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Current time:0:00Total duration:12:29

Object image and focal distance relationship (proof of formula)

Video transcript

we've been doing a bunch of these videos with these convex lenses where we drew parallel rays and rays that go through the focal point to figure out what the image of an object might be but what I want to do in this video is actually come up with an algebraic relationship between between the distance of the object from the convex lens the distance of the image from from the convex lens usually on the other side and the focal length and the focal length so let's see if we can do this and just to save you the pain of having to watch me draw straight lines I drew this ahead of time and so we can imagine this green thing right here is the object this is the object and these two little pink points right here are the focal points they're a focal length away and I did what we always drew I drew one parallel ray from the tip of that arrow to the actual convex lens and then it gets refracted so it goes through the focal point on the right-hand side and goes all the way over there and then I drew a ray that goes through the focal point on the left-hand side and then when it gets refracted it becomes parallel and then it actually intersects with that previous that previous ray up here right over here and so this gives us this gives us a sense of what the image what the image will look like it's inverted it's real in this case it is larger than the actual object but what I want to do is come up with a relationship with these values so let's see if we can label them here and then just do a little bit of geometry and a little bit algebra to figure out if there is an algebraic relationship right here so the first the first number the distance of the object that's this distance from here to here or we could just label it here since this is already drawn for us this is the distance of the object this is the way we drew it this was the this was the parallel light ray but before it got refracted it traveled the distance of the from the object to the actual lens now the distance of from the image to the lens that's this right over here this is how far this parallel light ray this parallel light ray had to travel so this is the distance from the image to the lens and then we have the focal distance the focal length and that's just this distance right here this right here is our focal length or we could view it on this side as well this right here is also is also our focal length so I want to come up with some relationship and to do that I'm going to draw some triangles here so what we can do is and the whole strategy I'm going to keep looking for similar triangles and then trying to find if I can find try to see if I can find relationships or ratios that relate these three things to each other so let me just find some similar triangles so the best thing I can think of to do is let me redraw this triangle over here let me just flip it over let me just draw the same triangle on the right-hand side of this diagram so this so if I were to draw the same triangle it would look like this it would look like this and let me just be clear this is this triangle right over here I just flipped it over I just flipped it over and so if we want to make sure we're keeping track of the same sides if this length right here is D sub 0 or D not sometimes we could call it or D 0 or whatever you want to call it then this length up here this length up here is also going to be this length up here is also going to be D 0 D 0 and the reason why I wanted to do that is because now we can do something interesting we can relate this triangle up here to this triangle down here and actually we can see that they're going to be similar and then we can relate and then we can get some ratios of sides and then what we're going to do is try to show that this triangle over here is similar to this triangle over here get a couple of more ratios and then we might be able to relate all of these things so the first thing we have to prove to ourselves is that those triangles really are similar so the first thing to realize this angle right here is definitely the same thing as that angle right over there there sometimes are called opposite angles or vertical angles they're on the opposite side of lines that are intersecting so they're going to be equal now the next thing and this comes out of the fact that both of these lines this line is parallel to that line right over there and you could look at you can I guess you can call it alternate alternate interior and if you look at the angles game or the parallel lines or the transversal of a parallel lines from geometry we know that this angle since their alternate interior angles this angle is going to be the same value as this angle right because they're both you could view this line right here as a transversal and D of two parallel lines these are alternate interior angles so they will be the same now we can make that exact same argument we can make that exact same argument for this angle and this angle and so what we see is this triangle up here has the same three angles as this triangle down here so these two triangles are similar these are both this is really more of a review of geometry than optics these are similar triangles similar there sim I don't have two right triangles they're similar and because they're similar the ratios of corresponding sides are going to be the same so D not corresponds to this they're both opposite they're both opposite this pink angle they're both opposite that pink angle so the ratio of D not to d1 let me write this over here so the ratio of D not let me write this a little bit neater the ratio of D not to d1 so this is the ratio of corresponding sides is going to be the same thing and let me make some labels here that's going to be the same thing as the ratio of this side right over here this side right over here we call I'll call that a it's opposite its opposite its opposite this magenta angle right over here that's going to be the same thing as the ratio of that side to this side over here to side B and once again we can keep track of it because side B is opposite the magenta angle on this bottom triangle so that's how we know that this side is its corresponding side and the other similar triangle is that one they're both opposite the magenta angles so we know D naught is 2 D 1 as a is 2 B as a is 2 B so that's interesting we've been able to relate these two things to sum to these kind of two arbitrary lengths but we need to somehow connect those two the focal lengths and what we could to connect them to a focal length what we might want to do is relate a and B a sits on the same triangle as the focal length right over here so let's look at this triangle right over here let me pick it let me put it in a better color so let's look at this triangle right over here that I'm highlighting in green this triangle in green and let's look at that in comparison to this triangle that I'm also highlighting this triangle that I'm also highlighting in green now the first thing I want to show you is that these are also similar triangles this angle right over here and this angle are going to be the same they are opposite angles of intersecting lines and then we can make we can make a similar argument alternate interior angle so there's a couple argues we can make one you can see that this is a right angle right over here this is a right angle if two angles of two triangles are the same the third angle also has to be the same so we could also say that this thing let me pick this do this in another color because I don't want to be repetitive too much with the colors we can say that this thing is going to be the same thing as this thing or another way you could have said it as you said well this is this line up over here which is kind of represented by the lens or the lens that the line that is parallel to the lens or right along the lens is parallel to the kind of the object right over there and then you could make the same alternate interior argument there but the other thing is just look I have two triangles two of the angles in those two triangles the same so the third angle has to be the same now since all three angles are the same these are also both similar triangles so we can do a similar thing we can say a is to be remember both a and B are opposite the 90-degree side they're both the hypotenuse of the similar triangles so a is to V as we could say this base length right here and it got over it in a little bit but this base length right over here is f that's our focal length as F as F in this triangle is related to this length on this triangle they are both opposite that white angle they're both opposite that white angle so as F is 2 this length right over here now what's this length so this whole distance is di all the way over here but this length is that whole distance minus the focal length so this is di minus the focal length so a is to be as f is to di - the focal length and there you have it we have a relationship between the distance of the object the distance of the image and the focal length and now we just have to do a little bit algebra if this is equal to this and this is equal to that then this blue thing has to be equal to this magenta thing and now we just have to do some algebra so let's do that so we got D naught or D 0 I guess we call it - di is equal to is equal to the ratio of the focal length to the difference of the the image distance to the dist the focal length to the image distance minus the focal length and now here we just have to do some algebra so let's just to simplify this let's cross multiply it so if we multiply D naught times this thing over here we get d not di I'm really just distributing it - d not f - D not F I'm just distributing this D not just cross multiplying which is really this the same thing as multiplying both sides by both denominators or multiplying with the denominators twice either way that's all cross multiplying is that is going to be equal to di is going to be equal to di times F is going to be equal to di times F and now we can add this term right over here to both sides of this equation now I'm just gonna switch to a neutral color so we get D not di is equal to I'm just adding this to both sides so you add to the left side it obviously cancels out is equal to di f this thing over here plus D not F and then let's see we could factor out a f of focal length so we get D not di is equal to F times di + D not and then what can we do we can we could divide both sides by F so this will become over F this becomes this becomes over F essentially essentially cancelling it out and then I just want to skip too many steps let me just rewrite what we have here so then so these cancel out so we have the not di / f / f is equal to di + do and now let's divide both sides by D not di so 1 over D not di divide this side by D not di cancels out over here and so we are left with on the left hand side 1 over the focal length is equal to this thing over here and we can separate this thing out this thing over here this is the same thing we just separate out the numerator is the same thing as di over D not di + D not over D not di but di over D not di the D is cancel out we just have a 1 here the d nots cancel out you just have a 1 so this is equal to 1 over the distance of the object and this is plus 1 over the distance of the image so right from the get-go this was a completely valid formula we actually had achieved what we wanted but this is even as much this is a neater formula you don't have the DI is repeated in the F for repeated right here we have an algebraic relationship for a convex mirror that relates the focal length to the distance of the object and the distance of the image anyway I think that's that's pretty neat how it came out to be at least a pretty clean formula