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# Mirror equation example problems

## Video transcript

mere equation problems can be intimidating when you first deal with them and that's not because the mere equation is all that difficult it's kind of easy it's just a few fractions added together the place where it gets tricky is deciding whether these should be positive or negative so there's a bunch of positive and negative sign decisions that you have to make and if you make even one of those incorrectly you can get the wrong answer so let's do a few mere equation problems and you can see how the signs work now I have to warn you everyone has their own sign convention there's a lot of different sign conventions when you deal with optics the one I'm gonna use is the one that I feel like most textbooks are using these days and it's the one where anything on the front side of the mirror so by front side your eye should be over here somewhere so let's say your eyes over here right you're looking at some object maybe it's an arrow or a crayon right here a blue crayon you're holding in front of this mirror right here so this is the mirror right here and the convention I'm using is that anything on this side of the mirror is going to be counted as positive so if it's a focal point on that side positive if it's an object distance on this side it's positive and if the image distance comes out positive you'll know that it's also on this front side of the mirror anything that comes out negative so if we get a negative image distance after we do our calculation we'll know that that thing is behind the mirror that kind of makes sense negative like behind positive like in front so this side over here is in front of the mirror positive and back here would be behind the mirror and that would be negative so I've got some numbers in here already let's just solve this one so what do we do we're gonna use this mirror equation we're gonna say that one over the focal length and already we have to decide on a positive or negative sign so this mirror the way it's shaped right here based on how we're looking at it is a concave mirror and with the sign conventions we just discussed and the signs I'm using in this formula concave mirrors always have a positive focal length so in other words since this focal point is 4 centimeters from the center of the mirror I'm gonna have to plug in the focal length as positive 4 centimeters notice I'm not converting that's okay if you leave everything in Center me you'll just get an answer in centimeters so it's okay you don't have to convert as long as everything's in the same units and we'll set that equal to one over the object distance just one other warning sometimes instead of do you'll see this as so4 object distance or you might see di as SI for image distance is the same thing it's just a different letter so the object distance again is on this side in front of the mirror so it's going to be positive 12 centimeters since it's located 12 centimeters from the center of the mirror so there's also going to be positive 12 centimeters and now we're gonna add to that one over the image distance the image distance we don't know I haven't drawn the image on here it's gonna be a surprise we don't know what this is gonna be but we can solve for it so we can solve for image distance I'll subtract 1 over 12 from both sides which will give us 1 over 4 centimeters minus 1 over 12 centimeters and that's gonna have to equal one over the image distance so 1/4 you could rewrite as three twelfths so three twelfths minus 1/12 is just gonna be two twelfths and that's gonna equal one over the image distance but two twelfths is just 1/6 so 1 over the image distance is just gonna be 1/6 but that's what 1 over the image distances sometimes people forget to flip this at the end we don't want 1 over the image distance we want the image distance so finally you take one over each side and we solve and we get that the image distance is going to be 6 centimeters and it came out positive that's important this came out to be positive 6 centimeters so where's our image gonna be so since this image distance came out to be positive our image is gonna be in front of the mirror so it's gonna be over here it's gonna be 6 centimeters from the mirror somewhere around here so at this point right here is where our image is gonna be but we don't know how big it's going to be or whether it's right side up or upside down to figure that out we have to use a different equation and that equation is called the magnification equation says that the magnification is equal to the height of the image divided by the height of the object and that's equal to negative the image distance divided by the object distance so it turns out this ratio negative image distance over object distance is always equal to the ratio of the height of the image over the height of the object so what's the height of our image gonna be let's just solve for it if we solve for the height of our image but you get that the height of the image is gonna be multiplied both sides by H Oh we'll have the negative signs already here so negative height of the object times this ratio of the image distance over the object distance and now we can just plug in numbers the height of the object it says that's three centimeters tall right here so it's three centimeters so negative three centimeters times the ratio of the image distance was six the object distance was twelve and so if you solve this you'll get one half of negative three which is negative one point five centimeters the negative sign means that this image got inverted so got flipped over it's gonna be upside down compared to what it was before and the one point five is how tall is going to be so what we end up with is an image six centimeters from the mirror and it's gonna be have a height half as tall so at one point five centimeters and it's gonna be upside down because of this negative sign so it's gonna be one point five centimeters tall and upside down that's what you're gonna see if you look into this mirror it's like a funhouse mirror it's a weird curved mirror you'd see an upside on image right here it might look like you could reach out and grab it but it's gonna be an optical illusion there's gonna be no object there it's just gonna be the image of this object here so that's an example using a concave mirror what would change what if we did this what if we took our object so say we don't put it here anymore we move it inside here so instead of being at 12 centimeters we moved it to like 3 centimeters what would we do differently everything would be the same we just would plug in instead of positive 12 down here we'd plug in positive 3 so the mirror equation works exactly the same you still plug in whatever that object distance is you solve for your image distance we're of course gonna get a different image distance but whatever you get that would tell you where the images and then you would take that plug it into the magnification equation if you wanted to decide how big the image is going to be and where it's gonna be right side up or upside down so these numbers gonna be different but you would use this equation in the exact same way now what would happen if instead of using a concave mirror we used a convex mirror let's say we use the mirror shaped like this so imagine our eye again is over here looking at this object inside of the mirror and we're gonna see an image of the object we're gonna see the object right here but we're also gonna see the image of the object this mirror this time instead of concave this is a convex mirror so its focal length is behind the mirror so what do we do now to figure out where the images we again use the mirror equation we're gonna use the same equation we're gonna have one over the focal length and again I immediately have to make a decision on the sign with the convention that I'm using this focal length is behind the mirror so this focal length for a convex mirror is gonna be negative so this would be negative 4 centimeters and that's gonna equal one over the object distance well again the object is in front of the mirror 12 centimeters so this object distance is gonna be positive 12 centimeters and then we add to that the image distance which we don't know this is what we're gonna solve for so this time if we solve we're gonna have 1 over negative 4 and again we have to subtract 1 over 12 and that's all gonna have to be equal to 1 over the image distance so now on the left hand side we have negative 1/4 but that's the same as negative 3 twelfths so negative 3 twelfths minus 1/12 is the same as negative 4 twelfths and that's got to equal 1 over the image distance but negative 4 twelfths is the same as negative 1/3 so 1 over the image distance has to be equal to negative 1/3 and if we flip that over we get that the image distance finally is gonna be negative 3 centimeters so in other words this here is negative 1/3 so when you flip that over you get that the image distance is negative 3 centimeters where is this image gonna be well since it came out negative that means it's behind the mirror because that's the sign convention we're using so it's gonna be 3 centimeters behind the mirror so it's gonna be like over here about at this point right around here somewhere three centimeters behind this mirror and again if we want to figure out how tall it's gonna be whether it's right side up or upside down we're gonna have to use the magnification equation and that magnification equation looked like this is said that the height of the image over the height of the object had to be negative image distance over object distance so let's just see what this right-hand side is gonna be if we just set this equal to it's gonna be negative of negative 3 is the image distance so I have to plug in the negative 3 and you you keep that negative in there this negative out here comes along always but now we have another negative inside of this image distance of negative 3 and the object distance again was 12 centimeters so what are we left with we get negative of negative 3 which is positive 3 over 12 which is positive 1/4 and the centimeters cancel so what this ratio tells you which is the magnification is that the image is not going to be inverted this positive means it's going to be right side up and the 1/4 ratio means it's going to be 1/4 as large the image I should say is going to be 1/4 as large as the object is going to be and the reason is this ratio is equal to height of the image over the height of the object so in other words if I can multiply if I want to multiply both sides I can say that height of the image is equal to positive 1/4 times the height of the object well the height of our object was 3 centimeters in this case but whatever your height of the object is you multiply it by this ratio of negative DRDO and you get what the height of the image is going to be so we're gonna go 1 4 3 so we get positive three fourths of a centimeter so gonna be it's gonna be tiny there's gonna be a little little image that's gonna be right-side up so it's gonna be right here but it's gonna be teeny it's only gonna be like this 3/4 of a centimeter tall that's what our image is gonna look like so recapping you can use the mirror equation to figure out where the images are gonna be located the sign convention we're using is that objects images and focal lengths in front of the mirror are gonna be positive anything behind the mirror is gonna be negative and you could use the magnification equation to figure out how tall the image is gonna be relative to the object by taking negative the image distance over the object distance