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# Multiple lens systems

## Video transcript

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 this is this we've got two lenses this first lens is going to create an image of this object over here so it creates an image of this object then the second 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 the second lens doesn't exist we're going to figure out what image does this 1st lens create then we'll pretend like the first lens doesn't exist and we'll treat that image that the first lens creates as if it's the object for the second lens we'll do another calculation figure out where this second lens creates an image of the thing it thinks is the object but it's actually the image from the first one so we're going to figure out first the object get turned into an image from the first lens then the second 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 what are I 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 so I'm going to pretend like the first the second lens doesn't exist so it doesn't confuse me we use the thin lens equation all right here we go 1 over F so 1 over okay here we go 12 centimetres do I make it positive or negative it's on the left maybe I make it negative note 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 all right object distance so one 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 I 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've given you got to plug in that distance and that's going to be 24 plus 12 it's actually 36 so I have 1 over 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 is 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 one 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're going to get 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 be 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 points only 12 so I'm going to be somewhere around here I'm going to label that that's going to be 18 centimeters and that's where the first image is so that's where the image is going to be that this first lens creates 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 second lens back we know the image that the first lens created is right here this second 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 as 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 exactly 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 the it's object is and it's object is over here which is not 18 that's why I gave you this distance between lenses the distance between lenses is 33 centimeters so if the distance from this first 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 quote unquote for the second lens is going to be so this 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 is 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 first lens would have created an image of this object way over here on the wrong side of the lens in other words if it if this 1st lens created an image that was closer to our eye then the second lens was let's just all messed up that's going to be a negative object distance this objects you're not supposed to hold an object between your eye and the lens so we'd count that you do your calculation just the same I mean everything would work out right it's not like the whole thing is 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 if you'd have to treat it as a negative object distance that didn't happen here our image is still farmed on this side we're going to pretend say we're in 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 I looking through the lens outer objects 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 all right focal length what do we look at for focal length we look at what type of lens this 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 second lens equals 1 over our object distance we just found it's not really an object 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 all right 1 over negative 10 centimeters minus 1 over 15 centimeters equals 1 over D I you solve that on the left hand side you flip it over you're going to end up getting that di is once you do that inversion negative 6 centimeters that's what our second image is going to be so what is this mean negative 6 centimeters this is always from the lens and negative number negative image distance means on the left hand side or the opposite side is 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 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 be and just to recap what happened whoops so this 1st lens created an image of the object over at this point here this 15 centimeter point and then the second 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 D oh all right 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 positive 18 up here imma leave off the unit's 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 one half negative one half what does that mean that means this first lens the image it created is going to be an image that's one-half 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 what I labeled the image distance going to be upside down and half as big as the object so if the object was that big my image 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 the second 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 second lens was it was negative six remember right here that's what we solved the second lens created an image distance negative six I'll leave off the units because they cancel and my object for the second lens was 15 centimeters that was what this was right here the distance from the lens to where its object quote unquote object was and that was 15 positive 15 and what do I get if you take negative of negative 6 over 15 you end up getting positive 2/5 so this is the second magnification the magnification of the first lens was negative 1/2 magnification of the second 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 first image so now this is a magnification of the image because that's what the second 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 the second 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 this second 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'd 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 well we had two magnifications here is there some way to figure out the total magnification some slick way of doing it 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 the object's height by in other words what factor could I multiply by to get the both orientation and height of this final image well you just take the magnification of the first lens times the magnification the second lens if I had more I 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 can be handy if you're 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