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## Class 10 Physics (India)

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

Lesson 15: Prepare for CBSE board exam# Convex/Concave - lenses & mirrors: CBSE board practice

Prepare for class 10 CBSE exam for the chapter Light, by recalling the concepts of concave/convex mirrors and lenses.

## Want to join the conversation?

- what are the difference between mirror and lens(1 vote)
- a mirror is an opaque reflective surface, and reflects all the light back, whereas a lens is a transparent object through which light passes, and refracts(bends)(3 votes)

- When will the Values of
**V**,**U**,**F**be negative and positive(1 vote)

## Video transcript

let's recall everything about lenses and mirrors to help you in your board exams on the right you can see the index so if you want to jump to any particular topic feel free to do that by going to that time alright so let's start with the basics the naming and the properties of the mirrors and lenses so can you quickly pause and recall what are the names of these mirrors and lenses go ahead give it a try all right so the names are dependent on the shapes so if you look at the first one when you shine a ray of rays of light onto this on the reflecting side notice that it hits the cave part of the mirror and that's why it's called a concave mirror on the other hand over here when you shine light on the reflecting side you will see it hits the bulged part of the mirror and that's why it's called convex mirror of X means bulge what about the lenses well here you can shine light from any direction so if you shine light say from the right again notice again hits the bulged part of the lens and so this is called the convex lens over here when you shine light on it notice it hits the caved part of the lens regardless of which direction you hit it no that's why it's called concave lens okay now let's think about what happens to these rays of light once they hit the mirror and the lens so again can you pause and think about this remember recalling is the best way to learn for exams so pause and think about what happens to beam all right so for a concave mirror after reflection we'll find they will the rays of light will be directed towards the center they'll be focused at a point call the principal focus so this is a converging mirror on the other hand if you look at a convex mirror since its bulged out you will see it will diverge the rays of light again these rays will appear to now come from a point not really doing that but they appear to come from a point again that point is what we'll call the focus what about the convex lens well here refraction happens not reflection there is of that we just go through and they will Bend and over here in the case of convex we will again find it will converge the beam of light to a particular point the focus so this is a converging lens and if you look at this one we'll find that this will die there is a flight again appearing to come from a single point and so you can immediately see concave mirrors and convex lenss are similar in the sense they're both converging devices converging optical devices and similarly if you look at convex mirror it is similar to concave lenss says that they're both diverging optical devices and therefore when you look at ray diagrams there'll be no wonder that these will be very similar to each other similar cases when it comes to images image formation and these two will have similar cases when it comes to image formation now of course there's one big difference which was a very common mistake I should do is that and it comes to Kahn you know it comes to mirrors will find that the focus lies right in between exactly in between the center of curvature and the poke and the pole center of curvature is the center of this imaginary sphere right so focal point lies right in between the two and the same thing before will be for a convex mirror as well but that's not true for lenses if I would take the center of curvature if there is no direct relationship between the center of curvature and the focus and so that's not true for lenses mainly because over here refraction is happening and how much the rays of light Bend depend not only on the curvature but also on the material so long story short here that we will not use center of curvature because there's no relationship with focus but over here we will with that in mind let's not jump to ray diagrams let's start with concave mirrors imagine we have an object kept beyond the centre of curvature where would its image be formed we can figure that out by drawing a couple of rays so one ray of light I like to shoot parallel to the principal axis because I know that Ray after hitting the mirror goes through the focus we just saw that all Pal rays of light should go through the focus another ray of light what I could do is shoot it through focus because I know that after hitting the mirror it's gonna go parallel to the principal axis now wherever these two lines these two rays intersect we know that's where the image needs to be formed but we can also have a third ray just in case so for example I can draw another ray of light passing through the center of curvature because I know then this is where the normal lies and so any ray of light along the normal will just bounce back and so all the three rays will now intersect at this point giving you an image over there but we don't need three ways so I'm just gonna erase that and we get an image over here this is a real image because the rays of light are being focused it can be captured on a screen it's real it's inverted it is diminished so when the operator is outside it or beyond the center of curvature we get a diminish real inverted image between F and C all right can you try using ray diagrams to figure out where the image will be in these cases go ahead pause the video and give it a try all right for the second case will have similar ray diagrams one rip parallel to the principal axis another ray passing through the focus and you will see they will meet somewhere over here giving us an image which is beyond see and now enlarge compared to the object so you can kind of see as the object comes closer to the focus the image ends up going farther away but most importantly the image starts becoming larger and larger okay what about over here here again I can draw one ray of light passing parallel to the principal axis passes to the focus after reflection but I can't draw another ray of light through focus because it won't hit the mirror this is where I am gonna draw the second ray along that normal so this is the Ray along the normal which is basically passing through the center of curvature I know now that ray is gonna just bounce back and I'm going to go over here now notice that these two rays after reflection are not focusing at any point they're just diverging away they're not converging they're diverging away and so they appear to be diverging from a point behind the mirror which means now I get a virtual image this image is virtual because you can't capture it on a screen all virtual images are erect and notice the virtual image is enlarged and so it concave mirrors you can get both real and virtual images diminished and enlarged images as well all right let's now jump to a convex mirror again can you try drawing ray diagram to see where the image of this is formed go ahead give it a try okay let's see we need to be a little bit careful over here now this ray of light will not go through the focus because it has to reflect remember it's a mirror so the reflected light will reflect in such a way it appears to come from the focus remember it's a diverging mirror so it appears to come from the focus doesn't go through the focus and another ray of light what we can do is we can we can pass it along the normal basically along this line joining the center of curvature we know it's gonna hit the mirror and just bounce back and again we get two diverging rays which appear to be coming from here giving us a virtual erect image notice that over here we always get virtual directly magical I saw where you keep that object you can always test that you can try that yourself you always an image between P I forgot to mention the P over here Paul between P and F it'll always be virtual and diminished so what big difference you see between concave and convex well we see that concave can do both real and virtual convex can only always give were to the images what's the difference between the virtual images of concave and convex mirrors the concave is always enlarged as you can see we don't have to remember this ray diagrams help you but for convex mirrors it's always diminished okay now with that in mind let's jump to lenses what happens if we have an object which is kept outside or beyond to F notice remember for lenses we don't have center of curvature it's we don't use that we use twice the focal length where will the image of that object be formed now you may be wondering why do we have two values of f over here that's because you can keep objects on either side if the object is on this side the Rays of that will be coming this way and they will be converged over here so this will become the focus if the object was this side then this will become the focus and so I put this as green so this is the focus and so one ray of light we can pass parallel to the principal axis we know after reflector refraction this will refract it will pass through this focus okay a secondary of light what we can do for lenses will very something very similar to the mirrors we can pass it through the focus and we know after refraction it will go parallel to the principal axis but you know what instead of doing that what what I like I like to pass a ray of light straight through the center of the lens because the ray of light through the lens goes undeviated okay and so now I know that these rays are intersecting at this point this point over here and so that's my image and so notice we end up with a real inverted but diminished image between F and 2f okay your turn can you draw a diagrams for these two cases of convex lensses all right hopefully you have tried so for the second one it's gonna be very similar we'll end up with the real in what it image this time it will be enlarge it will be a little bit farther away from the lens but it'll be enlarged what happens over here well if you draw the two rays this time you'll find they are no longer converging at any point so we don't get a real image anymore instead they're diverging and appear to come from a particular point and so we end up with a virtual image why is this virtual because you can't capture it on a screen it's erect all virtual images are erect now you can take a moment and just compare these two you see they're very similar concave mirrors are similar to convex lenses no surprise because you already saw they're both converging devices so lastly this brings us to concave lens what happens over here again pause and try alright we'll start with the usual one ray parallel to the principal axis where does it go does it go through this focus no because over here this is the focus a concave is a diverging lens and that's why this is the focus and so that's that ray of light will diverge appearing appearing to come from this point alright so need to be very careful when you're dealing with concave lenses that's one way another way we can just shoot straight through the optic Center just like before you go on deviated and you can now see these two rays of light up again diverging and they appear to come from this point and as a result we end up with a tiny tiny virtual image and regardless of where you keep this object you will find you will always get a virtual diminished image so we saw all the cases let's put it all together in the converging case we find that there are three possibilities one we can get real image 1 we can get virtual image you can also get no image at all if the object is kept at F we will get no image because the rays of light will be parallel but for diverging diverging devices will always get a virtual image only one case and this actually is a bonus question for you to think deeply about why is it that for converging we have three cases but for diverging me how only one case I won't answer that but something for you to think about anyways since there are three cases for converging mirrors and lenses or real virtually no image we need to know when each of them happen so when does when do we get real images we get real images when the object is outside the focus outside the focus outside the focus real images when your object goes inside the focus you get virtual images so the focus is the dividing point what about the size well again if the object is closer to the focus you'll end up with bigger image notice closer you go to the focus bigger the image closer to the focus bigger the image same case even here if you go closer to the focus you get big image farther from the focus small image farther from the focus small image and of course at C the object and the image you will be having exactly the same size I have not run that over here here at 2f object and the image will end up having exactly the same size and remember the difference between these two is that the converging mirrors and lenses they can form virtual images but they'll be always enlarged these will always form virtual images which are diminished now that we have put all of this together let's see if we can recollect by asking some questions so here are four questions for you we have given the nature of the image and you need to find out which mirror can do that which mirror and which lens are capable of doing that so again great idea to pause and see if you can recollect what we just learned don't refer back but please recollect please see if you can answer these questions alright let's start with the first one real diminished image which mirror can do that well I look at real know that real images can only be found by converging devices so this is a converging mirror which is our converging mirror concave mirror real enlarged image again I just look at real I know real can only be formed by converging devices you need to focus it at a point only converging devices can do that so I need a converging lens which is a converging lens convex lens is a converging lens the third one virtual enlarged image well virtual images can be found by both but virtual enlarged and the virtual image can only be formed again by a converging device okay that's what we saw and so which is our converging lens again convex lens finally virtual diminished image this is formed by a diverging object diverging mirror which is our diverging mirror well convex mirror is the diverging mirror in fact it's easy for me to remember this because I've seen convex mirrors on parking lots and everything and I always see virtual erect diminished small-sized images so I just remember this everything else is con-con converging devices okay ready for a challenge now comes follow up questions for each of these cases where should you keep the object for example if you want to get a real diminished image where exactly should you keep an object in front of concave mirror similarly over here where should you keep the object again can you try and remember all right let's see for the first one I want a diminished image for a converging mirror we just saw that the farther I go from the focus the smaller the image becomes right so I need to be outside F for real and for diminish I need to be far away so basically beyond C can you imagine that can you visualize that we need to go beyond C for the second one a real enlarged image for convex lenss where do I get enlarged image but again for real I need to be outside F but for enlarge need to be close to F remember closer you are to have more bigger the image so outside F but close to FA that's between F and 2f not see because a lens so between F and 2f between all right third one I need a virtual enlarged image virtual image for convex lens for converging when do I get virtual images for converging lenses we saw we need to be inside F right so clearly it should be between F and the lens or we can say between F and the centre of the lens which is usually represented as au the last one I want virtual diminished image in a convex mirror well for convex mirror like a diverging mirror we only have one case so regardless of where you keep the object you will always end up with this and so you can keep this anywhere you want now if you are trouble-free collecting all of this no worries you can always jump back to the previous section recollect and come back over here and see if you can answer them on your own if all is well we can now move on to the next part where we will start talking about the mirror and the lens formula these formulae connect the object distance usually denoted by u the image distance we and the focal length same here as well object distance image distance and the focal length and the something called magnification formula which connects the height of the objects and the image with their distances again the height of the object on the image with their distances so can you recall what the mirror formula and the lens formula are and what the magnification formula for the mirror and the magnification formula for the lens are again good idea to pause and see if you can remember alright here we go for the mirror the mirror formula is 1 or V plus 1 over u equals 1 over F and the magnification formula for the mirror where m which is magnification defined as the height of the image divided by height of the object is given as minus V over u ok what is it for the lens it's pretty similar but important for lens lens formula becomes one or we minus one over u equals 1 over F so the negative comes in the lens formula but for the magnification we see no negatives at all now clearly when we have to remember the signs for all the formulae it's very easy to get confused so here's what I do I only remember the mirror formula and I remember that the middle formula is all positive right all the values are positive then I know magnification formula becomes negative okay once I know this I know for lens its opposite I know for lens this becomes negative and for magnification this becomes positive hopefully this will help you remember and get less confused with the signs when you're applying them to numericals but before we apply them to numericals we need to understand sign conventions because all these values can take both positive and negative values so what are the signs when do they become positive when do they become negative again can you recall that so here's how it goes we first consider the center of the mirror or the center of the lens as the origin either the pole is the origin or the optic center is the origin once we do that one side of the origin is positive another side is negative ones that is positive another side is negative how do we remember which is faucet which is negative here's how I like to remember always incident direction is chosen as positive incident direction so over here incident direction or incident light comes this way so the incident direction of of the pole over here to the right becomes positive values positive positions opposite direction becomes negative positions same thing over here the incident direction is to the right so all the in all the positions to the right side of the origin or the optic center becomes positive all the positions on the left side in this case becomes negative so these are all the positions right we also have signs for the heights that's a little bit little bit more straightforward anything above principal axis we call it as a positive height anything below the principal axis colorize the negative height so just to give an example in this particular drawing the object distance is negative notice on the negative side image is also on the negative side so image distance is also negative focus is also on the negative side so focal length is also negative object height is positive because it's erect image height is negative because it's inverted below the principal axis what about over here well object is on the negative side so object distance is negative image is on the positive side image distance is positive focal length is on the positive side because focus on the positive side focal end is positive object height is above the principal axis positive image height is below the principal axis negatives with this we can now solve numerical what kind of new miracles do we get well most of the new miracles one of the two will be given to us and we'll be asked to find the third and the fourth one for example we might be given U and F and we might be asked what is V and then we might be asked what is the magnification or sometimes we might be given V and U and we might be asked what is M and we might also be asked what is F let's take some examples so here's the first one usually when it comes to lenses and mirrors I like to first draw a diagram then see if I can write the data from that and then see if I can solve the problem so great idea for you to pause and read this problem and see if you can try this yourself first alright let's see it's given the image of the candle flame kept 30 centimeter from the mirror forms on the screen 60 centimeters from it so this means we have a candle flame which is kept 30 centimeters from the mirror and the image of it forms on a screen 60 centimeters from it so can we immediately tell which mirror it is we can write because it's given that it forms on the screen that's the clue it's a real image and which mirror gives us real image there's only one way I can do that so it's concave converging mirror concave mirror can do that that's something that we already saw in our previous section so that's the first answer we're done with that all right now to calculate focal in the magnification next let's immediately draw a diagram so here's our diagram here's a concave mirror that is an in front of it is the object sixty centimeters behind it is the real inverted image remember Emily real image is always inverted so from this we can now immediately right out of the data so we know U is 30 centimeters object distance V the image distance is given to be 60 centimeters and remember to use sign conventions remember signs that everything on the incident direction of the pole is positive so everything to the right over here is positive everything to the left is negative so this would be negative this would be negative and now we have to calculate what the focal length is what formula do we use for focal length we use the mirror formula and what's the mirror formula and what about the science I remember that mirror formula has everything positive so 1 over V plus 1 over you use U 1 over F and if we substitute in this we will now find F turns out to be negative 20 centimeters the negative sign is basically saying that the focal length is also the focus is also on the negative side which makes perfect sense for a concave mirror focalin is always negative so it again confirms it's a concave mirror if we if we had gotten positive that would be wrong so something would be wrong and the focal length is 20 centimeters so we got that what about the magnification how do we calculate magnification well magnification formula remember the magnification formula since the mirror formula has positive mind efficient formula has a negative in it so M is going to be negative V by U and again if we substitute the values of we and you we will get m to be equal to negative 2 what does that mean it means that the image is twice as big as the object but it seems real and inverted that's why the negative sign as the magnification is negative 2 that's our answer we've solved our problem all right let's go to second question again pause and see if you can try and solve this yourself all right a two centimeter tall pin is 10 centimeters away from a concave lens of focal length 10 centimeters so immediately we can draw a diagram over here we are given there is a concave lens for cooling for ten centimeters and the object is also kept ten centimeters away from its height is given two centimeters we need to find its where the image is positioned nature and the height of the image so quickly we can go ahead and draw a ray diagram what we have seen before and we'll see we'll get end up with a diminished image but now let's go ahead and calculate that so let's go ahead and write the data we have you to be given as in 10 centimeters and again if you look at the sign convention everything on the positive side on incident direction is positive on the opposite direction is negative so you get negative 10 focal length is given to be negative 10 as well and we're given the height of the object which is positive why is that positive that's because the height is above the principal axis anything about the principal axis is positive I'd able to find the image distance first to do that we're gonna use the lens formula remember the lens formula for mirror formula has everything positive lens formula has a negative sign image so I remember one or V minus one over U equals 1 over F and if we substitute we will now find we to be equal to negative 5 centimeters and if you look at the diagram you will immediately see the negative sign makes sense because the object is the image is on the negative side so we found the image distance or image position now I define the nature and the height well we immediately know from the diagram it is virtual concave lens always gives you a virtual but let's calculate it to calculate we go back to the magnification formula which does not have a negative sign remember the lens formula has negative magnification formula for lens has no negative if you now substitute in this you will get magnification to be half this means that the image is having half the size of the object it's positive because image is erect compared to the object that's what it's saying and that makes sense so immediately we know it's a virtual image it's height is half half of 2 centimeter is 1 centimeter so immediately we understand the height of the image is 1 centimeter but again you can use the form for magnification which is HOH oh that's the definition of magnification that is half and if you substitute you to get H is one centimeter and there we have it that is the height it is virtual and it is diminished we have found everything okay since you are loving it so much here is one last question before we wind up give it a try it's a little different but do you better try okay this is given a lens produces an erect five times and larged image of a pin kept 20 centimeter in front of it what is the lens which lens focal length and image distance yeah let's look at the lens it's producing an erect five times and larger means it's a virtual image erect means virtual and enlarged image which lens produces erect and large image that's a convex lens something we saw before and in such case the four four lenses we will find that the virtual image will be on the same set as the object so immediately we can kind of draw this and say look this is the kind of drawing that we have and so we can now write the data what is the data say well we know a pin is kept 20 centimeters front of it so U is negative 20 centimeters and by the way we already know it is a convex lens that's done this is convex U is negative 20 why is it negative well the incident direction if you go that's positive this is negative we know now magnification is given five times and large so in this question magnification is given to be plus Phi why is it plus because it's erect because though the the image is have the same orientation as the object so that's plus five and we need to find the focal length any resistance how do we do that well since magnification is given this traffic in totality for the magnification formula of the lens magnification formula is M equals V over u and if you substitute you will get V equals negative hundred centimeters so there you have it that's him that's your image distance so this distance is negative 100 that makes sense it's on the negative side so imagine is hundred centimeters and now we need to calculate the focal length to the focalin you can go back to lens formula so if you use the lens formula and if you plug in value of U and value of we we will now get f to be plus twenty five centimeters and that makes sense because for the folk for the convex lens focal length is on the positive side so this turns out to be 25 centimeters and that means again we found the answers to all of them now if you're struggling to answer these questions don't worry I struggled as well when I was learning this the key is to practice and practice and practice and so that's the whole idea behind this video to recall and give you more practice so that you would be more ready for your exams all the best