 We use sign convention once when we are deriving the general mirror formula and then we use it again while solving problems. In this video we will understand why exactly do we use sign convention toys. We will do that by deriving the general mirror formula for two cases for a concave and a convex mirror and then see why do we need to apply sign convention toys that is once in derivation and once while solving problems. So let's begin. We start by deriving the mirror formula for a concave mirror. So here is a concave mirror with pole P and this is where the object is placed. Let's say the height of the object is HO. One ray parallel to the optical axis passes through the focus upon reflection. So here is our focal point and the other ray can be incident at the pole P itself where it is incident at pole P it reflects back intersecting with the original ray. The point of intersection is the point where the image is formed in this case the height of the image is HI. Let's also label the object at the image distance and also the focal point. So there you go. This is U that is the object distance, V is the image distance and F is the distance of the focus. Now for deriving the mirror formula we look at similar triangles in this situation. So we can begin by picking up these two triangles the triangle in blue and the triangle in in light red. These two triangles are similar because one angle is just 90 degrees in both of them. This angle right here is 90 degrees and these two angles are equal. This is the angle of incidence and this is the angle of reflection. From the law of reflection they have to be equal. This ray was incident at the pole and the optical axis is just perpendicular to this point so these two angles turn out to be the same. Now because the triangles are similar corresponding sides will be of the same ratio. So H0 divided by HI. This is H0 divided by HI. This can be equal to this entire length that is the object distance U divided by this distance right here that is V the image distance. We can pick one more pair of similar triangles. So when we hide this one there is one more pair and that is right here. If we drop a perpendicular from this point on the optical axis these two triangles will again be similar. Why is that? Because these two angles are 90 degrees we have a 90 degree and we also have these angles which are opposite to each other and therefore they are equal. So we have this side this side divided by HI and this side if you see this is just equal to H0 this is the height of the object this side right here. So let me write H0 divided by HI. This is equal to this much distance distance of the focus from the pole P. Now we can see that the perpendicular is not really touching at the pole P but we can assume that these two points are extremely close to each other. So we can take this distance as Fp the distance of the focus that is and that is F. So writing F over here this is F divided by divided by this much distance and that is that is V the total distance minus F this is V minus F. Now we can equate U divided by V with this ratio right here and when we do that we can cross multiply. So on cross multiplying we will have to multiply U with the denominator here V minus F and we will multiply V with F. So on multiplying U with the denominator V minus F we get UV minus UF and this is equal to V into F we have cross multiplied the RHS of these two ratios. Now we can divide this entire equation with UVF so dividing by UVF when we do that this is 1 upon F minus 1 upon V this is equal to 1 upon U and when we take 1 by V to the right hand side this becomes 1 by V plus 1 by U equal to 1 by F. So let's let's write that let's write that when we take this to the right hand side and let's let's just write it in this way. So this this basically becomes 1 by V plus 1 by U equal to 1 by F alright so this is the equation we get for a concave mirror with a real object and a real image. Now let's bring in a convex mirror and see what do we get. So for a convex mirror here it is we have the object on the left and you have a parallel ray to the optical axis when it's reflected it bounces back but we can extend this ray and say that it is passing through the focal point it is passing through the focus and the second ray can be can be incident on the mirror at an angle of 90 degrees. So when it is at an angle of 90 degrees it will just bounce back entirely but when we extend it it will be passing through the center of curvature. These two extended rays wherever they appear to meet that is the that is the position of the image and in this case the image is virtual and the height is h i. Let's again label U V and F for this case here we have the object distance the U the image distance V and this is the distance of focus from the pole that is F. For deriving the mirror formula even for a convex case again we look at the pair of similar triangles so first pair can be can be this right here the big blue triangle and this is a smaller light red triangle. These two are similar because you have these angles in 90 degrees and there is this angle which is common okay now this would be H naught divided by H i and this is this is H naught divided by H i this can be equal to this entire length this entire length of the blue triangle and that is U this is U plus F plus F because the center of curvature is at a distance of 2F from the pole so this entire distance is 2F so this is U plus 2F this is U plus 2F divided by the corresponding side that is this one for the light red triangle and let's see what the length of this side is so this distance right here this will be F this will be F minus V and we're adding one more F we are adding one more F to F minus V to get the entire side length so this is 2F minus V this right here is 2F minus V okay let's look at one more pair of similar triangles now so let me hide this one one more pair can be can be this one right here if you drop a perpendicular from this point to the optical axis you can call this point P we can assume that these two points are extremely close to each other and they almost coincide so we can we can ignore this much distance just like we did for concave mirror so we can just call this point P so now in this case the two triangles are this bigger the bigger blue triangle and you have the smaller light red triangle these two are similar because you have 90 degree and one angle in common and this distance right here you can see that this is equal to the object length the object length that is H naught so writing this in terms of ratio this is H naught divided by H I and this is equal to we can say we can take this distance right here this is from the pole we are assuming that this point is extremely close so from the pole to the point F this is just the distance F so F divided by this small distance that is F minus V F minus V now the RHS of these two ratios are same so we can equate them again we can cross multiply so in this case U plus 2F will be multiplied with F minus V and 2F minus V gets multiplied with F because we are cross multiplying so when we cross multiply this comes out to be equal to U into F UF minus UV U into V plus 2F square 2F square minus 2F V this is equal to 2F square this denominator gets multiplied with a numerator here 2F square 2F square minus minus V into F okay 2F square gets cancelled you can see it gets cancelled right away and we can take minus 2FE to the right-hand side then it becomes plus 2FE plus 2FE minus VF is just VF so when we write this this is UF minus UV equal to equal to VF now again dividing by dividing by UVF throughout when we do that this becomes equal to 1 upon V minus 1 upon F this is equal to 1 upon U and let's take 1 upon U to the left-hand side let's take 1 upon F to the right-hand side when we do that when we do that this becomes 1 upon U equal to 1 upon F now we see that we got a different formula for the convex case and there can be many more cases we took a real object and a virtual image case with a convex mirror they can also be cases with a virtual object and turns out that the mirror formula for that case is different it would be 1 by U minus 1 by V equals to 1 by F so for different cases we are getting different mirror formulas and it becomes difficult to remember which formula belongs to which case that is a problem to solve that problem we apply sign convention we apply sign convention once so when we apply sign convention once that is we consider that the incident direction is positive when we do that let's see let's see what happens what changes occur to these two formulas for the first case for the concave mirror case this is the incident direction which is considered as positive the incident direction is to the right which is taken as positive so that means that everything to the right of this pole P will be positive and everything to the left will be negative here the object image and the focus they are on the left-hand side of the pole P so everything would be negative in place of V we write minus V in place of U we write minus U and in place of F we write minus F but turns out all the negative signs get cancelled all the negative sign gets cancelled and finally the equation that you get is 1 by V plus 1 by U equal to 1 by F and for the second case again the incident direction that is the same direction this is positive again because the incident direction is to the right everything to the right-hand side of this pole P will be positive and everything on the left-hand side will be negative so over here V is positive U is negative F is positive but notice when you take U as negative there is already a negative sign over here so it becomes positive so finally you land at the same equation you land at the same equation which is this 1 by V plus 1 by U equals to 1 by F and that is the benefit of using sign convention once because for different cases we got different formulas that is very inconvenient that is very difficult to remember so to work out this problem to solve this problem we apply sign convention once so we only have to remember this formula the general formula we only have to remember the general the general formula and nothing else and that is why we use sign convention once in derivation but we again use it in numericals we use it the second time when we are solving problems we need to do that because this was never really the equation the original equation were these this was for this case if there would have been a virtual object if this would be 1 by U minus 1 by V those were the original equations for for this one this was the original equation this was never really the original equation you want to go back to the original equation that is that is these equation right here so we are using the sign convention the second time in numericals so that we go back to these equations and how do we do that we write everything in terms of coordinates we treat this pole as the origin so everything on the right hand side of this pole is taken as positive everything on the left hand side is negative similarly in this case everything on the right hand side of this pole would be positive and everything left hand side would be negative and notice what happens when you write things in terms of coordinates we have this formula right here for the second case in place of U whatever U is we will write minus of let's say 10 centimeters 20 whatever the distance that is given minus 10 minus 20 and V and F would be just written as positive plus 10 plus 20 or whatever the values are so notice what happens unknowingly you are using the original equation when we write U V and F in terms of coordinates in terms of positions U is negative V and F is positive so this equation just becomes 1 by V minus 1 by U equals to 1 by F so unknowingly you go back to the original equation which was supposed to be used but we could not do that because it's just so difficult to remember different equations for different cases and for this one for object image and focus point because it is on the left hand side of the pole P we can treat this as an origin everything on the left hand side is negative so everything would be negative here they will just get cancelled and you will use the same original equation so we use the sign convention the second time so that we go back to the original formula and turns out the same idea is also applicable for lenses even in lenses we use sign convention toys once when we are deriving the lens formula and secondly when we are solving problems the same idea is true even for lenses we use it once to get the general formula so that we don't have to remember different formulas for different cases and then we use it the second time when we write U V and F as positions or coordinates so that we can go back to the original formula