Cartesian sign conventions mirrors

in a previous video we derived the expression for calculating the image distance and the image height the only problem with this formula is they may not work for all the cases because we derived it for a concave mirror for a real image they may not work for a virtual image we don't know they may not work for a convex mirror again we don't know in this video what we'll do is we'll use a trick and convert these specific formulae which works for specific case into a general formula it'll work for all the cases for any mirrors that we want convex concave virtual real anything we will not explore why the trick works because it's just a convention will not explore that would we'll see what that trick is and how to use them in problems and solve them now the trick is this let's must get rid of this rays we don't need them all right we'll imagine that this whole setup is on a graph sheet okay and whenever we have a graph sheet you will have an x-axis and a y-axis our x-axis is going to be the principal axis over here and wherever the x and y axes meet we call it as the origin right so we're going to always choose our pole as the origin so let's write that down so that's one thing to always remember the pole is going to be the origin and in this formula instead of thinking of these as distances will imagine these are the values of the positions so f can be thought of as the value of the position of the principal focus similarly V is the value of the position of the image and u similarly is the value of the position of the object the reason is if we think of it think of these as positions then in a graph positions can have both positive and negative values so these numbers become sign-sensitive so if we build this formula with appropriate signs that we have over here then we'll end up getting the general formula now to get the signs we have to choose one direction as positive and one direction as negative right well we're always going to choose in any problem we'll always choose the incident direction as positive alright so over here knot is the object is here and you know the middle is over here and therefore the rays of light the incident rays are going to be towards the left which means in this drawing we will choose the left direction as positive and what this means is all the positions to the left of the origin will be positive positions and all the positions to the right of the origin will be negative positions now all we have to do is write this formula with their appropriate signs and that formula will then be the most general formula it can be used anywhere all right so let's include signs over here let's start with the principal focus this focal length represents the value of the position of the principal focus but notice the principal focus is on the right side which is negative so this value would be negative so this is a negative value similarly what can you say about we V is the position of the image which is over here again that's negative U is the position of the object which is again negative so now that we have included the signs over here this now represents the most general formula but of course if you have negatives everywhere you can just multiply the whole thing with minus sign and get rid of the negative sign but whatever it is we have now taken care of the sign and therefore this now is the general formula so we're going to remember this as our mirror formula and we can use this formula for any cases and we'll look at a couple of examples mirror formula similarly they'll do the same thing over here when it comes to Heights we're now dealing with the vertical direction right so these are the vertical direction so we need another sign convention for the vertical we're always going to choose all the positions a bow or all the heights above the principal axis is positive so let's do write that above the principal axis we'll call it as positive Heights and below the principal axis we'll call it as the negative Heights okay so let's now put signs over here and make this formula general H I is the height of the image that's below the principal axis so that's negative H o is the height of the object which is Bhalla principal axis so that's positive I'll not write that V is the image position and when it comes to position we have to come back to our horizontal axis and that's negative image position is negative U is the object position which you already know again it's negative we can this negative this negative cancels so this goes and if you multiply the whole equation by minus one we can bring this negative sign over here and there we have it this now is the general formula which can be used for any mirrors and any cases and by the way this ratio of height of the image to height of the object we give a name we call that as the magnification so if the height of the image is five times the height of the object then magnification will say is five makes sense right five times more and so we call this formula as the magnification formula which can be used now for any any general case so this is magnification formula alright let's take a couple of examples to get the hang of this let's look at the first one so we have an object in front of a concave mirror we are gonna solve this we'll just see how to you know how to use this formula with sign conventions so we have the object distance we know the focal length we need to figure out the image distance so let's use the mirror formula with science to do that first we have a treat we first were to find our origin our origin is always going to be the poles at this point is our origin the next is to figure out which direction is positive and which direction is negative well we'll always choose the incident direction to be positive over here the incident direction is towards the right and therefore the right side is positive so all the positions to the right over here are going to be positive positions and all the positions to the left of this are going to be negative positions so these are going to be negative positions all right now let's substitute over here and see what we get well 1 over the focal length the focal length is 20 centimeters but notice that the principal focus on the left side so that's negative 20 that's going to be equal to 1 over the image distance well we're not going to put any signs for the image because we don't know what it is we'll get the signs in the answer finally plus 1 or you use the object position notice the object is over here it's negative so the object position is 5 centimeters minus 5 and that's it if once we have this it's all algebra you just have to figure out what V is and we can do that but will not do that over here and if we turns out to be a positive number it means it's on this side and if we turns out to be a negative number it means on this side and that's how we can interpret the signs of our V and once we get the value of V we can plug it in this equation and we can calculate what the value of the height of the images similarly if the height of the image turns out to be positive it means it's above the principal axis it means it's erect and if the height so it's a virtual image and if the height turns out to be negative that means it's below the principal axis that means it's inverted and that's a real image so just by using science we can understand all the properties of the images okay let's take one more example this time we have a convex mirror and again an object is kept in front of it it'll be a great idea to pause the video and see if you can try to write the mirror formula with the signs substitute over here with the signs yourself all right let's do this again here's our pole so that's going to be our origin then incident Direction has to be positive so over here the object is here and the mirror is over here so the incident direction too will be towards the left therefore all the positions to the left of the origin are going to be positive positions and all the positions to the right of this is going to be not going to be negative positions okay let's substitute 1 over F 1 over the focal length well focal length is the position of the principal focus now the principal focus on the positive side so this focal length is going to be positive number so 1 over 15 that's going to be 1 over V again I don't know what the value of 3 is plus 1 over U or not is U is the object position it's on the near side so that's going to be negative position so minus 10 centimeters and there we have it again we can solve for V and depending on the values of the V we can understand whether it's on the positive side this side or on the negative side and similarly we can calculate the height and do the same thing so we'll solve more rigorous problem in another video