 Let's solve a couple of questions on image formed by multiple lenses. For the first one we have two convex lenses L1 and L2 having focal lens F1 which is plus 7.5 and F2 plus 10 and they are placed along a common axis. These lenses they are separated by a distance D over here that is 40 cm. And it also says that an object O is kept 30 cm in front of L1. The question is to figure out how far from L2 will the final image be formed. And there is a note which says to use the proper sign convention while reporting the answer. Alright as always why not you pause the video first and give this a try. Alright hopefully you have given this a shot. Now we see convex lenses L1 and L2. So first let us look at the lens equation there it is. This will be used throughout the question. Now the question is asking us to figure out how far from L2 from this lens will the image be formed. So it is interested in figuring out the image distance from L2. And for L2 we know the focal length. The question is asking us to figure out the image distance. So we could do that if we knew the object distance. So then using the lens formula we can plug in the values and find this V. But we don't know the object distance for L2. We know that for L1. We also know the focal length for L1. So using the lens formula we can figure out the image distance for L1. We can figure out how far from L1 will the image be formed. Then that image the image formed by L1 that image can act as an object. It can act as an object for the second lens. And turns out for multiple lenses that is a common strategy that we use. The image formed by the first lens that acts as an object for the second lens. Let's write that down. Image formed by first lens acts as an object for the second lens. So we will need to use this relation toys. First for figuring out the image that is formed by the first lens. The distance of the image that is formed by the first lens. And then use it again to find the image distance for the second lens. So let's use it once. We know there is an object in front of L1. So this distance would be U1. And there will be rays coming out from this object which will pass through the convex lens. They will undergo refraction and because it's a converging lens they will converge to a point. And that is where the image is formed and this distance is V1. Now we know what U1 is that is the object distance. We know what the focal length is. We do not know what the image distance is. So let's use this relation to figure out the image distance. So this becomes 1 upon V1 minus 1 upon U1 that is equal to 1 upon F1. If you put in the values this becomes 1 upon V1 that is equal to 1 upon 1 upon 30. Because U1 is 30 plus 1 upon 7.5. Now we need to think about the sign convention as well. We need to think about the sign convention. Using the sign convention in the numericals it involves treating the optical center of the lens as the origin. So for this one we can take that this is the origin. Everything on the left hand side of this origin could be negative and on the right hand side could be positive. So U1 here would be minus 30. This would be minus 30. Now we can work out this. So 1 by V1 this is equal to we can take an LCM of these two. We can get 30 itself and in the numerator we can get minus 1 that's just this one plus 7.5 into 430 so plus 4 and this is 3 by 30 and this is 1 by 10. So when you invert it finally you get V1 as plus 10 centimeters. The sign convention fits what would really happen. It's plus 10 and everything to the right hand side of this lens should be plus and it's forming at a distance of 10 centimeters from L1. So this V1 is 10. Now the image created by L1, the image created by L1 it acts as an object for L2. So this distance right here that is U2. This is U2 this is the object distance for the second lens. And after going through refraction it meets at the optical axis and an image is formed at a distance of V2. So now we can use the lens formula again, the thin lens formula again to figure out this distance V2 and that is what the question is interested in. So when we do that this is 1 upon V2 minus 1 upon U2 that is equal to 1 upon F2. Now here we know what F2 is that is plus 10. U2 would be this total distance D minus V1. So U2 is D minus V1 and that is we know what D is that is 40. 40 minus 10 this is 30 centimeters and when we are placing it in a thin lens formula we do need to follow the sign convention for the lens L2 this point can be treated as origin everything on the left hand side would be negative on the right would be positive. So 1 upon V2 this becomes equal to 1 upon minus 30 plus 1 upon 10. The LCM could be 30 and this is minus 1 plus 3. So this is 2 divided by 30 that would be 1 by 15 and V2 comes out to be equal to 15 centimeters. So the answer is plus 15. All right let's move on to the next question. The second question says that a person who cannot see far away objects clearly is said to be myopic. We can correct this condition by using a concave lens in front of a convex lens. So again we have multiple lenses now we have concave in front of convex lens of the eye. This combination can focus the image on the retina. A concave lens of focal length minus 4 is being used in front of an eye lens of focal length plus 2.1 and the final image is formed on the retina. The distance between the two lenses concave and convex is 0.5 and the question is to figure out the distance of the retina from the eye lens. One important note it says assume that the object is far away so the incoming rays are parallel and we need to report our answers correct to two decimal places. Again pause the video maybe try to draw the entire setup over here and attempt this one on your own first. All right hopefully you have given this a shot. Now for multiple lenses I always find it helpful to draw what the question is describing or or stating. So here we here we have a convex lens a concave lens and also there is a retina where the image is formed. So I would start with drawing where the retina is we don't need to really draw what exactly retina looks like but maybe the retina is at this place over here and there is a convex lens in front of it so there could be a convex lens here and there is a concave in front of the convex so concave lens could be over here and the question is asking us to figure out the distance of the retina from the eye lens so let's let's label that distance that distance would be a distance would be this right here this distance D right here that is what the question is asking us to figure out we know what the focal lengths are and the question also tells us the distance between the concave and the eye lens so let's let's write that also this distance is 0.5 centimeters okay one more thing that a question is saying is assume that the object is far away so incoming rays are parallel okay so the rays that are falling on this concave lens they are parallel first let's draw an optical axis for this lens and the rays are parallel like this that is all that the question is telling us along with the focal lens also now this is again a question of image formed by multiple lenses we will be using the thin lens formula so here it is and even in the previous case we devised a sort of a common strategy for approaching these kind of problems for that obviously image that is formed by the first lens that image itself acts as an object for the second lens and then the final image is formed so in this one there are rays that are parallel and they are incident on the concave lens after undergoing refraction they will actually diverge this is a diverging lens and the rays could look somewhat like this they are diverging if we extend them we can have a virtual image being formed at this point so we can we can write this point as the image formed by the first lens and this image itself this image itself acts as an object for the second lens so before we go over there first let's try to find this distance right here this is this is v1 if we are able to figure out v1 we then know that this image formed due to the first lens will act as an object for the second lens we would then get some idea on u2 the object distance and we already know the focal length maybe then we should be able to figure out what the image distance should be so first if we use if we use the thin lens formula once we can write this as 1 upon v1 minus 1 upon u1 that is equal to 1 upon f1 so we need to figure out what v1 is 1 upon v1 this is equal to 1 upon u1 now u1 here is infinity so let's write that infinity plus 1 upon f1 f1 is minus 4 1 upon infinity would be 0 so therefore v1 just becomes equal to minus 4 centimeters this distance right here is 4 and it came out to be minus 4 because if we think about the sign convention this point should be considered as the origin everything on the left hand side would be negative now this image this image acts as an object for the second lens so the object distance for the second lens that is u2 that would be v1 that is 4 plus 0.5 so u2 would be 4 plus 0.5 that is 4.5 centimeters and when we use the thin lens formula again to figure out the image distance that is formed at the retina so let's see how that looks like these light rays after they undergo refraction they converge and the image is formed at the retina this is a distance d that is basically this is really just v2 so let's write that this is 1 upon v2 minus 1 upon u2 now u2 is the distance is 4.5 but it is on the left hand side of the optical center of the lens so we will write we will write minus 4.5 and this is equal to 1 upon f2 that is 2.1 and you can then write this as 1 upon v2 equal to 1 upon 4.5 plus 1 upon 2.1 i encourage you to do that calculation and when you do the calculation the answer that you will get is 3.94 centimeters so this distance right here is 3.94 this is 3.94 centimeters all right you can try more questions from this exercise and if you're watching on youtube do check out the exercise link which is added in the description