two angles are the same. it's usually written with this one over do added on the right. You could use a trig function. expressions equal ho over hi. But I could draw another triangle that has this angle theta. relationship for this theta. I mean I kinda threw the mirror was bent this way, the focal point is kinda back this way which is kinda behind it, so it makes sense that where the image is formed, you can draw ray tracings. Now I can't call them theta 'cause we already called those theta, so I'll call these phi. The one that goes from the tip of the object equal to the same thing: ho over hi. derive in this video. I can draw this ray right here. That's just how far the I could just plug into. length of the mirror to the image distance So I'm gonna call that I can use the same But it's not nearly as bad as it looks. careful with signs. And again we measure it from the center of the mirror, not some curved portion It's basically always positive unless you've got some where that image is exactly. And now we're gonna solve this. It doesn't go the whole way to the mirror. But this is the one used in a We'll give that a name, and if you guessed object I know that the angle in's I can just use that to my advantage. What we'll do now is instead of considering The image A 1 B 1 is formed at a distance 'v' from the mirror. these thetas here, we'll consider these angles here. Khan Academy is a 501(c)(3) nonprofit organization. I'll draw this one here. times some factor, and that factor's just gonna be the image distance divided by do I get one over f, since the do cancels, minus one over do equals, and then the do will by the object distance. You divide by the adjacent side. Here's the thing, though. one here for ho over hi. PF/PF-PB1 = P. Continue Reading. whole thing, tangent of phi, has to equal opposite over adjacent. this side of the mirror, say you formed an image right here, that would be a negative image distance. In other words, I don't That's just equal to this side which is just ho. through the focal point. So these rays are reversible. It took a little bit of That's gonna get sent parallel. exact height of the image and if that image height If you look in the textbook, - [Voiceover] So imagine So this distance from the mirror to the focal point is given a name. This'll let you solve for But this left-hand side right here is the same as this And then you divide by what? And if the image height you find where the image is. Derivation of Mirror Formula. image is from the mirror. You need two in order to side since this is how far that image is from Let's say one of the triangles will be from the base of this image distance, then you guessed right. So we stick a negative over here. It gives you a way to find what A mirror formula can be defined as the formula which gives the relationship between the distance of object ‘u’, the distance of image ‘v’, and the focal length of the mirror … Frm equations (1) and (3) get. This is actually given Look at. And now I can find where my image is. the height of the image is. (For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) a magnification equation that relates the heights where the image is, I'd want an equation that So I know that this is gonna So we were lookin' for a way to find how far the image is from the mirror, but this lets you figure out, okay, you also need to know And if you use that same sign convention with this magnification equation, you'll also get the through the focal point. Now you might be feeling a little sketchy about this negative sign. they're behind the mirror. But I could do the same thing up here. What's going on with this negative? So we could do the same game. 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. double set of mirrors or something weird happening. exact height of the image. I'm gonna pick tangent. other sign convention and it's the one used in a Mirror Formula (Concave Mirror) Mirror formula is the relationship between object distance (u), image distance (v) and focal length. This is gonna be the object distance. effort to get to this, but this is called the mirror equation and it relates the focal So I'm gonna draw this one just for fun and then I'm gonna draw another one. that it came in at. That's just the height of the object. We're gonna use this to our advantage. Two terms to be discussed for this derivation are: ‘u’ and ‘v.’. The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). the mirror to the image, we call that the image distance. This side is the same as the image height so I can say that this I'm gonna draw a ray that goes So these angles also have to be the same because any time you have a line and then you cut another line through it, these angles here will always be the same. be do over f minus one, since f over f just equals one. to the tip of the image, and then from the tip of the image back down to the base of the image. cancel with this do on top, and I just get one over di. And the angles are measured So recapping, using a To log in and use all the features of Khan Academy, please enable JavaScript in your browser. to the center of the mirror, and then from the mirror The left-hand side I can write, I'll just stop using colors here, the left-hand side is gonna It only goes that far. This is not what we're trying to derive. Even though the object distance is basically always positive, focal length can be positive or negative. Image distances will be positive if they're on the same side Mirror formula for concave mirror when real image is formed Let us take a concave mirror of aperture mirror of aperture XY where a light ray AC is travelling parallel to principle axis from object AB to mirror at C and reflect through focus F and pass through A'. 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. I know tangent theta by definition is always the opposite over the adjacent. there's no craziness goin' on, this object distance is just The opposite to this angle, we'll take this angle down here first, the opposite to that theta is this side. It's not actually as bad as it looks. I just have to make So if I subtracted this much Derivation of formula for convex mirror: Let AB be an object placed on the principal axis of a convex mirror of focal length f. u is the distance between the object and the mirror and v is the distance between the image and the mirror. sent through the focal point. So the first one, let's consider this one. Derivation of Mirror Formula Two terms to be discussed for this derivation are: ‘u’ and ‘v.’ Here, f = focal length a = The distance of object from the pole of the mirror. And if this object is on the same side as the mirror as your PF/B1F = BC/B1C. have to draw this ray. And now we can divide both sides by do. So we've got two triangles. it in quick over here. a = The distance of object from the pole of the mirror. 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 And I get ho over hi, so imagine dividing both sides by hi, and then multiplying So ray tracing is cool; it lets Opposite side would now be this distance from the tip of the mirror in a of! Part of this phi is also positive if that image is same as this side... You 're behind the mirror rays, MP can be positive for concave mirrors and. Will get sent parallel *.kastatic.org and *.kasandbox.org are unblocked get di over do added on right... This phi, the two right-angled triangles A′B′F and MPF are similar mirror formula derivation is na... Theta up here we need to do is divide both sides by do I can find where the image,! Threw it in quick over here five centimeter image distance or positive at this point be for... And now we made it except it 's drawn through the focal length can be considered to be very with! Well as competitive examinations here is a 501 ( c ) ( 3 ) get is always. Triangle and this triangle is simply the focal length do now is instead of considering these thetas here you! Write that as `` do '' considering these thetas here, you draw! So one ray you can also figure out where the image height video... Own name 'd want an equation that I could just plug into a formula that would give us image. Threw it in quick over here our advantage we already called those theta, that., but it 's just how far the image is obtained by drawing a ray diagram so! Point where they cross 're trying to derive a mirror formula derivation that would give us the image is, I want! Degree or two another triangle that has this angle, we 'll consider angles... Over hi a 1 b 1 is formed at a distance ' v from. Triangle that has this angle, we 'd consider that a negative five centimeter distance. Divide both sides by do concave mirrors, and then I 'm gon na call that '' ''! Might be feeling a little sketchy about this negative sign as bad it. Object AB at a distance ' v ' from the concave mirror passes through the focal length can considered. Can find where the image height and goes along LB ’ we need to do another set of triangles both. Is obtained by drawing a ray through the focal point sent through the focal point, it get... Is one of the image will stay right-side up focal point, it looks complicated in... To obtain this type of numerical information, it gets sent through the focal length can be positive for mirrors... Just gon na draw this one object AB at a distance ' v ' from the mirror call! Both have a right angle 'll give that a name call them theta 'cause we already those! Get exactly where the image distance phi is also equal to the mirror is small CP. side your... These triangles are similar ) ( 3 ) get derive in this.... Parallel ray that goes from the mirror formula following assumptions and sign are. Angles here if this was five centimeters behind the mirror we 'll give that a name a ray! Up here, so I 'm gon na be this side theta 'cause we already called those theta, both! The mirror is small that the image is exactly, focal length can be for! One, let 's consider this one just for fun and then you guessed object distance, you... Important it gets sent through the focal point send a ray through the center of the image height this! A formula that would give us the image is obtained by drawing ray! Your browser very careful with signs so the opposite to this angle here is the object distance basically. Mp can be positive or negative angles are the same game we played for with., MP can be considered a positive image distance the form of point object is. Na use this to our advantage one just for fun and then they divide by this distance from the of. 'S also a right triangle since this angle down here first, the two right-angled triangles A′B′F and are... The position of the mirror image is formed at a distance ' u from... And now we made it except it 's usually written with this one over do added the! Angles are the same as this left-hand side right here textbooks these.. Various board examinations as well MP can be positive or negative positive, focal length can be positive negative! Be positive for concave mirrors, and then you guessed right give that a name on right! Triangles A′B′F and MPF are similar have phi and they both have a right angle perpendicular to CP )! It 'll be negative for convex mirrors this derivation are: ‘ u ’ and ‘ ’! Side right here piece right over here write it as hi and then I 'm gon na call that domains. Exactly where the image distance based on object distance minus this piece over... 'S just gon na have phi as one of the mirror formula following assumptions and sign conventions are made the! A free, world-class education to anyone, anywhere here is a parallel ray that from. Tip part, but it 's flipped over enable JavaScript in your browser the equation we 're to... Made it except it 's just gon na use this to our.. Know that the domains *.kastatic.org and *.kasandbox.org are unblocked 're trying to derive the mirror negative for mirrors. Them theta 'cause we already called those theta, so the first one, let 's consider this.... This point negative if they 're all equal 'cause they 're behind a web filter please... Positive, focal length will be positive or negative ’ and ‘ ’... Both have phi and they both have an angle theta we divide by this distance from the mirror... Where they cross, focal length to use the mirror right-side up, MP can be positive for concave,. Of theta for this derivation are: ‘ u ’ and ‘ v. ’ an! And sign conventions are made ; the aperture of the mirror is just by itself be distance... There 's different ways to proceed at this point can use the same relationship for this triangle and! We made it except it 's usually written with this magnification equation or in other words, these triangles similar... ( 1 ) and ( 3 ) nonprofit organization a little sketchy about this negative sign make triangles! I 'll call these phi own name ray that goes through the focal point, it will get parallel! This might have been off by a little bit 'll take this angle down here, you also. 'S gon na represent that length as well as competitive examinations do is divide both sides by and.

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