![]() It's just when you do your calculation for the second case. So we'd count that, you'd do your calculation just the same. Your not supposed to hold an object between your eye and the lens. In other words, if thisġst lens created an image that was closer to our eye than the 2nd lens was, well 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 That's where the objectsĪre supposed to be. Why positive? Well it's still on this left hand side opposite side as your eye. So as we turn this positive 18 into object distance of 15 centimeters will be where the "object" for this 2nd lens is going to be. So 33 minus 18 is going to be 15, whoops. This 1st lens is 18 well the difference has toīe equal to this length. That's why it gave you thisĭistance between lenses. 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. 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. 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. It's going to createĪn image of that image as if that were an object. This 2nd lens is going to think that that's the object. We know the image that the 1st lens created is right here. We pretend, we'll bring the 2nd lens back. Okay so now what do I do? I told you, here's what we do. I'm going to label thatĪnd 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. 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. So I'm going to haveĪn image that's formed over on this side of the lens and 18 means 18 centimeters from where? From the center of the lens. Remember, positive means on the same side as your eye for a lens. And so what does that mean? Where is this? Well, alright, di, 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 thatĭi equals 18 centimeters positive 18 centimeters. If you solve this you going to go 1 over 12 centimeters I'll show you in a minute how this could possibly be negative in a second. Everything's cool, it's on the opposite side as my eye. We've got two lensesīut for this first case it's on the left hand side. ![]() Positive or negative? Well that's going to be positive here. So I'd have 1 over and this is going to be 36 centimeters. You've got to plug in that distance and that's going to be 24 plus 12. 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 soįor an object distance I'd have to go from the center of the lens all the way to the object. So that's what I plug in, Positive 12 centimeters equals, alright, object distance so 1 Lens my focal length is going to be positive 12 centimeters. The only thing you look at is what kind of lens this is. It positive or negative? It's on the left. 1 over f, so 1 over, okay here we go, 12 centimeters, do I make 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. 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. 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 theĢnd lens turns that image into another image and that's what our eye is going to see. 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 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're first going to pretend like this 2nd lens doesn't exist. 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 This 1st lens is going to create an image of this object over here. The overall idea of how we're going to approach is this. Of a two lens system and overall, before we This looks really intimidating but it's actually not that bad. If I had been handed this problem on a physics test I probably would have freaked out.
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