33. Geometric Optics

Thin Lens And Lens Maker Equations

# Image Formation by a Biconcave Lens

Patrick Ford

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Hey, guys, let's do an example. Ah, by Con Cave lens has two different radio of curvature. If the radius of curvature of one piece of glass with the refractive index of 1.52 is four centimeters and the radius of curvature of the other pieces seven centimeters, what is a focal length of the lens? If in objects placed five centimeters from the lens, where will the image be formed? And is this image real or virtual? And finally, if the option is one centimeter tall, what's the height of the image? Okay, so let's start all the way at the beginning. What's the focal length of this lens Now? The lens maker equation tells us as we know that it doesn't matter the orientation of this lens. We're gonna get the same focal length, so I'm just gonna choose an orientation so we can assign a near radius in a far radius. So I'll say the near radius is four centimeters and the far radius of seven centimeters. Now the lens maker equation tells us that one over f is in minus one times one over R one minus one over R two. Okay, the index of refraction is 152 The near radius is four centimeters, but the center of curvature appears in front of the lens. So by convention, it's negative. The far radius is seven centimeters, and the center of curvature appears in Sorry behind the lens. So by convention, that's positive. Plugging this into a calculator, we get negative 0.393 But that isn't our answer. We have to reciprocate this because this is one over F. So the focal length is going to be negative. 25 centimeters. Okay, One answer done. Now, if we place an object five centimeters from the lens once again, it doesn't matter the orientation of the lens because it's the same focal length on but either side. If we place it five centimeters from the lens, where will the image reformed? Now we need to use the thin lens equation. That one over s O plus one over s. I equals one over F one over s. I is therefore gonna be won over f Sorry. Minus one over S O, which is one over negative to five minus 1/5, which is negative. 06 Okay, so if I were sip a Kate. This answer, because once again, negative six is not the answer. It's the reciprocal. Then I get an image distance of negative. 1.7 centimeters. Okay, two answers down. Two more to go. Is this a real image or a virtual image? You guys should know this instantly by now. This is a virtual image. Okay? Why? Because the image distance is negative. And finally, we want to know if the object is one centimeter tall. What is the height of the image? So for that, we need to use the magnification s I over eso once again not gonna mess with the negative sign because we know that since this is a virtual image, it's gonna be upright. That negative sign will just tell us whether it's upright or inverted. And we don't need that information. So this is 1.7 centimeters over the object distance, which was five centimeters and thats So the height of the image is the magnification 0.34 times the height of the object, which is one centimeter. So the height of our images 0.34 centimeters. All right, so we know our focal length of this lens. Negative 2.5 centimeters. Image distance negative. 1.7 centimeters Which means it has to be a virtual image. The question wasn't asked, but this image is therefore upright. The magnification 0.34 which means that the images roughly one third the height of the object or 0. centimeters. Alright, guys, that wraps up this problem. Thanks for watching.

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