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### Course: Optics (Essentials) - Class 12th > Unit 3

Lesson 4: Lens formula to quickly find where the image will be formed- Thin lens formula
- Solved example on lens formula
- Using magnification formula for lenses
- Using the lens formula
- Convex and concave lenses
- Virtual Object
- Thin lens formula - virtual object
- Image formed by multiple lenses
- Worked example: Image formed by multiple lenses

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# Solved example on lens formula

Using the lens formula can be tricky because we need to substitute using signs. In this video, we will take one example and discuss how to use the lens formula. Created by Mahesh Shenoy.

## Want to join the conversation?

- what is lens formula(1 vote)

Here,`1/f = 1/v -1/u`

f= focal length of lens

v= position of image

u= position of object

Hope it Helped!!(2 votes)

- At1:40of this video

I do not understand. Do you just estimate 15 and 6 cm or with a ruler?(1 vote) - We already applied sign convention while creating the formula. Then why are we assigning negative values of terms to apply the formula. Isn't it already done?(1 vote)
- I recommend checking out Sir HC Verma's video on this question (YouTube). He has explained it in a wonderful manner.(0 votes)

## Video transcript

let's solve a numerical on lenses so we have an object of height six centimeters is kept 15 centimeters in front of a concave lens its focal length is 30 centimeters find image distance image height and the nature of the image meaning whether it's a real image or virtual image the first step to solving any physics problem is always to think about what is the situation what's given over here and then try to gather data from that and then see how we can use that data to find out what's asked so what's the situation so let's read this problem one more time to figure out the situation we have been given the height of the object we know the height to be 6 centimeters it's kept 15 centimeters in front of a concave lens which means we're dealing with a concave lens and there's an object kept 15 centimeters in front of it and it's focal length is given to us the lens has a focal length of 30 centimeters all right the next thing I will immediately do is draw this because drawing things will immediately help us understand what's going on all right so pause the video and see if you can make a rough sketch of this entire situation all right let's see we have been given a concave lens so the first thing I draw is a concave lens it's focal length is 30 centimeters what that means is the principal focus on either sides is 30 centimeters far away from the lens so this distance is 30 and this distance is 30 centimeters then we have an object of height 6 where it's kept 15 centimeters in front of it so let's try an object here is our object it has a height this height over here this height is 6 centimeters and it's kept 15 centimeters in front of it I know it's inside the principal focus because this is 30 centimeters now you don't have to draw it with such precision but there it is all right now the next step is to figure out what we need to calculate we need to calculate the image distance image height and the nature way to basically figure out all the properties of the image so let's let's now make some space we don't need to read the question anymore all the data is there in this diagram that's why I love to draw diagrams when it comes to lenses so we need to figure out image distance and image height now when it comes to distances and heights in lenses there are only two formula that comes to my head so the only two formulae I remember when it comes to lenses and that is the lens formula and magnification formula let's first go ahead and write that down so lens formula is 1 over F equals 1 over V minus 1 over u and the magnification formula is M is the height of the image divided by the height of the object equals V over u V or you okay so in this image distance is V that means we need to figure what V is figure out what B is image height is H I so we need to figure out what H is and then somehow we need to figure out the nature of the image whether it's real or virtual as well all right so let's let's write down the data object distance is given to us that's you that is 15 centimeters focal length is given to us that's F that is 30 centimeters and height of the object is given to us height of the object is given to us that is 6 centimeters let's see if we have everything I have U and F U and F I can substitute that I can get V once I have V I can substitute here I know H oh I can calculate H I I'm pretty much done but there's one thing we are forgetting and that is sign conventions remember when we're substituting the values over here we have to substitute with signs and what is the sign convention let me quickly tell you the sign convention we choose is we are always going to first draw an incident ray I like to draw an incident ray this is the incident direction so once you figure out what is the incident direction the right side we're going to choose that direction as positive our incident direction is to the right which means we choose right side as positive which means all the distances to the right side of the lens is positive all the distances to the left side of the lens is negative so from that since our object is on the left side I should have really put down over here negative 15 so please don't forget to put your signs over here what about the focal length this is always tricky for me focal length because there are two principal force I should I use the positive one or the negative one in such cases I always like to draw ray diagram so I remember that when you have a parallel ray of light that after hitting the lens it diverges away and it diverges in such a way that it appears to come from here so clearly it's very it's very clear to me after I draw this that is this principal focus that we're dealing with and as a result the focal length will be calculated in this side not this side and therefore we're going to use the negative focal length so the focal length is going to be minus 30 and then when it comes to height also we have sign conventions we choose the upwards as positive and downwards as negative since our object is above the principal axis it's upwards it's positive and by the way almost all the time you will get object to be positive because almost all the time you're gonna keep object above the principal axis all right so now that we have this widgets can go ahead and substitute and plug in we're pretty much done with all the physics now all we have to do is plug in and do the algebra so great time to pause the video and see if you can do this yourself all right let's do it again I now don't need to look at that image anymore because the physics is done all laugh to do now is plug in so let's go ahead and plug in these values we get 1 over F which is minus 30 is 1 over V which I don't know and the things which I don't know I don't even care about their signs I just keep it as V okay only the values you know that you have to substitute with signs so you is minus 15 so minus fifteen comes here you get one over minus thirty one over V plus 1 over fifteen the negative negative times Kansas and becomes fifteen and you know what I'm gonna put this negative sign on the numerator we can do that right and now since I want to isolate one over V I'm gonna subtract one or fifteen on both sides I'll just write that down take the entire equation this is algebra this is pretty much algebra and subtract one or fifteen on both sides so what do I end up with well let's start I'll end up with a one or minus one over so minus one over thirty and then this additional minus one or 15:01 or fifteen and that will give me one over V and the one over fifteen 1 over fifteen cancels out okay now we're gonna take a common denominator I'm gonna stick to purple now the common denominator is what 30 and 15 that's 30 so I get minus one here 15 goes 15 is 2 times 30 so I'm gonna multiply the numerator and denominator by 2 so I'll end up with minus 2 over here so that gives me minus 3 over 30 and 3 goes 10 times so that gives me minus 1 over 10 and this is not the final answer because this is 1 or V we want to find out what V is right so V is the reciprocal I have to take reciprocal on both sides so if I take reciprocal of this I get minus 10 don't forget the minus sign minus 10 and what's the unit well note is everywhere the units are centimeters centimeters centimeters centimeters so the units over here is also going to be centimeters most of the time I'd like to actually substitute the unit's over here but since I didn't want to crowded that over there I didn't want to crowd it over there I just kept it as it is so interest so this is our image distance so we found our first thing that we wanted to find image distance is done so next step is to find the image height and that's over here so let's do that over here image height hatch i / h o is six head Jo is 6 6 centimeters okay let me just substitute over here I should have turned that over here as well but anyway the substitute is equal to V / u if you substitute with units will have no confusion finally what the units will be so substituting with units is good so let's substitute over here you get minus 10 centimeters / u U is minus 15 centimeters so from this what did I get what do I get so centimeters cancels out I can multiply the whole equation by 6 centimeters because I want to get rid of 6 and so that will give me 6 centimeters here so I would get minus 10 times 6 divided by minus 15 and we have a centimeter centimeter okay so let's simplify that Phi goes 2 times 5 goes 3 times 3 goes 2 times and we are done so height of the image is mm little minus sign the minus cancels for centimeters the negative negative Kansas negative negative cancels so 4 centimeters and we are done over here as well so we got two things we found out the image image distance and we also know the height of the image so let's go back and let's see what that means let's go back over here we found the image distance to be minus 10 centimeters what does that mean well so let me just write that over here image is minus 10 and we got the image height as plus 4 minus 10 minus is on this side that's telling me that the image is somewhere over here because this is 15 10 centimeters from here so that's going to be somewhere over here and the image height is 4 centimeters which means it's smaller so it would be I don't know I'm just is kissing so somewhat like this I know the whole diagram looks a little bit messy but that's what it's get telling me and let's see if that let's see if this makes sense to us first of all because we know we can we can figure this out by drawing ray diagrams a secondary I would usually draw is the one that goes through the principal focus and sorry the optic center and the two rays of light are appearing to come from here so it makes sense we would expect the image to be smaller than the object that's exactly what we're getting we also expect the image to be closer to the lens from the ray diagram and it's also we're getting over here so things make sense over here all right the last question that was asked is what is the nature of the object of the image now the nature is whether it's a real image or virtual image and when it comes to this I like to think of it in different ways one way is we can just we can just look at this and we can tell it's a virtual image how do I know that it's a virtual well one reason is because it is erect all erect images are virtual remember that another reason is I know that the rays of light are diverging away and when the rays of light are diverging they well then they appear to come from a point and that point is always a virtual image so that's another reason this is virtual image but even without the ray diagrams we can think about whether it's a real image or virtual image if we don't look at the diagram you just look at the height the height is positive positive height means it's above the principal axis and above the principal axis means erect so just by looking at this answer we can also say that it's a virtual image so anyways you like to think about it the image is virtual all right so whenever we have lenses and objects kept in front and we are asked to figure out the properties of the images this is how we usually solve the problem