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## Class 10 Physics (India)

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

Lesson 13: Lens formula & magnification# Magnification formula for lenses

Let's explore the magnification formula (M= v/u) for lenses and see how to find the image height and its nature (whether it's real or virtual). Created by Mahesh Shenoy.

## Want to join the conversation?

- is this formula true for both concave and convex lenses? This video seems like its only talking about convex lenses(1 vote)
- What is the width of the image of an object which subtends an angle of 8 degrees at the lens which is 0.5m from the image?(0 votes)

## Video transcript

in a previous video we took a convex lens of focal length five centimeters and in front of it we kept an object six centimeters in front of it and our goal was to figure out exactly where the image would be without having to draw any ray diagrams and what we did for that is so we introduced a formula called a lens formula which basically connects the three things the focal the image distance and the object distance says that if we know any two of this we can figure out the third one just like what we have over here and so we substituted these values over here using science and then we figure out that the image was 30 centimeters from the lens in this particular video we want to figure out what is the height of this image and we also want to know whether this would be a real image or virtual image and again we don't want to draw any ray diagrams all right since we're not using any ray diagrams over here we might need a formula to tell us what the height of this image is going to be and if you look at the lens formula the lens formula has nothing to do with Heights so we might probably guess that there is another formula out there that helps us calculate the height of the image and there is and this formula is called the magnification formula we'll talk about why it's called magnification in a while but if you look at this formula notice on the left hand side we are dealing with the heights and the numerator has the height of the image I for image the denominator has the height of the object so the left-hand side is the ratio of the height of the image to the height of the object and that equals the ratio of the image distance V is the image distance and the object distance now before we go ahead and substitute let's talk about why this is called the magnification formula and to see why we just have to look at the left hand side so let's ignore the right hand side for a while so the reason why this is called the magnification formula is because this ratio itself is the magnification M which means this number tells us how big the image is compared to the object and by the way if you've studied mirrors then the concept over here is exactly the same that we discussed over there but if you are not familiar with this don't worry we'll take a couple of examples and make this thing super clear so let's take one example let's say M is 2 so this ratio is 2 what would that mean well that would mean that the height of the image so if this this value was 2 that means height of the image H I is 2 times the height of the object which means the image is twice as big as the object so doesn't that mean that image is twice magnified compared to the object and that's why this number this ratio is called the magnification it tells us how big the image is compared to the object but not only that this number can also tell us whether we're dealing with a real image or a virtual image and all you have to do for that is look at the sign of the magnification let's quickly recall the signs that we use for heights whenever we have Heights above the principal axis we usually call it as a positive height and anything below the principal axis is called as a negative ID so in our example the magnification is a positive number which means that the height of the image and the height of the object have the same signs notice that if this is positive this will also be positive if this was negative this will also be negative that means that both the object and the image are either both above the principal axis or below the principal axis so they have the same orientation or the image is erect and remember or recall that reduct images are always always virtual images so the moment we saw that this is positive this immediately tells us that this is a virtual image virtual image so let's take another example let's say in our second example M had a value of minus 0.5 could you pause the video and think about what this means to the size of the image the nature whether it's real or virtual think about this alright now if this ratio if this number was minus 0.5 then that means that the height of the image is going to be minus 0.5 times the height of the object and point five is half so this is telling us that the image has half the size of the object so that's that and look at the negative sign this is telling us that the image has the opposite sign of the object height so if this is positive this is negative if this is negative negative times negative becomes positive in other words the image and the object object have opposite orientation one is above the principal axis the other one will be below the principal axis the image is inverted and we may recall that inverted images are always real so this means that whenever the magnification is a negative number this always means it's a real number real image so long story short this number this magnification tells us two things it tells us how big the image is compared to the object and it also tells us the sign tells us whether it's a real image or a virtual image and if you think about it that's exactly what we want to figure out over here we want to find out what's the height of the image and we want to know whether this is a real or a virtual image so this means all we have to do is figure out what's the magnification so how do we calculate this magnification well I'm pretty sure you might have already guessed it we use this magnification formula and just to be clear this is not a formula you see this ratio itself is called the magnification that's a name that we're giving it that's not a formula so what we would need is a connection between this ratio this magnification and these things maybe the object distance and the image distance and the focal length because using these values I want to know what the magnification is so we need a connection between em and maybe you V or F or something like that and that's where the right-hand side comes into the picture so we can now look at the right hand side and see what the magnification formula tells us turns out that the magnification for lenses equals the ratio of the image distance and the object distance and again by the way if you have straight mirrors this is very similar to what we got for mirrors the only difference is for mirrors we had a negative sign here we don't have that and so all we have to do now is go ahead and substitute the values over here and figure out what we want but before we do that remember one small thing that these values are also sign sensitive and so how we how we take care of science for distances well we start from the optic center and we move in the incident direction and we call that as positive and over here notice the incident rays will be towards the right so in certain direction would be towards the right and so all the positions to the right side become positive positions that we hear positive positions and all the positions to the left side will become negative positions ok so now we have everything ready we can substitute figure out the magnification and then figure out the height of the image and whether it's real or virtual so great idea to again pause the video and see if we can try this yourself all right let me do this over here let's get rid of the lens formula so we need to calculate the magnification M and that equals V over u V is the image distance and since the image is on the positive side V will be plus 30 I'm directly substituting over here divided by U U is the object distance that's over here and since the object is on the negative side this object distance would be negative 6 centimeters so if we simplify centimeter cancels out 6 goes 5 times and as a result notice that we end up with a negative 5 because there's a positive and negative and so magnification is minus 5 so this means that the height of the image oops let's use the same color the height of the image is negative 5 times the height of the object and so the negative sign is telling us that this is a real image and the five is telling us that it is five times magnified compared to the object so we can now draw the image it's gonna be below the principal axis it's negative it's real its inverted and this length is gonna be five times more than this so if this is one centimeter this would be five centimeters if this is ten centimeters this would be fifty centimeters and we have now sold our problem so to quickly summarize what we learned we saw that if you want to figure out the height of the image or its nature whether it's real or what true then all we need to do is figure out the magnification the number tells us how big the images and the sign tells us whether it's real and virtual or virtual and four lenses this magnification turns out to be equal to the ratio of the image distance and the object distance