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### Course: Class 10 Physics (India) > Unit 1

Lesson 13: Lens formula & magnification# Lens formula

Let's see how to use lens formula (1/v-1/u= 1/f) to locate images without having to draw ray diagrams. Created by Mahesh Shenoy.

## Want to join the conversation?

- why is the mirror formula and the lens formula totally different and just opposite of each other's formula and even the magnification formula. pls clarify(1 vote)
- Yo,

check out

https://youtu.be/PgQJCm_sXVY

*mirror formula derivation*

https://youtu.be/x9F0AQCGkbc

*lens formula derivation*

After you see those 2 videos (*by khan academy*)

I am sure you would understand,

But the main reason (spoiler to those videos) to the difference is that

when we derive the formula,

we shall obtain the following**Concave mirror - along with sign convention**`v = -ve, f = -ve, u = -ve`

1/-f = 1/-u + 1/-v

-(1/f) = -(1/v + 1/u)

1/f = 1/u + 1/v**Convex lens - along with sign convention**`v = +ve, f = +ve, u = -ve`

1/f = 1/v + 1/u

1/f = 1/v + 1/-u

1/f = 1/v - 1/u**We apply sign convention to make the equation obtained by similarity of triangles to make it general**

as the signs for f and v are opposite with respect to concave mirror and convex lens the difference arises

Now try out for*the magnification formula*as well

Hope this helps,

If I'm wrong do let me now

Ciao*for now*(5 votes)

- But in our school,they don't teach us with a minus before the 1/u, it's basically just like the mirror formula. I'm kinda confused.(1 vote)
- I am little confused while using sign convention for focal length of lenses .

Let say focal length is 6 cm, since we have f1, f2; which sign convention will be used for this focal length.(-ve or +ve).(1 vote)- look at where the image is forming

if the image is on the other side of the lens it is positive(1 vote)

- *It has been taught to us before...*(0 votes)
- in the previuos videos we have learnt mirror formula which is diiferent thing(1 vote)

## Video transcript

let's say we have a convex lens of focal length five centimeters and we keep an object six centimeters in front of it how would you find where its image is going to be formed well four lenses there's only one thing we know to do that is to draw ray diagrams so maybe we draw this whole thing on a piece of paper by doing measurements using a ruler and then we draw a couple of rays wherever these two rays meet that's where our image is going to be and then we can use the same ruler to figure out exactly how far that image is from the lens but this is pretty tedious because we have to draw this whole thing accurately and then do measurements and especially if we have multiple problems it's going to take a long time so the question is if object distance and focal length is given to me how do I figure out the image location without having to draw anything and that's what we're gonna find out in this video so to locate our image without having to draw the ray diagrams you may guess that we might have to use a formula and that formula is called the lens formula and it looks somewhat like this so we have F which is the focal length V is the image distance so let me just write that this is the image distance and u represents the object distance object distance and all these distances are measured always from the center of the lens or optic center oh and if you've studied mirrors and mirror formula then you would see this is very similar to that the only difference is in mirrors it is 1 or V plus 1 over U for lenses it's 1 over V minus 1 over u so when we use the formula if you know any two of these then you can find the third for example if we are given the focal length and the image distance then we can substitute and find the object distance if we know the emails distance and the object distance then we can sufficient find the focal length notice for us the object distance is given to us and the focal length is given to us so which means we can substitute and find out the image distance but before we substitute very important to remember that this formula is sign-sensitive which means while substituting the numbers we need to substitute with appropriate signs and the sign convention that we are going to use is going to be exactly the same as we do it for mirrors so let me quickly recall what that is the first thing we do is figure out exactly what direction the incident rays will be so over here the object is here and the lens is over here so if you draw any rays of light we know that the rays of light will go to the right that means our incident direction is to the right once you find that you start from the optic center and you move in the incident direction for us it's to the right and we call that direction as positive so I'm going to call this as positive which means all the positions and all the distances you measure on this side of the lens will call it as a positive distance and if you measure any distance on the left side of the lens we're gonna call that as negative distance so having said this we are now ready to substitute with proper signs so let's quickly write down the data given to us we know the object distance that's given to us that's 6 centimeters but since we're measuring it on the left side of the lens it's negative so we call that as negative 6 centimeters what about the focal length and here we need to be careful because for lens's notice we have a focus on this side and we have a focus on this side one is a positive other one is a negative which focus are we going to choose and this is where you know drawing ray diagrams help us so we don't have to actually draw it but if we just imagine just imagine as a parallel ray of light from here I will draw it over here but you know we can usually just imagine it if we draw a parallel ray of light or a ray of light that's parallel to the principal axis after refraction where would it go would it go through this focus already go through this focus well we know that since it's a convex lens it's a converging lens so the rays of light will get conver over here isn't it and since this ray of light is passing through this point of focus we know that it's this focus that we're interested in and as a result we're gonna call the focal length as positive because this focus lies on the positive side so f is going to be positive five centimeters and if this was just as an example if this was a diverging lens like a concave lens then this ray of light would have diverged and would have appeared to come from this point that's when we consider the negative focal length so for diverging lenses you will get negative focal lengths you don't have to remember this just if you remember how the ray diagram works then you can you know assign appropriate signs to this alright and what we need to find out is we need to find out what the image distance is going to be so we just have to substitute and do the algebra so great idea to pause the video and see if you can try this yourself alright let's do this so if we substitute we get 1 over plus 5 plus 5 that would be equal to 1 over V V I don't know minus 1 over U which is minus 6 and everything is in centimeters so I'm not gonna write the unit our V will also be in centimeters and after this all that follows is algebra physics is pretty much done so let's just do the algebra let's simplify this so we get 1 over V this negative times negative becomes positive we get plus 1 over 6 and so I want to figure out what 1 over V is so we'll subtract 1 or 6 from both sides so we will get 1 over 5 minus 1 over 6 equals 1 over V and the 1 by 6 and minus 1 over 6 will cancel and we just have to now take the common denominator the LCM for this is 30 so we multiply this by 6 and I multiply this by six so I get six here - to make this third day multiply this by five and I multiply this by 5 I get five and that gives me one over 30 and so there we have it so what are we is 1 over 30 which means V is the reciprocal of this and that is 30 centimeters and that's our answer that's the image distance so this tells us our image is going to be 30 centimeters from the optic center but which side will that image be because the images can be on either sides is it on the left side or the right side well for that we look at the sign of the image if the image distance was negative if this was negative 30 then it means that the image is on the left side of the lens but since we are getting it as a positive 30 that means the image distance is to the right side of the lens and so this is 5 so that it would be somewhere over here and therefore we know the image is somewhere over here this is not drawn to scale this is not very accurate but somewhere over here 30 centimeters is our image distance and we have solved our problem so to quickly summarize what we learned in this video the lens formula connects the focal length image distance and the object distance so whenever we are given any two of these we can find the third one using this formula but while substituting we have to substitute the numbers with appropriate signs and the sign convention which chooses we always start from the optic center and then in the incident direction all the distances are considered positive and finally when we get our answer the sign of that answer is going to tell us which side that measurement is