this is going to be well, it's not going to be equal to do because remember, you have to measure everything from where? From the center of the lens so for this second calculation, my object distance is going to be from the center of the lens all the way to, let me not use that color. Let me see, let's pick this one here. From the center of the lens all the way to where it's object is and it's object is over here which is not 18. That's why it gave you this distance between lenses. The distance between lenses is 33 centimeters. So if the distance from this 1st lens is 18 well the difference has to be equal to this length. So 33 minus 18 is going to be 15, whoops. 15 centimeters will be where the "object" for this 2nd lens is going to be. So as we turn this positive 18 into object distance of positive 15 centimeters. Why positive? Well it's still on this left hand side opposite side as your eye. That's where the objects are supposed to be. Now here, this is where I told you there's one case where you can get negative object distances if our 1st lens would have created an image of this object way over here on the wrong side of the lens. In other words, if this 1st lens created an image that was closer to our eye than the 2nd lens was, well that's just all messed up. That's going to be a negative object distance. Your not supposed to hold an object between your eye and the lens. So we'd count that, you'd do your calculation just the same. I mean everything would work out right. It's not like the whole thing's going to break. You'd get an image. It's just when you do your calculation for the second case. If your image ended up on the wrong side of the lens for the first image that was created you'd have to treat it as a negative object distance. That didn't happen here. Our image is still formed on this side. We're going to pretend like this 1st lens didn't exist. We got an object over here, we're going to treat it like our object. Everything's fine. Eye looking through the lens at our object. So we still treat this as a positive 15. So let's do our calculation. We'll have for this lens it's going to be 1 over. Alright, focal length. What do we look at for focal length? We look at what type of lens it is. This lens is a diverging lens. Diverging lenses always get contributed, always have a negative focal length that they contribute into this equation. So a negative 10 centimeters. That's the focal length of the 2nd lens, equals 1 over our object distance we just found. It's not really an object but that's okay. It's an image but this lens doesn't know that. It thinks it's an object so 1 over positive 15 because it's still on the side that that object was supposed to be on, plus 1 over our image. This is going to be a second image. This is the second image that this lens system is going to create and so I'll just put di. Alright, so we do the math, alright 1 over negative 10 centimeters minus 1 over 15 centimeters equals 1 over di, you solve that on the left hand side. You flip it over, you're going end up getting the di is, once you do that inversion negative 6 centimeters. That's what our second image is going to be. So what does this mean? Negative 6 centimeters. This is always from the lens and negative. Remember negative distance means on the left hand side or the opposite side as your eye. So I'm going to come up here and I'm going to draw a line that goes from the center of my lens, 6 centimeters to the left somewhere around there. So that's going to be 6 centimeters right here. Whoops, I'll label that. That's going to be 6 centimeters because that's our image distance and that means our final image is right there. Our eye is going to see an image of something our arrow right at that spot right there that's where our image is going to be and just to recap what happened. So this 1st lens created an image of the object over at this point here, this 15 centimeter point and then the 2nd lens created an image of that image. It treated it like it was an object and it created an image at this white point right here that's where we're going to see the image but we don't know what it's going to look like. These calculations, this thin lens formula only shows us where the image is going to be. If we wanted to figure out how big it was we have to use the magnification formula so let's do that. Over here, magnification equals negative di. Remember it's negative di over do. Alright, so what was di for this first case? We're going to do it one step at a time. For this first case, the di was positive 18. So I'm going to put a positive 18 up here. I'm going to leave off the units because they're going to cancel. Do was not 24 it was 36. So what do I get? I get negative 18 over 36 that's 1/2, negative 1/2. What does that mean? That means this 1st lens, the image that it created is going to be an image that's 1/2 as big as my object and it's going to be negative magnification. That means it's upside down so I'm going to draw this first image. I'm going to do it in orange because that's what I labeled the image distance. It's going to be upside down and half as big as the object. So if the object was that big. My image is going to be upside down and about half as big. So this was my image one. I'll label it image one. That's what I'm treating as my second object. So that's image one that's object two essentially because that's what this 2nd lens is going to think is the object. What magnification does it provide? Well let's do that one. Magnification equals negative di over do so negative of my image distance for this 2nd lens was, it was negative 6 remember right here? That's what we solved. The 2nd lens created an image distance negative 6. I'll leave off the units because they cancel and my object distance for the 2nd lens was 15 centimeters. That's what this was right here. The distance from the lens to where it's "object" was and that was positive 15 and what do I get? If you take negative 6 over 15 you end up getting positive 2/5. So this is the second magnification. The magnification of the 1st lens was negative 1/2. The magnification of the 2nd lens was positive 2/5. That means at this point over here, I'll make it white because that's where I labeled this image distance. I'm going to get an image that's 2/5 as big as what? As the 1st image. Now this is a magnification of the image because that's what this 2nd lens is doing. So I don't look back at my object over here. I'm done with this guy for now. Now I'm just looking at what this 2nd lens thinks is the object. So 2/5 as big as this object and positive means right side up but in this case this really just means not inverted. So this started off our object for the 2nd lens started off upside down. This positive means that you maintain the direction, you don't invert it. So I leave this thing upside down. That's what it means I leave it upside down. If it would have started right side up I would have left it right side up but in this case, I leave it upside down and I make it 2/5 as big. 2/5 is a little under 1/2 and so I'm going to draw this about like that. That's where my image would be, right there. That's what I would see. I would see a very little image that's upside down and it would be right around there and so that's an example of where you can do a two lens system. You treat each lens separately and you use these formulas accordingly. The last thing I'll say is, if you wanted to know why we had two magnifications here, is there some way to figure out the total magnification? Some slick way of doing it? Yeah and it's really easy it turns out. You can just take your total magnification. If all I wanted to know was if I had this object what my total magnification was going to be. What could I multiply this object's height by, in other words, what factor could I multiply it by to get both the orientation and height of this final image? Well you just take the magnification of the 1st lens times the magnification of the 2nd lens. If I had more I'd just multiply them all. That would give me the factor. That's the overall magnification that I can multiply this object's height by to get the height of the final image and the orientation of it. It doesn't give you the position. This just gives you the overall magnification, That could be handy if your making a microscope or something but it doesn't tell you the position. For the position you've got to use these thin lens formulas. So that's how you treat a multiple lens problem.