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Current time:0:00Total duration:8:53

Video transcript

let's say we have a concave mirror of focal length 2 centimeters and in front of it we're going to keep an object and let's say that the object is kept 6 centimeters in front of the mirror where will its image be in this video we're gonna find out exactly where that image would be all right now in some of our previous videos we actually discussed about this we've seen that there are a couple of ways you can solve this one way would be to actually you know draw this on a piece of paper and then use your ruler to figure out exactly by drawing ray diagrams figure out exactly where the image is going to be but if we don't want to do all that all those drawings then another method would be to build a master formula a master formula which connects the object distance the focal length and the image distance and the one which will work for all cases for concave and convex mirrors and we can just substitute in that and get our answer so in this video we're not going to build the master formula if you are interested in figuring out how to derive that master formula you can there's a bonus video and you can just go there and watch that but we will see what that master formula is we'll understand it and then we'll use it to solve the problem all right so that one formula which is going to help us figure out so our solve problems like this that formula looks like this and it's called the mirror formula for obvious reason because it works for mirrors and let's see what these things are there are three things you can see f stands for the focal length so that's something we already know V is the image image distance so this is image distance and that's something we don't know in this problem and U is the object distance these are the letters that we usually use for mirrors object distance and one super important thing to remember is that this is a sign sensitive formula which means when we are substituting the numbers into it we need to take care of their science they can be positive values and negative values and to quickly recall why we do that that's because we need to differentiate between measurements done in front of the and measurements don't be high in the mirror we assign one of those measurements as positive the other measurements as negative and so to quickly recall how we how we figure that out which is positive which is negative we always choose the incident direction as positive so notice that here is the object and here's the mirror and the incident ray would be this way so incident direction right side is positive this means that all the positions to the right side of the mirror will be positive positions anything on the right side of the mirror will be positive and anything to the left side of the mirror will be negative positions and if this thing is not super clear to you then it'll be great to pause this video and go back to previous videos we have discussed this in great detail watch that and then come back over here alright now that we have our signs ready let's let's look at what we have we need to find the image distance that's what we need to figure out so this is what we need what do we know we know the object distance that is 6 centimeters right object is 6 centimeters from the mirror and so you is 6 centimeters but remember we need to take care of the signs notice that the object lies on the negative side because all the positions here are negative so you will be minus 6 centimeters then what else do we need we need focal length we can see the focal length the principal focus also lies on the negative side and therefore this focal length will be negative 2 centimeters hope that is clear anything that lies on the negative side will get negative distances all right ok so we have two things f nu and we need to figure out what V is and so now then whatever follows is just going to be substitution and algebra so great idea to pause the video and see if you can do the algebra yourself now before you pause a quick thing I want to emphasize on what these negative signs mean in this particular example so once we have these sign conventions when we substitute the negative six over here we are telling the formula that the object lies on the negative side which means in front of the mirror if we had substituted plus six my mistake then the formula would assume that the object lies on this side which is behind the mirror now I know that sounds silly but that's what the mirror that that's what the formula would assume and it will probably get the wrong answer similarly when we substitute the negative two centimeters over here we're telling the formula that we're dealing with a concave mirror how because only if it is a concave mirror the focal point will lie on the negative side at least in this example and if we have substitute + 2 centimeters over here then it will assume that the focal point is on this side meaning it would assume it's a convex mirror again it would give us a wrong answer just telling you how important it is to you know to include proper signs in our formula all right pause and compose and try to do the algebra yourself now all right let's do this so if we substitute we'll get 1 over F which is negative 2 that will equal 1 over V plus 1 over U which is negative 6 now I'm not gonna substitute the centimeters because everything is in centimeters so my image distance will also be in centimeters I like to put them negative side negative sign on the numerator circle as minus 1/2 equals 1 over V negative times positive is negative so minus 1 over 6 excellent now I need to isolate 1 over V because that's what I want to figure out so I'm gonna try to get rid of this negative 1 over 6 by adding 1 over 6 on both sides so if we add 1 over 6 on both sides on the left hand side we are adding 1 over 6 we will also add 1 over 6 on the right hand side and when we do that that will cancel with this one or negative 1 over 6 so we will end up with 1 over V ok so let's solve this we have to add two fractions we're gonna make the common denominator 2 & 6 LCM is 6 and so I'm gonna multiply this by 3 multiply this by 3 I'll get negative 3 plus 1 plus 1 that's gonna be negative 2 so we can just write that down somewhere okay minus 2 over 6 and that's 1 over V I know it's getting a little crowded but you know this it's almost done - then we can just cancel this 2 goes 1 times 2 goes 3 times and so we have 1 over V equals negative 1 over 3 so we want V so we'll do the reciprocal V will be negative reciprocal of this 1 over 3 reciprocal is 3 and so we get negative 3 centimeters there we have it we've found what V is so what this means is that the image is 3 centimeters away from the pole from the center of this mirror remember whenever we are using this formula all the distances are from the pole so 3 centimeters but which side well the sign tells me it's on the negative side and so I know it's in front of the mirror right so it would be somewhere over here because this is 2 it's a given so this would be somewhere over here 3 centimeters and there we have it we have solved our problem so to quickly summarize we learned what the mirror formula is which connects focal lengths object distance and image distance and to use this formula we have to remove to use our sign conventions and the sign convention that we use is we consider the incident direction is positive and then the other direction becomes negative now before we wind up one thing I need to mention is something that we should always remember is that a sign convention that we are using is just a standard if you ever forget that don't worry you will still get the right answer so what I mean is if by mistake when you're solving this problem if you chose the left side is positive and the right side is negative but you substituted your values accordingly then you would still get the right answer you would eventually get that the image is 3 centimeters in front of the mirror but usually we like to choose the incident direction as positive and we like to stick to this sign convention