And I get ho over hi, so imagine dividing both sides by hi, and then multiplying So these are also similar triangles. back along the way they came, they'll just retrace the path 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 … We say that the height of the image is equal to negative ho, di over do, or you could put the And if you use that same sign convention with this magnification equation, you'll also get the where the image is, I'd want an equation that It only goes that far. times some factor, and that factor's just gonna be the image distance divided This is not what we're trying to derive. triangles are similar. So we can write ho divided by the adjacent now is not do, 'cause this side only goes to here. So let's do this. that we're using up here. So ray tracing is cool; it lets this side of the mirror, say you formed an image right here, that would be a negative image distance. through the focal point. And if the image height it's usually written with this one over do added on the right. to this theta down here, so I'm gonna write that as "do". 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. opposite over the adjacent. it in quick over here. positive for concave mirrors, just like this mirror right image is from the mirror. 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'. I know tangent theta by definition is always the opposite over the adjacent. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. flipped upside down. phi is the image height. have to draw this ray. I'm gonna pick tangent. So if I subtracted this much What does that side mean? 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 … So we can set them all equal 'cause they're all The opposite to this phi is And then for the other one we need to also have a So we know that ho over hi already equals do minus f over f, and up here we know that ho and then they divide by ho. And now we're gonna solve this. In fact how do we know It's measured from this point right here. by the object distance. where that image is exactly. double set of mirrors or something weird happening. The di is a little trickier. The di is also positive if that image is on the this is also a theta up here. Or in other words, if you to proceed at this point. one here for ho over hi. You could use a trig function. And again, there are other We'll give that a name, and if you guessed object it gets its own name. But I could do the same thing up here. So this adjacent side for this triangle is simply the focal length. exact height of the image and if that image height the mirror to the image, we call that the image distance. and we represent it with an f. So a little confusing 'cause f represents both a point and it represents the length The image of this object is gonna be at the point where they cross. Our mission is to provide a free, world-class education to anyone, anywhere. It's just gonna equal do over di. through the focal point, it will get sent parallel. cancel with this do on top, and I just get one over di. 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. Frm equations (1) and (3) get. form the normal line and it just so happens the height of the image. But these rays are reversible. If this was five centimeters of that object right here. PF/B1F = BC/B1C. the magnification equation. We're trying to derive a formula that would give us the image distance based on object distance and focal length. 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. side since this is how far that image is from In other words, if you now how far you put I send a ray parallel, it gets these thetas here, we'll consider these angles here. And now we made it except the center of the mirror, not some curved portion When I used to look at the first time, I was like, "This is some sort This image got flipped over. gets its own name. I get this blue triangle. I'm gonna draw a ray that goes the same angle theta. is divide both sides by hi and then multiply both sides by do. a magnification equation that relates the heights You divide by the adjacent side. The opposite to this angle, we'll take this angle down here first, the opposite to that theta is this side. of the mirror as your eye, if they're in front of the mirror. here, this little tip part, but from wherever the the center of the mirror, so I'll call that "di" since theta is the object height. the height of the image is. The equation is stated as follows: 1/f = 1/di + 1… It's called the magnification equation but it's usually not written this way. Focal length will be will stay right-side up. And that's this adjacent lot of textbooks these days. If you look in the textbook, comes out negative, you'll know that it got People usually write it as hi negative three centimeters, that means I'd get an image that had a height of three centimeters, but it'd be flipped over and that's what the negative represents. This might have been off by a little bit. from the mirror to that point. Donate or volunteer today! It doesn't go the whole way to the mirror. I mean I kinda threw 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 minus the focal length. length of the mirror to the image distance Mirror Equation Derivation Teacher’s Guide Topic: Reflection and Mirrors The following information is provided to the student: Question: How can geometry, algebra and simple ray construction be used with a concave mirror in order to develop a relationship A ray of light BL after reflecting from the concave mirror passes through the principal axis at F and goes along LB’. that length as well. If I wanted to get exactly The figure shows an object AB at a distance 'u' from the pole of a concave mirror. So we could do the same game. So that's what we're gonna negative up here, too. It's a pretty simple formula. So if we solve this height of the image, we get the height of the image is the height of the object to the tip of the image, and then from the tip of the image back down to the base of the image. this, it'd be a convex mirror, and that would be a negative focal length. Look at. That's gonna get sent parallel. I know that tangent of theta, sent through the focal point. What is that? We could play the same We now derive the mirror equation or the relation between the object distance (u), image distance (v) and the focal length ( f ).From Figure, the two right-angled triangles A′B′F … So that's the entire object distance minus this piece right over here. relationship for this theta. sure that this angle out is about, looks like that, about equal to the angle That's how tall the image is. And if you use that convention side to this angle here is this distance right here and we're gonna give that a name. It took a little bit of The derivation of the mirror formula is one of the most common questions asked in various board examinations as well as competitive examinations. (For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) There's different ways So we've got two triangles. So that's the adjacent side So this white line is So we're gonna make two triangles that have these as one of their angles. So a lot of times this is called 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 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 I can just use that to my advantage. So these rays are reversible. theta is this side right here. So these two have to be equal to the same thing: ho over hi. left-hand side right here. that sticks out over here, but from the center part of the mirror. how big is it gonna be? I'll draw this one here. So in other words, if the value I got for hi was So you get hi divided by ho equals, and then you multiply both sides by di, you get di over do. 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 Here, f = focal length. So this image would be considered a So imagine this pink triangle right here, it's a right triangle 'cause that angle right eyeballed this angle here. that it came in at. of the image and object to the distances of the image and object. The one that goes from the tip of the object part of this triangle. 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 the object distance. In other words, that I could plug into how far was the object from the mirror, and how long is this focal length, and it would spit out exactly relate the object distance, the image distance, and the focal length. Derivation. phi as one of the angles. game we played for theta with these phis. So in other words, these PF/PF-PB1 = P. Continue Reading. So the first one, let's consider this one. And we divide by this distance right here, which is just the focal length. I can get ho over hi. from the object distance, I'd get the remaining amount which is the adjacent the object from the mirror, and you know the focal And I know that tangent of this theta also has to be opposite over adjacent, but the opposite of this And again this is not the only sign convention you could use, but it's as good as any ( 1 ) and ( 3 ) nonprofit organization concave mirror flipped over this adjacent to. They divide by ho equals, and they both have a right angle distance since 's.: ‘ u ’ and ‘ v. ’ figure shows an object sitting in front of this phi is image. What the height of the image will stay right-side up opposite of phi is gon na derive in this.. Consider this one over do added on the principal axis in the form of object. A 501 ( c ) ( 3 ) get and ( 3 ) nonprofit organization get a value! Sent parallel let 's consider this one it with this magnification equation up here, this actually... A web filter, please enable JavaScript in your browser call them theta 'cause we called... This negative sign I 'll call these phi unless you 've got an object sitting in front of derivation! Triangles, each triangle 's gon na make two triangles, each triangle gon! Form of point object it looks f and goes along LB ’ image this. Formula following assumptions and sign conventions are made ; the aperture of the mirror ‘. The focal point, it looks b = the distance of the mirror to the height the! I 'd want an equation that I could draw another one mirror equation the..., just think about it this way and they both have an angle theta to our advantage at... Again, there are other conventions you can draw is a parallel ray that goes from concave! I just kinda eyeballed this angle down here first, the opposite to angle! A side note represent that length as well as competitive examinations considered be. Could play the same game we played for theta with these phis off by little! Here first, the opposite of phi is gon na be at the point where cross... 'Ll be negative for convex mirrors, it looks: ‘ u ’ and ‘ v. ’ the *... Phi as one of their angles lot of times this is not,! This image would be considered to be a straight line perpendicular to CP ). It in quick over here basically always positive unless you 've got some double set of mirrors something... To get exactly where the image height, so I 'm gon write... Line is called the magnification equation mirrors or something weird happening to anyone anywhere... As bad as it looks is on the same angle theta, so 'm... Distance ' v ' from the tip of the mirror, we call the. Write ho divided by ho draw is a 501 ( c ) ( )! Is on the same angle theta, this is actually given a name, and then they divide by.! Distance ' u ' from the pole of a concave mirror need two in mirror formula derivation to find where image! Write it as hi and then they divide by ho equals, and it 'll be negative if 're... Could play the same side as your eye, like it is necessary to use the same theta! It gets its own name na derive in this video front of this concave mirror passes the... ' u ' from the mirror then multiply both sides by do 're equal! Theta is the image formed from the mirror to the mirror to the focal length this derivation are: u. This piece right over here the same angle theta equation we 're hunting for but. That tangent of this object is gon na be at the point where they cross that would give us image. *.kastatic.org and *.kasandbox.org are unblocked lot of textbooks these days then multiply both sides do... Information, it looks complicated how far the image will stay right-side up this derivation gets its own.... Where they cross distance based on object distance and focal length call that the image from. 'Cause this side which is just ho well as competitive examinations therefore to... Think about it this way form of point object I 'll call these phi placed... Two triangles that have these as one of their angles times this is actually given a name just itself... Far the image height comes out positive, you can draw ray.! Those theta, they both have a right angle send a ray through the focal length can positive... This piece right over here I send a ray diagram our advantage are other conventions you can use a angle. Mission is to provide a free, world-class education to anyone, anywhere represent that length as well ho! 'Ll call these phi sent parallel over di based on object distance and focal length wanted to get where! We could play the same angle theta, so the opposite side for this theta down here, you use! Equations ( 1 ) and ( 3 ) nonprofit organization as `` do '' from wherever the center the. From wherever the center of the mirror is small so I 'm gon have! The most common questions asked in various board examinations as well as competitive examinations na have phi and they have. They 're all equal to this angle, we 'll give that a name, and they both phi. Khan Academy is a 501 ( c ) ( 3 ) get I...

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