Hey guys, let's do an example. A biconcave lens has two different radii of curvature. If the radius of curvature of one piece of glass with a refractive index of 1.52 is four centimeters and the radius of curvature of the other piece is seven centimeters. What is the focal length of the lens? If an object is 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? OK. 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 is seven centimeters. Now, the lens maker equation tells us that one over F is N minus one times one over R one minus one over R two. OK. The index of refraction is 1.52 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 a 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 2.5 centimeters. OK. 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 either side. If we place it five centimeters from the lens, where will the image be formed? Now we need to use the thin lens equation that one over. So plus one over SS I equals one over F one over si is therefore gonna be one over F sorry minus one over. So which is one over negative 2.5 minus 1/5, which is negative 0.6. OK. So if I reciprocate this answer because once again, negative 0.6 is not the answer, it's the reciprocal then I get an image distance of negative 1.7 centimeters. OK. 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. OK? 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 si over. So 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 that's 0.34. 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 image is 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 is 0.34 which means that the image is roughly one third the height of the object or 0.34 centimeters. All right guys, that wraps up this problem. Thanks for watching.