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Current time:0:00Total duration:14:32

Derivation of the mirror equation

Video transcript

so imagine you've got an object sitting in front of this concave mirror if you wanted to figure out where the images formed you can draw ray tracings and so one way you can draw is a parallel ray that goes through the focal point but these rays are reversible I don't have to draw that one turns out if you send Ray's back along the way they came they'll just retrace the path they came along the other way so these rays are reversible I send a ray parallel it gets sent through the focal point but if I send a ray through the focal point it will get sent parallel in other words I don't have to draw this Ray I can draw this Ray right here the one that goes from the tip of the object through the focal point that's going to get sent parallel so I'm going to draw this one just for fun and then I'm going to draw another one you need to in order to find where the image is I'll draw this one here so this this white line is called the principal axis it's drawn through the center of this curved mirror I'm going to draw a ray that goes from the tip of the object to that center point because I know the law of reflection says that the angle in has to equal the angle out and the angles are measured from the normal line and it just so happens that this principal axis that's usually drawn here anyways is serving as a perfectly good normal line for this center of the mirror since it's passing through that center of the mirror in a perpendicular way I can just use that to my advantage I know that the angle ends got to equal the angle out I just have to make sure that this angle out is about looks like that about equal to the angle that it came in at so these two have to be the same angle theta and now I can find where my images the image of this object is going to be at the point where they cross so the tip of this object gets mapped to this point right here and we get an image that's upside down and it looks like that so ray tracing is cool it lets you find where the images but I mean I just kind of eyeballed this angle here this might have been off by a little bit it might have been off by a degree or two if I wanted to get exactly where the images I'd want an equation that I could just plug into in other words that I can plug in to how far was the object from the mirror and how long is this focal length and it would spit out exactly where the image is going to be so that's what we're going to derive in this video it's called the mirror equation and it'll tell us how to relate the object distance the image distance and the focal length so let's do this how do you derive it if you textbook it looks complicated it's not actually as bad as it looks when I used to look at it the first time I was like this is some sort of mathematical witchcraft I don't want to deal with but it's not nearly as bad as it looks we're going to start with drawing triangles so we'll notice that these two angles are the same we're going to use this to our advantage so we're going to make two triangles that have these as one of their angles so the first one let's consider this one let's say one of the triangles will be from the base of this image to the center of the mirror and then from the mirror to the tip of the image and then from the tip of the image back down to the base of the image so imagine this pink triangle right here it's a right triangle because that angle right here is a right angle and it's got theta as one of its angles but I could draw another triangle that has this angle theta I can go from the tip of the object to the center of the mirror and then from the center of the mirror to the base of the object and then from the base of the object to the tip of the object and I get this blue triangle that's also a right triangle since this angle here is a right angle so in other words these triangles are similar they both have an angle theta they both have a right angle or in other words if that if you don't like similar triangles just think about it this way you could use a trig function pick your favorite trig function I'm going to pick tangent so let's take tangent theta I know tangent theta by definition is always the opposite over the adjacent the opposite to this angle we'll take this angle down here first the opposite to that theta is this side what is that side mean that's the height of the image so I'm going to call that H I for image height that's how tall the images I know that this is going to equal height of the image divided by the adjacent side to this angle here is this distance right here and we're going to give that a name that's just how far the images from the mirror so we give this a name we call this distance from the mirror to the image we call that the image distance and it's measured from the center of the mirror so not from like this end right here this little tip part but from wherever the center of the mirror is its measured from this point right here and that's this adjacent side since this is how far that images from the center of the mirror so I'll call that D I since it's the image distance but that was for this theta down here I know that tangent of theta this is also a theta up here I use the same relationship for this theta and I know that tangent of this theta also has to be opposite over adjacent but the opposite of this theta is this side right here and what is that side that's just the height of the object so I'll call that H oh that's the object height so the opposite side for this theta is the object height and then you divide by what you divided by the adjacent side that's just going to be this distance from the mirror to the object we'll give that a name and if you guessed object distance and you guessed right there's going to be the object distance and again we measure it from the center of the mirror not some curved portion that sticks out over here but from the center part of the mirror so that's the adjacent side to this theta down here so I'm going to write that as do and so this is an important relationship this is actually given a name just by itself this isn't the equation we're hunting for but it's so important it gets its own name just part of this derivation gets its own name it's called the magnification equation but it's usually not written this way people usually write it as H I and then they divide by H oh so you get h I divided by H Oh equals and then you multiply both sides by di you get di over do so a lot of times this is called the magnification equation it gives you a way to find what the height of the image is so we were looking for a way to find how far the images from the mirror but this lets you figure out okay you also need to know how big is it going to be so if we solve this for the height of the image we get height of the images the height of the object times some factor and that factor is just going to be the image distance divided by the object distance here's the thing though look at this image got flipped over so if we define this image distance as positive if it's on the same side as the object we get a flipped over image I'd be like a negative height so since we want to represent flipped over images with a negative value we actually write this equation down with a negative inside of here we say that the height of the image is equal to negative H ODI over do or you could put the negative up here too so we stick a negative over here that way we know that if you get a negative value for the image height you know it's flipped over so in other words if the value I got for H I was negative 3 centimeters that means I'd get an image that had a height of 3 centimeters but it'd be flipped and that's what the negative represents okay so that's kind of a side note this is not what we're trying to derive we're trying to derive a formula that would give us the image distance based on object distance and focal length so we needed to do another set of triangles what we'll do now is instead of considering these Thetas here we'll consider these angles here I can't call them theta because we already called those theta so I'll call these Phi so these angles also have to be the same because anytime you have a line and then you cut another line through it these angles here will always be the same so we can do the same game we could play the same game we played for theta with these Phi's will form two triangles each triangle is going to have Phi as one of the angles so for the first one we'll do this object height as one side over here to the base of Phi and then back up over to here and then for the other one we need to also have a triangle that has fine it so we'll do this from here to here down to there and then back up to Phi so we've got two triangles this triangle and this triangle and they both have Phi and they both have a right angle so these are also similar triangles in other words we'll play the same game will say that tangent of this Phi is going to have to equal the opposite over the adjacent the opposite to this Phi is this side which is just H 0 so we can write H Oh divided by the adjacent now is not do because this side only goes to here it doesn't go the whole way to the mirror it only goes that far so that's the entire object distance - this piece right over here so if I subtracted this much from the object distance I'd get the remaining amount which is the adjacent part of this triangle so this distance from the mirror to the focal point is given a name it's called the focal length and we represent it with an F so a little confusing because F represents both a point and it represents the length from the mirror to that point so f is going to represent that length as well so this adjacent side we could write as the object distance minus the focal length since this remaining part right here is the adjacent side which is object distance - the focal length but we know that this Phi is also equal to this Phi so I can do tangent of theta for this Phi the opposite side would now be this side what is that what is this side of the Tri that's just equal to the height of the image the side is the same as the image height so I can say that this whole thing tangent of Phi has to equal opposite over adjacent this time the opposite of Phi is the image height and we divide by this distance right here which is just the focal length so this adjacent side for this triangle is simply the focal length so I'll just divide by F and so what I've got are two equations that I'm going to put together and we will get the mirror equation out of this there's different ways to proceed at this point what I'm going to do is I don't want h o or H I and either of these so I'm just going to solve this one here for H o / H I and I get h o / H I so imagine dividing both sides by H I and then multiplying both sides by do minus F and I get h o / H I equals do minus F the focal length divided by the focal length but I could do the same thing up here I can get h o / H I is just going to equal do over di since all you have to do is divide both sides by H I and then multiply both sides by do but this left-hand side right here is the same as this left-hand side right here so we know that H over H I already equals d o- f / f and up here we know that H over H I equals D o / di so that means that do minus f over F has to also equal do over di since both of these expressions equal H o / H I so we can set them all equal because they're all equal to the same thing H o / H I and now we're going to solve this we're just going to clean it up the left-hand side I can write I'll just stop using colors here the left-hand side is going to be D o / f minus 1 since F / f just equals 1 and that's got to equal do over di and now we can divide both sides by do if I divide both sides by do I get 1 over F since the do cancels minus 1 over do equals and then the do will cancel with this do on top and I just get 1 over di and now we made it except it's usually written with this one over do added on the right if we add 1 over do to both sides we finally get the expression we wanted which is 1 over the object distance plus one over the image distance equals one over the focal length it's a pretty simple formula it took a little bit of effort to get to this but this is called the mirror equation and it relates the focal length of the mirror to the image distance and the object distance in other words if you know how far you put the object from the mirror and you know the focal length of the mirror that lets you figure out exactly where the image distance is instead of just kind of eyeballing it and using a protractor this will let you solve for where that image is exactly and if you couple it with this magnification equation up here you can also figure out the exact height of the image now you might be feeling a little sketchy about this negative sign I mean it kind of threw it and quick over here what what's going on with this negative in fact how do we know if any of this stuff is negative or positive will the convention that I use that a lot of textbooks use now is the one where focal lengths will be positive for concave mirrors just like this mirror right here this is a concave mirror but if it was bent the other way if the mirror looked like this it'd be a convex mirror and that would be a negative focal length and it kind of makes sense if the mirror was bent this way the focal point is kind of back this way which is kind of behind it so it makes sense that it's sort of negative and the object distance over here if it's on the same side as your eye if you're using this correctly your eye should be over on this side you would see an object right here but you'd also see an image of that object right here and if this object is on the same side as the mirrors your eye which it should be then this object distance is positive it's basically always positive unless you've got some double set of mirrors or something weird happening if you got a single mirror there's no craziness going on this this object distance is just always defined to be positive using the conventions that we're using up here and again there are other conventions you can use but this is the one used in a lot of textbooks these days the DI is a little trickier the DI is also positive if that image is on the same side as your eye like it is right here so this image would be considered a positive image distance since it's on this side if the image got formed on this side of the mirror so you've formed an image right here that would be a negative image distance if this was five centimeters behind the mirror we'd consider that eye negative five centimeter image distance and if you use that convention with this mirror equation you'll get a correct relationship between the object distance image distance in the focal length and if you use that same sign convention with this magnification equation you'll also get the exact height of the image and if that image height comes out negative you'll know that it got flipped upside down and if the image height comes out positive you know that the image will stay right-side up so recapping using a bunch of similar triangles we were able to derive a mirror equation that relates the object distance image distance and focal length and along the way we derived a magnification equation that relates the heights of the image and object to the distances of the image and object and you have to be very careful with signs even though the object distance is basically always positive focal lengths can be positive or negative focal length will be positive for concave mirrors and it'll be negative for convex mirrors image distances will be positive if they're on the same side of the mirror as your eye if they're in front of the mirror but there'll be negative if they're behind the mirror and again this is not the only sign convention you could use but it's as good as any other sign convention and it's the one used in a lot of textbooks these days