mirrors going to be the topic of this lesson in my new General Physics playlist which when complete will cover a full year of University algebra based physics now in the last chapter we dealt with reflection and refraction we're going to pick up right where we left off in this chapter on mirrors and lenses we'll cover mirrors in this lesson and lenses in the next now we're going to talk about both plain mirrors as well as spherical mirrors in this lesson and the two types of spherical mirrors both concave and convex mirrors my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat so now we're ready to deal with plain mirrors here and what we want to do is draw some Ray diagrams so we can discover where the actual image is being formed so now we're going to find a couple things now the distance from the mirror to the object is going to be called the object distance and then the distance from the mirror to the image is going to be called the image distance now it turns out for a plain mirror we need two rays to kind of identify where that image is going to be so and it turns out it could be any two rays you draw to the mirror but there's two that are pretty convenient here and the first one is going to be one directly headon here so and if we hit a plain mirror head on then it's going to get reflected right back as well and there's a reflected ray now again we could pick any other reflected array here so but the one I'm going to choose is going to be one that goes kind of of the midpoint of this mirror here because it's going to be able to identify that law of reflection that the angle of incidence is going to equal the angle of reflection as well all right so now we've got two rays drawn and these two reflected rays in blue if you look the further we get away from the mirror the farther and farther apart they get they're never going to really intersect and we're not going to form a real image and that's true for a plain mirror for a plain mirror you never form a real image your imag is always going to be virtual as we'll see so these light rays don't really intersect but they do appear as if they had come from a common point back here behind the mirror right back here so and that point right there is the image and in this case again the light rays aren't really intersecting in fact there's no light rays behind the mirror at all the reflective surface is all over here so but it appears as if they came from that point and so that's a virtual image so it turns out again for a plain mirror that the image is always virtual now a couple other things we can identify here so one so when the image forms on the same side of kind of this midpoint of the mirror as the object itself so the object here is above that midpoint so is the image that's evidence you're going to have an upright image as well so we'll find out in a little while that for any single mirror or any single lens virtual images are always going to be upright and real images are always going to be inverted and upside down and we'll kind of see why that is with some of the other Ray diagrams but for a plain mirror the images will always be virtual and always be upright now a couple other things we can identify here so again we talked about the image distance and the object distance so again the the straight line distance between the object and the mirror that's called the object distance we often use P or some books do o I'm going to use P so for the rest of this lesson and then the distance the straight line distance from the mirror to the image location that's going to be Q or some books will use di now personally I kind of like doo and Di because it's not difficult to remember what those stand for but p and Q is a real common convention and the ones I'm going to use here because probably the majority of textbooks use that so p and Q object distance image distance cool now for a plane mirror it turns out that the object distance and the image distance are always equal as well so if you're standing in front of a mirror and let's just say you happen to fall asleep while standing in front of the mirror and you wake up and your initial incident is that somebody who looks a whole lot like you is standing a little ways away now if you're three feet in front of a mirror what it looks like is that there's somebody who's 3 feet behind the mirror looking back at you because again this image distance is always going to equal the object distance for a plain mirror and so for a plane mirror you're always going to have an image behind the mirror it looks like it's the same distance away as uh the object is to the mirror so and it's always going to be a virtual and upright image now one other thing we should talk about is the magnification is also going to be one and what that means is that the object is I'm sorry the image is exactly the same size it appears to be the same size as the actual object so we'll find out that magnifications that are positive are going to be for upright images whereas magnifications that are negative are going to be for upside down or inverted images we'll find out that magnifications that have an absolute value that's bigger than one look bigger than the actual object so in magnifications that are the absolute value is smaller than one means the image is going to look smaller than the actual object so positive versus negative and then absolute Valu is bigger or smaller than one are the keys there now one thing to note and you've probably noticed this every time you look in the mirror is there's this Left Right reversal if you're standing in front of the mirror and you raise up your left hand in the mirror it look looks like the person staring back at you is actually raising up their right hand and so on and so forth so again it's still an upright image there's no top bottom reversal if you would but that is a left right reversal for a plain mirror so the rest of this lesson we're going to focus on spherical mirrors and again there are two types concave and convex and again the concave refers to the fact that the reflective silvery finish is on the concave surface the convex the opposite so here's an example of a concave mirror and let's see if we can actually get the camera in view there and you can see there's my camera it's on the tripad and it's inverted we'll come back and talk to about that in just a little bit so and here is an example of a convex mirror and you kind of uh might recognize seeing these in stores and stuff like this to prevent theft and they're kind of up and give a wide field of view and stuff like that so we'll come back to that as well now it turns out with a concave mirror there are two different scenarios that we need to talk about and so one of them is when your object distance is greater than the f FAL distance one of them was when your object distance is less than the focal distance and you might be like what was that the focal distance or focal length that's called and yeah we got to talk about that for just a second it turns out that the focal length or focal distance is equal to 1 12 of the radius of curvature and so we're drawing Ray diagrams for these lovely mirrors we often draw what's referred to as the principal axis right down kind of the horizontal middle of the mirror right to where the mirror is perfectly vertical if you will so it is perpendicular to that uh that vertical point and there's our radius of curvature where we've got this lovely radius all the way around well halfway between the radius of curvature in the mirror is where you're going to find this focal point we'll kind of highlight it on both of these right about here somewhere and again it's exactly half of the radius of curvature now what that focal distance is you can kind of think of this if you had an object we're going to use this little lovely red rectangle here but let's say it was infinitely far away from this mirror well what you'd find out is that the light rays reflecting off the mirror would all intersect right at that focal distance that's exactly what would happen we can see why that happens if we take a look at the other piece of math we're going to be using quite a bit here and that's called the mirror equation so one over the uh object distance plus one over the image distance equals one over the focal distance we're going to use p's and q's but again some uh textbooks like to use D for the object distance and Di for the image distance and personally that's my preference but this is probably used a little more commonly so I'm going to stick with it but if you look if the object is infinitely far away so effectively p is infinity here well 1 divid infinity is0 and that term effectively goes out of the equation and the only way this the remainder of this equality would hold true is if Q equal F if the image distance equal the focal distance and so when the object is infinitely far away so the image will form at the focal point that's kind of the idea there so let kind of an idea give you an idea of what that focal point is but it turns out it's a valuable piece of information for doing calculations with mirrors and we'll find out for thin lenses in the next lesson as well all right so we're going to set up drawing some Ray diagrams here and so again with a concave mirror there's two scenarios when your object is beyond the focal distance like in this first one here we're beyond that focal distance away from the mirror and then when your object is closer to the mirror than the focal distance which is this scenario here so we'll find out that with the concave mirror so when you're beyond the focal distance the image is going to be inverted in real and that's why when I held up the concave mirror the camera appeared to be upside down it was inverted because right now we're you know from where I was standing we were beyond the focal distance of that mirror now the other scenario is when we're going to be the object is placed within the focal distance like this one here and in that case we're going to have an upright but virtual image instead and then we move on to talking about the convex mirror it turns out there's no two scenarios for that one regardless of where you place the object you were always going to get an upright and virtual image and so I've kind of given you kind of where this lesson's headed we're going to explain why these happen but I kind of wanted you to upfront know these three different situations you might encounter because qualitatively you can say some things about what you should expect about the image long before you ever do a calculation okay so we want to start by drawing some Ray diagrams and so it turns out we want to draw some rays and uh again we're trying to see if the Rays actually or really intersect and if they do these reflected Rays then we're going to have a real image but if they don't really intersect we'll end up with a virtual image a point where it looks like the Rays actually originate but they don't all right so let's see how this works so there's common to draw three different rays and it turns out if you actually can draw two of the Rays that's usually enough to identify where the image is going to be located so first one we're going to draw here is going to be parallel to the principal axis and so I'm going to choose the very top point but it turns out I mean every point on this lovely object is going to be reflecting rays and stuff like that but I'm just going to show the the Ray diagrams coming off the top Point here the first Ray we're going to have is going to be parallel to the principal axis and you might recall the law of reflection that the angle of incidence equals the angle of reflection and if we kind of looked at the angle of the mirror right here and we could draw then the normal to it we could draw an incident angle and then the reflected angle and it turns out you could work it out and find out that the reflected angle is going to go right through here now again we're eyeballing this a little bit but rather than kind of having to examine it that way what you're going to want to do is probably remember where these Rays go and the first one parallel to the principal axis is the incident Ray and the reflected ray is going to come out through that focal point that's Ray number one now the the second one is going to kind of be the opposite of that so and that Ray is going to go through the focal point so and then it's going to the reflected array is going to come out parallel to the principal axis and so it's kind of the opposite all right well if you notice these two rays are really intersecting right at that point right there and that's going to be where the image is located and because the light rays are really intersecting that's what's going to make it a real image as we'll see here in a second so now again two rays is really sufficient but most textbooks and professors going to recommend that you draw a third Ray as well so there's a couple different options but the one I choose is to draw right to the principal axis and so we're going to go right to the principal axis so and in the pr you know right to the point at least where the principal axis intersects the mirror and because the principal axis is perpendicular that is the normal so and the incident angle is going to equal the angle of reflection in this case as well and so it's going to be the exact same angle and there it is where the incident angle equals the angle of reflection and you can see that all three of the Rays we've drawn intersect at a single point and that is a real image so because the light rays the reflected rays are really intersecting that's a real image now here's the deal though so if we draw kind of where this is going to be located here so it's going to be something like that and what you're going to find out though is that in this case so the top point of our object the light rays are reflecting to form kind of the bottom point of the image what you find this thing is completely inverted and and what you find out as well is that all the time a real image is always going to be inverted and so in this case inverted and real always go hand in hand and we're going to find out that upright and virtual always go hand in hand as well and so it's convenient that we can kind of remember this with ir and UV both parts of the electromagnetic spectrum so inverted images are real real images are inverted upright images are virtual virtual images are upright those go together now it turns out we can say that they always go together if you're dealing with a single mirror or a single lens now when you start dealing with combinations of mirrors and lenses all bets are off and this may or may not hold true but for any single mirror lens which is what we're going to deal with in this chapter so largely that's what's going to hold true real images are inverted so and virtual images are upright okay and now we can see why that is so if we go and compare this now to our second scenario where the object distance now is closer than the focal point so we can draw our light rays once again and the first one again same same diff we're we're going to draw one coming in parallel to the principal axis and it's going to get reflected back out through the focal point right there so no difference there the second one is really going to be the same thing as the second one we drew over here as well but it's going to feel just a little bit different because we want to draw through the focal point and then come back out reflected parallel to the principal axis but we can't actually go through the focal point because we're closer than it and so what you actually have to do is kind of draw as if you were going from that focal point instead and then that light Ray is going to come back parallel to the principal axis and we can see here that these two reflected rays are only getting farther and farther apart they are never going to intersect there's no real image here whatsoever now if we draw one more of these here so in this case we can draw one more we're going to draw right to where the principal axis intersects the mirror again and once again that angle of incidence is going to equal that angle of reflection and again these light rays are only getting farther and farther apart they're they're never going to intersect at a common point but what you can do is kind of like extrapolate back and see that they look like they all came from a common Point behind the mirror now did the light rays really originate from that point no they didn't really and that's why they call this a virtual image the light rays don't really intersect there uh in this case all right so a couple of things here when you go to solve for Q the image distance in the mirror equation here so it turns out when you get positive values for Q that's going to be an indication you've got a real image when you get Negative values for Q that's going to be the indication that you have a virtual image and so we can see it with the ray diagrams or we can see it mathematically with the mirror equation for the values of Q being either positive or negative all right now we also want to take a look at this equation for magnification here and so magnification is the the relative uh ratio of the height of the image relative to the height of the object if the image is bigger than the object you get a magnification that's bigger than one if the image is smaller than the object you get a magnification that's less than one now I say greater than one and less than one but what I really should say is that the absolute value of the magnification is larger or smaller than one because it turns out these magnifications can also be positive or negative so so here's the way it works if the absolute value is bigger than one the image is bigger than the object okay if the absolute value is less than one then the image appears smaller than the object but now positive versus negative if the magnification is positive that's going to be an upright image here so we'd see over in this case now we haven't talked about it we just said it was virtual but it's also for virtual image they're going to be upright as well and we'll see why here so in this case the object is above the principal axis well so is the image and that image is going to be right side up if you will as well and so both the object the image right side up that's an upright image whereas in this case the object was above the principal axis the image would be below the principal axis that's why it's an inverted image all right and going back to magnification then if your magnifications are positive that's a real image if your magnifications are I'm sorry if magnifications are positive that's going to be an upright image if your magnifications are negative that's going to be an inverted image and we'll find out now why inverted and real always go together and upright and virtual always go together because it turns out if you work out the GE here this ratio of the height of the image to the height of the object can also be solved with Q over P the image distance over the object distance we you kind of make this make a little sense of why the magnification should be proportional to the image distance so if you look at uh putting a projector and showing a big projection up on the wall and stuff like that you can make that projection bigger and bigger if you move the projector further and further from the wall or if instead of it being a wall let's say it was just a Mobile screen you could move the screen further and further away from the projector in either case the image that's being projected on the screen so is traveling farther and farther to get to the screen and that's the key the farther the light has to travel to form that image the larger it gets and if it travels far the light travels farther to form the you to where the image is located relative to where the object located you get a magnification that's going to have an absolute value bigger than one so that's kind of why this makes sense in this case but it's q overp and again if the magnification overall comes out negative it's an inverted image if it comes out positive it's an upright image and so now we can kind of put this all together so Q we said if that's positive that means you've got a real image well the object distance is always positive so for any single lens or mirror it's always going to be positive now again if you use some certain special cases where you've got combinations of lens or mirrors you might end up with some situations where that is not defined as being positive but for anything you're going to see in this course so that object distance is going to be positive so if Q is positive I.E it's a real image and P is always positive by the time you factor in that negative sign the magnification is going to come out negative and your real image with a positive value for Q it's going to end up with a negative magnification and therefore be inverted and that's why real images are always inverted that works the opposite as well so let's say you've now got a virtual image well a virtual image is indicated by the fact that you've got a negative value for Q and A negative time a negative all divided by a positive object distance is going to give you a POS positive magnification and so the negative value for Q again tells you have a virtual image the positive value for the magnification tells you it's an upright image and that's why virtual images are always upright and that's why these go together cool so those are our two scenarios for concave mirrors so let's kind of see how this plays out with a cool little demo here all right so we're going to use the current book I'm reading Called Peace Child by Don Richardson which I'm really enjoying it's kind of a way to look at this and you can see that it currently is inverted because we are beyond the focal point of our concave mirror but as we move closer and closer together well it turns out we're going to approach that focal distance when you reach that focal distance it turns out that the image gets infinitely large and is unrecognizable kind of somewhere in that ballpark and then when you get past that point so all of a sudden now the image is no longer inverted it's upright so and it gets smaller and smaller as you then get closer and closer to the mirror so from this point on this is kind of where we use like vanity mirrors that are kind of used for magnifying the so we can see all the blemishes in our face because apparently we like doing that for some reason so but once you get to that focal point again the image gets infinitely large and on the other side of it we're now back to being an inverted image instead so now let's just briefly explain what's going on mathematically to match our observations with the book and the concave mirror so in this case when you're infinitely far away when your object distance is infinity again your image is going to form right where the focal point is as we kind of have already stated now but if you look at the magnification equation so when p is infinity effectively as P approaches Infinity your magnification is going to approach zero because dividing by Infinity gets you zero but as you bring that object closer and closer and closer together p is going to go down and your magnification the ab absolute value anyway is going to go up now as long as you're beyond the focal distance with that object then when you solve for the values of Q you're going to get positive values indicating that's a real image and if you plug in a positive value for Q the magnification is going to be negative showing that it's inverted just like the book was inverted when we were further away but once you actually have P approaching a value of f you're approaching that focal distance so when the object right gets onto that focal distance that's when we couldn't resolve the image of the book anymore and so if you look if 1 over p and 1 over f are equal well the only way that's true is if 1 / Q equals 0 well if 1/ Q equals 0 that means Q is going to be Infinity the image is forming at Infinity well you can't really get to Infinity right so there's no real image that's viewable there but also if Q is approaching Infinity the magnification is approaching Infinity as well so the object is getting infinitely large but the image is forming infinitely far away to where you can't really resolve it so that's what's going on but once you get to the other side of that focal distance when the object's inside the focal distance now all of a sudden you're going to get Negative values for Q when you use the mirror equation indicating that it's it's a virtual image and when you plug in the negative value for Q in the magnification you get positive values for the magnification indicating that you had an upright image as well and it turns out once you cross that focal point so as close to the focal point as you can be but having crossed it closer you got the biggest image you're going to get and it's magnified it's actually the image is bigger than the object so but as you get closer and closer and closer to the mirror that image is going to get smaller and smaller and smaller but it will still be magnified relative to the original object that was magnification will be larger than one so now we're going to take a look at our convex mirror and again you might look at these as kind of being security mirrors but sometimes you'll see these uh kind of a on the passenger side uh rear viw mirrors in your car and things of where it says things may be larger than they appear so and that's because in a convex mirror so your magnification is always going to have an absolute value that's less than one your image is always going to appear smaller than the actual object now it turns out there's not two scenarios here so it doesn't matter where you're located it doesn't matter if if you're within the focal distance or beyond the focal distance or anything of that sort so with a convex mirror the image is always going to be upright and virtual and it's always going to be uh appearing smaller than the actual object so again it doesn't matter how close or far you get so you can kind of see a my camera there on the far wall there and it's upright the whole time so because we have an upright virtual image the entire time that always appears to be smaller than the actual object all right so a couple things we should know here so one we're not going to really form an image so there's no real image the reflected rays are not going to intersect uh over here on this side of the mirror and so if you recall in the last example with the concave mirror when the image formed behind the mirror so to speak it was a virtual image the light rays didn't really intersect there and same thing is going to be true here in a different respect so there's our radius of curvature and halfway to our radius of curvature between the Cur in the mirror is going to be the focal distance and just like we saw when the image formed on the back side of the mirror well in this case the focal points on the backside of the mirror and it turns out with a convex mirror you need to make sure that when you plug your value for the focal length or focal distance into the mirror equation that it is a negative value even if you're not provided with a negative value if it's for a convex mirror it's a negative value and you've got to plug it in as a negative in the equation super important all right that's the first first thing so if we draw our light rays now so it turns out we can draw exactly the same set of light rays first one we're going to draw is going to come in parallel to the principal axis so and if you recall before we said it gets reflected out through the focal point well that's not possible the focal Point's on the other side of the mirror but it's just as if it came from the focal point in this case and there's our reflected ray so next one we're going to draw we're going to draw it towards the focal point we'll never get there because we'll bump into the mirror first but we're going to draw it just as if we were going towards that focal point point right there and then that one's going to get reflected back out parallel to the principal axis and we can definitely see that these two reflected rays are definitely not going to intersect but they do look like they came from a common Point behind the mirror again and one more we can draw towards the principal axis itself or where the principal axis intersects the mirror so and the angle of incidence equals the angle of reflection so where this angle equals this angle and once again we can kind of work our way backwards here there's that one that one that one and they intersect that a common point right there and that is our virtual image the light rays don't really intersect there they just look as if they all originated from that point and once again when you've got a virtual image that's going to be a negative value for Q when you solve for it in the mirror equation and when you plug it into the magnification a negative value for Q is always going to end up with a positive magn ification indicating again that your virtual image is always going to be upright cool and again didn't matter where we place the object if we're closer than the focal distance far than the focal distance it totally doesn't matter one thing you should also know though is that no matter how you work this out your image distance is always going to end up less than your object distance and because your image distance ends up less than your object distance your magnifications are always going to have an absolute value less than one and in this case we could just say less than one because they're always going to be positive as well for a convex mirror but in this case they're always going to less than one because the image always appears smaller than the actual object so now we're going to do some math with this and we're going to use the mirror equation and the magnification equation in every one of the following four examples the first we're going to be dealing with the concave mirror the last one with the convex mirror and so the concave mirror we're just going to be moving an object uh slowly closer and closer and solving for the image distance every time so first question says an object is placed 12 CM in front of a concave mirror in in parentheses it tells us that the focal distance is 4.0 CM where is the image located is it real or virtual upright or inverted and what is the magnification of the image and so in this case we go to plug this into the mirror equation we're going to have one over the object distance which is 12 CM so plus 1 over Q which is what we're going to want to solve for and the focal distance was given as 4.0 CM so we can solve for Q now a couple things here so it is a concave mirror and we are at an object distance that is greater than the focal distance well if you recall we said that's when you're going to get a real inverted image and so before we even solve for Q if you qualitatively understood what we did with that concave mirror and be like oh if you're beyond the focal distance real and inverted if you're closer than the focal distance then it's upright and virtual then you could have answered at least the qualitative parts of this question already but where is the image located well that's what is you know that's what Q is and so in this case we'll have have 1/ q = 1 over 4 - 1 over2 and one thing to note you might be like hey Chad you need to convert centimeters to meters to get SI units turns out as long as you're consistent in your units you can use any units of length you want to and it's going to be a lot easier to do the math here in centimeters than it would if we plugged in like 0.12 MERS and 0.04 met so I'm going to leave this in centimeters and if you notice 1 over 4 - 1 over2 my lowest common denominator here is going to be 12 so in this 1 over4 I'm going to multiply by 3 over3 that way I get a total of 3 over2 - 1 over2 and so 1 over Q is going to equal 2 over 12 and so flip both sides back over and Q over one or just simply Q is going to equal 12 over 2 or 6.0 CM all right so we've solved for Q 6.0 CM we can also solve for the magnification magnification equals Q over P which in this case is going to be negative and in this case 6.0 CM the object distance was given as 12 CM we can see this is going to equal negative2 all right so if we didn't already realize this is a real inverted image we'd now know this because Q is a positive number that's a real image so because magnification is a negative number that's an inverted image so and there we go the fact that the absolute value is uh less than one also tells me that the image is smaller than the object had it been asked all right so the second question should look nearly identical the first one except we're placing object at a different distance from the mirror so it says an object is placed 6.0 cm in front of a concave mirror in parentheses FAL 4.0 CM where is the image located is it real or virtual upright or inverted what is the magnification of the image so it's pretty much the exact same set of calculations except in plugging in for p it's not going to be 12 CM anymore it's now going to be 6 cm instead so 1 over 6 + 1 Q = 1 4 so if I solve for Q 1 Q is going to equal 1 4us 1 over 6 get your common denominator which in this case your lowest common denominator would be 12 you could also technically use 24 but I like 12 and so to get that I'm going to multiply the first term by 3 over 3 second term by 2 over 2 we're going to have 3 over2 - 2 over2 which is 1 over 12 so in this case 1 Q = 1 over 12 and easy to see that Q is going to equal 12 CM your magnification is equal to Q over P which in this case is going to be 12 CM over 6 cm which is going to get us -2 all right so once again we are still at 6 cm further than the focal distance on a concave mirror so we should still get a real inverted image so we can see that so with Q being positive that's a real image with the magnification coming out negative that's an inverted image so the big difference here is with it being -2.0 so because the absolute value is bigger than one that means the object I'm sorry the image is now larger than the object in fact it's double the size of the original object so this is something that's now being magnified overall all right third question is going to be really similar to the last two but we're going to move it in a little bit closer and this time it's going to be within the focal distance question says an object is placed 2.0 cm in front of a concave mirror focal distance equals 4.0 CM where is the image located is it real or virtual upright or inverted what is the magnification of the image and so once again so what's going to change is p and p is now going to be 2.0 CM + 1 / Q = 1 over 4 still and now things have changed we're now closer than the focal distance and for a concave mirror if we're closer than the focal distance we no longer going to have a real inverted image we're going to have an upright virtual image as instead and so we could answer the qualitative Parts this question before ever doing the calculation if we remember that pattern but let's solve for this one more time and so in this case 1 / Q is going to equal 1 over 4 - 1/ 2 lowest common denominator here is going to be four multiply this term by 2 over 2 and we're now going to have 1 over 4 - 2 over 4 and so 1 / Q is going to equal 1 over4 and so Q equals -4.0 CM your magnification which is equal to Q over p is going to equal negative * -4.0 all over and again your distance here is always positive so 2.0 but a negative times a negative means we're going to get a positive magnification in this case positive 2.0 and we can see things have changed here so first off your Q value came out to be negative so that means your image here is forming behind the mirror not in front of the mirror so it is a virtual image in this case having a negative value of Q also your magnification came out positive which means it's going to be upright not inverted but again once you knew it was virtual you knew it was going to come out up right those always go together and then again your absolute value of magnification is larger than one so this is an image is going to be larger than the object and in fact it's going to be twice as large as the object all right so our final mirrors calculation here uh question says an object is placed 2.0 cm in front of a convex mirror focal length is given as -4.0 CM where is the image located is it real or virtual upright or inverted what is the magnification of the image so same set of questions now one once you realize this is a convex mirror couple things you should know one the image is going to be upright and virtual every time and that your image is going to be smaller than the object every time as well so those are always true for a convex mirror one other thing is you should always remember for a convex mirror the focal distance is going to get plugged into this equation as a negative number even if they don't give it to you as a negative and sometimes what they'll do to trick is they'll give you the radius of curvature and they'll just say the radius of curvature equals 8 centimeters and you'd have to remember like oh well focal distance should be half that but half that and negative even if they didn't give you the radius of cature is negative which they kind of probably should have but if they didn't you still have to know the focal distance should be negative for a convex mirror all right so let's do some pluggin and chugging here so one over p and once again it should look a pretty similar to this last calculation here so 1 over 2.0 CM + 1 / Q = 1 not over 4.0 CM but over 4.0 cm and so in this case we're going to have 1/ q equals -4 - 12 so 1 over Q you got to get your common denominator again which is still going to be four got to multiply the second term here by 2 over 2 and so you're going to have - 1/4 - 24s which is going to get you - 34s which which means that Q is going to equal -4 over3 if we go to get our magnification it's NE Q over p in this case is negative of -43 all over our object distance of two which is going to get us positive 46 or positive 2/3 okay so Q came out negative again confirming what we already uh knew should be the scenario with a convex mirror that we're going to have a virtual image so if you have a virtual image or magnification came out positive which means the magnification or the image is going to be upright so and again the magnification has an absolute value less than one which means the image is indeed smaller than the original object confirming all of our suspicions regarding a convex mirror if you have found this lesson helpful consider giving it a like happy studying