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Anderson Video - Convering Lens Example

Professor Anderson
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And in this one, they say a small insect is placed 6.4 centimeters from a seven centimeter focal length lens. Calculate the position of the image. Follow the sign conventions. Okay, so let's see if we can draw the picture. And let's put down the givens. They tell us that d0 is 6.45 cm. They tell us F is seven centimeters and we need to find di. Okay, let's draw a picture of this thing. Here's our lens, they told us that it is a seven centimeter lens. And that is a positive number. So it's a converging lens. So let's put the focus right there, f on either side. Now, they tell us that the insect is placed 6.45 centimeters from the lens. And so, the insect is in fact closer to the lens than that focal spot F. So, we have to follow our rules again for determining where the image is. And then we'll use a thin lens equation to do the calculation. So, parallel rays go through the focus. That's rule number one. Rays through the center don't bend. That was rule number three. Do those two rays ever meet over here? Dave says no. Anybody else? They don't, right? They are diverging still, so they're never gonna meet over here. So there cannot be an image located on this side. Alright, so how do we determine the image? Well, you trace this one back, you trace this one back, and where they meet, that's where the image is located. What does this mean? It means if I stood here and looked at this thing, where would I perceive the image to be? I would perceive it to be right there. Now let's make sure that this jives with the thin lens equation. So if we go to the thin lens equation, we have the following: 1/ d 0 plus 1/ di equals 1/F. And we're looking for di, so we can solve this for di. We get 1/di equals 1/F minus 1/D0. And so we get di is equal to 1/1/ F minus 1/ D0 And now we can put in some numbers. So let's try it, if you have a calculator, pull out your calculator. And tell me what you get. We've got 1 over 7. We'll just keep everything in centimeters, that's fine. Minus 1/6.45. And if you try that in your calculator, tell me what you get. I'll see if I can verify it here. Anybody get an answer for that one? Okay, that's what I got: 82.1. Okay, -82.1 cm. So what does it mean to be negative? Remember, if the image is over here, then it is positive. But if the image is over here, then it's negative. So that's what the negative means. Negative means it should be over there. Now, it also says -82.1. Doesn't really look like it from our picture, but we got a little creative with these lines because I knew we're gonna run out of learning glass here. So we could believe that -82.1 would do it. So let's try that. Let's plug in -82.1 and see if we get the right answer. It says correct. Alright, good. Second part says: calculate the angular magnification. And remember what magnification is. M is equal to the height of the image divided by the height of the object. This is the height of the object, this is the height of the image. And you can do a little trig here to show that this is the same as the image distance over the object distance. Except there's a negative sign right there. Okay, and now we have all those numbers. Negative di means positive 82.1 cm. d0, they told us 6.45. So, if I divide 82.1 by 6.45, let's see what we get. Let's add one more parentheses, here, so we know we're doing it right. Okay, I got 12.72. Yeah, it's a negative here because I forgot to add the extra negative sign. So let's just verify. Yeah, okay. 12.72 for our magnification. We know it's got to be bigger, right? Clearly, it looks bigger. 12 sounds like a reasonable number.