 So, this is the lens suppose ok, notice this lens what is this, this is what aperture that is it is aperture ok, a lens will look like a you know circular thing have you seen lens how you burn piece of paper using lens right, so it is a circular thing ok, usually what is this principle, this is principle axis ok, but how you construct the principle axis, what is the method of identifying principle axis what it is, you have two radius of curvature suppose this is mid up of sphere of radius r 1 and that is mid up of sphere of radius r 2 ok, so you can say that this is a part of this is sphere and this surface is a part of that sphere are you getting it ok, so radius of curvature r 2 is for this, so centre of curvature is suppose here, so this is let us say c 2 ok and for this surface centre of curvature is suppose here this is c 1 let us say this is c 1 that is c 2 ok, so a line which joins both the centre of curvature is principle axis and now notice ok, what we call this point as this is optical centre suppose here refractive index is mu 1, here also we will assume it to be mu 1 and inside the lens let us say refractive index is mu 2 and you know what is this distance, this distance that distance or this distance, does it matter no because our assumption is it is a thin lens it is like a straight line, so when you are measuring horizontal distance along the principle axis I just need to touch one of the points of the lens this distance is no this is r 2 this centre of curvature is for this radius are you getting it ok and this one will be r 1, situation anything else if you think of yeah I mean what can affect the light other than these, no focus can be derived can focus these things yeah, so a light that comes of course parallel to it how much it will deviate does it depend on variables ok, so whatever you have written here they are geometrical or deflection so these are the causes what are the causes mu 1 mu 2 kinds of deviation are you getting but I need to find out a unique type of deviation which I can say this is how much powerful this lens is this is how much it can deviate, so you need to choose a particular type of range, so just like you consider the parallel range with special ones and see where it will intersect and that becomes a focal point here also what happens to the parallel range so what happens to the parallel range it will deviate and this is a number is less can I say distances so can I say deviation is more, so I will track this distance this is also called focus a focal length are you getting it fine, so the deviation can be quantified in terms of how much is the focal length are you getting it okay, so you will need the principal axis is that it that point distance from the optical center here okay, actually what I am doing I am stating the obvious things in a more elaborate way are you getting what I am saying okay, so let us try to find out how much is the focal length in terms of mu 1 mu 2 so you have spherical interface to find out where this this parallel range of lines will hit the principal axis, so basically what you are doing you are trying to find image of an object which is kept at question of image that is your focal length okay try to do this see first you use with sorry for the first interface between mu 1 and mu 2 then you use that second interface to find the next image for let us say for interface 1 you have this interface formula n2 by v minus n1 by u equals to n2 minus n1 by r, what is the value of n2 for this interface this is the first interface what is n2, n2 is mu 2 right, so this is mu 2 divided by I don't know where is the image let us say v1 minus n1 is what mu 1 u is what infinity so I put infinity here this is equal to mu 2 minus mu 1 divided by r1 is this in here here for interface 2 object for second interface what is u for the interface v1 because second interface will see as if line is coming from v1 why because the image of the form is what the second interface will see okay the observer is seen here is what is the image after second interface are you getting the point observer sees what is the image after second interface but second interface sees the image of first interface so image of first interface is the object of second interface okay and what is n2 for the second interface mu 1 good this is mu 1 by v1 let us take it as v ruling because that is a final image minus n2 is mu 2 now sorry n1 is mu 2 now mu 2 by object distance is v1 this is equal to mu 1 minus mu 2 divided by what why plus r2 r2 is like this only right so this is r2 but it is plus r2 only you are not using sign convention have you use sign convention had you use sign convention it would have been minus r2 but you are not using sign convention why because they are variables once you know the value suppose r2 is 10 centimeter then you put value of r2 but when you keeping it as a variable you keep it as r2 only because the lens need not be like this lens can be you know like this also then you will say minus r2 sorry plus r2 write down the same minus r2 right so I will use sign convention when I am putting the value of r1 sorry value of radius okay so let us put this as r2 now when you add these two equations what do you get this is 0 is it okay and then you add this thing will go away in and out so you will get mu 1 by v is equal to mu 2 minus mu 1 1 by r1 minus 1 by r2 what is v v is f that is how focal length is defined is it so I can say that 1 by f is equal to mu 2 minus mu 1 divided by mu 1 divided by 1 by r1 minus 1 by r2 okay so basically the left hand side is the effect and right hand side is the cause so you got a relation between cause and effect how much infractive index or how much radius of curvature will cause how much deviation something like that you got the relation here fine okay and that effect is quantified as focal length so basically you have defined focal length in terms of radius of curvature and refractive index okay fine now for fixed r1 and r2 if you fix r1 and r2 what happened this bracket thing becomes constant okay you fix r1 r2 and you also fix mu 2 you are just changing mu 1 okay so if mu 1 is less than mu 2 f will have some sign plus or minus okay and if you then increase mu 1 to more than mu 2 f will change the sign okay so just by changing the refractive index of the surrounding you are able to make a converging lens a diverging lens and a diverging lens a converging lens you are able to switch the sign of the focal length itself fine so there is no direct correlation just like you had in case of mirror what you had in mirror f is equal to r by 2 that kind of correlation is no longer for case of lens because it depends on not only radius of curvature but also depends on refractive index not only refractive index of the lens but refractive index of the median forms okay and again I am reiterating here that our assumption is both sides of the lens have same refractive index okay and that is usually true because whatever like lens you are wearing you have both sides air only and when you are using microscope telescope then also both sides same refractive index okay but if suppose there is a numerical that comes in which one side is filled with water suppose this side is filled with water and this side is air then you can't use all this then you use spherical interface formula one interface then other interface are you getting it okay this formula is also called lens maker formula so what we have done now is derived we have derived the lens maker formula so it may come it basically relates focal length with the parameters of the lens but now let us derive lens formula we derive lens formula then we take a few numerical the previous one was lens maker formula this one is lens formula you might have already learned about lens formula in class 10 1 by v minus 1 by u is equal to 1 by f okay we are going to derive it now so draw a lens like this this is a principal axis this r1 radius r2 radius okay and then suppose you have an object here this is refractive index mu 1 and of course that is also mu 1 and this is mu 2 okay try to use spherical interface formula and write two equations you don't need to solve them just write down the two equations okay so this is the spherical interface formula okay fine now object distance will be given usually so I will assume that would be u okay so what is n2 for the first interface mu 2 and first image let us say is v1 I don't know where it is and I am not even interested in v1 it is just an intermediate image that is an object of second interface okay so mu 2 by v1 minus mu 1 by u is equal to what mu 2 minus mu 1 by r this is for the first interface okay n2 is what mu 1 divided by final image minus n1 is mu 2 so mu 2 divided by object which is v1 image of the first is the object of second mu 1 minus mu 2 divided by r2 this is r1 so you have first interface 2 like this okay now you add these two equations when you add it these two will go away so you are left with mu 1 by v minus mu 1 by u is equal to mu 2 minus mu 1 1 by r1 minus 1 by r2 this is what you get okay so 1 by v minus 1 by u is equal to mu 2 minus mu 1 divided by mu 1 into 1 by r1 minus 1 by r2 now what is this right hand side is it 1 by f right so we have derived the length formula 1 by v minus 1 by u equals to 1 by s okay sign convention for this remains exactly same that you have followed for the mirror okay so you got a formula for horizontal distances along the principal axis any doubt fine so this is for horizontal distances right now and you find out formula for vertical distances and formula for vertical distances takes you know that they will try to find out through multiplication okay so if you know that image will be if you know exactly maybe here then how many rays you should know you should have to look at the image only one you already know here with the image somewhere along this line you just need to know one ray wherever it interested this line that is the image but wherever is the image it should be around this line and one line if you pass through the optical center it will go undiluted that that's how we take in case of lens that passes through the optical center goes undiluted okay fine actually in real case scenario it will be slightly shifted but then we assume that the lens is very thin so there is an assumption because of that we can see that passes through optical center goes straight any doubt on this let us say this is E this is Q this is R S and T usually it is similar triangle this triangle and that triangle they are similar so I can use ratio sites why I am using ratio sites because this and that they are one object and image distance this I am getting from here so in terms of horizontal distances I will try to find vertical distances okay let us say height of object is h o and height of image is h i so I can say that h i divided by h o is equal to Q S divided by Q it is correct it need not be it could be that correct image gets formed so I need to consider a sign convention here also to have a generalized distance so when I write h i I will use it as minus because I have to assume upward as positive then downward becomes negative if one is positive other become negative so with sign convention I can write height and with sign convention I write value of Q S also if Q S is V it is V P Q with sign convention becomes minus U okay so h i by h o will become equal to V by U and this is what the magnification formula is magnification in case okay so we have got a formula for horizontal distances and this takes care of vertical distances and also it is going to be like this