Class 10 Physics (India)
Mirror formula derivation
In this video, we will derive the mirror formula (expression connecting u,v, and f) and magnification formula. We will use a concave mirror to do it. Created by Mahesh Shenoy.
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- If we are supossing sperical mirror to be a plane mirror the why not is showing the properties of plane mirror(1 vote)
- we are taking a small portion of the spherical mirror and saying that , the small portion exhibits properties of a plane mirror. We aren't saying they are the exact same(11 votes)
- *"At8:07, why do we divide the equation formed only by uvf?"*(5 votes)
- Since the object is moved 2cm, the image will also move 2cm
Hence, 2cm is the difference(1 vote)
- An object is placed 20cm infront of a plane mirror. The mirror is moved 2 cm towards the object. The distance between the position of original and final image seen in the mirror(1 vote)
- Hey Keshab!
Since the object is moved 2cm, the image will also move 2cm
Hence, 2cm is the difference
Hope it helps. Feel free to comment if your doubt persists...Have fun learning :)(3 votes)
- at4:02What are similar triangles?(2 votes)
- This only works for the cases where real images are formed by the concave mirror though, right? What about the case where the object is kept between the principal focus and the pole of the mirror? Is there a proof for that too?(1 vote)
- Works for convex mirrors too. For the case of object placed between principal focus and pole, we don't have the statistics for the image to even prove it. But If I am not wrong, the image will follow the formula.(1 vote)
- at3:48, how is that a triangle? it has a curved side.(1 vote)
in this video we're gonna figure out mathematically how to calculate the image distance and the image height when the object height the object distance and the focal length of a mirror is given to us so mathematically speaking we are giving the values of H oh the height of the object object distance U and and the focal length F so this these things are these things are given to us so these are given given these we need to figure out the image distance and the image height so we need to find out what V is and what hatch is and so if you think about it carefully this looks like a map problem it's longer a physics problem because you can treat it treat this whole thing like some kind of a geometrical figure where some of these sides are given to us and we need to find some unknown sides all right so let's do that so let's focus on what we want to calculate we need to calculate V that is this distance and hatch I that is this length over here and notice what's given to us let's focus on these two things H always given to us and you that is this distance is given to us can we somehow connect them that's the question can we somehow connect this distance and this height this distance and this height if you look at this figure carefully and in fact if you look at these two triangles carefully we can figure it out so let me let me highlight those triangles we're focusing on these two triangles because their sides are what we are interested in now if you look at them carefully notice that they're similar triangles let's see how notice that they have one of these angles equal because they're both 90 degrees they also have these angles over here equal the reason they are equal is because one is the angle of incidence and one is the angle of reflection and laws of inflexion tell us that this should be equal to that which means the other angle the other angle should also be equal making these two triangles similar to each other and what is the speciality of similar triangles well their sides are in the same proportion therefore by looking at these two triangles we can write down that the height of the image that is this side divided by this side the height of the object divided by this side that has to be equal to the ratio of these two sides have to be equal to ratio of any other two sides that has to be equal to ratio of this side and this side you get that this side is V so has to be equal to ratio of we divided by this side you and so notice what we have done we have found a connection between the quantities that we want but we're not done yet because not is that H I and V are both unknowns and we have one equation with two unknowns we can't solve it so we could say well this is one equation for us if we're gonna get another equation in H I and V and then we are done so to find the second equation let's just get rid of this triangles first alright now to find the second equation we're going to bring F into the picture because F is also given to us and what I want you to do now is pause the video and see if we can come up with a set of similar triangles just like what we did over here but which has F as one of its side lengths just look at this just just explore that all right let's see F is over here this side P F that is the focal length F all right and now you can see this one triangle over here a little bit weird looking which has that side but that's weird because it doesn't have a right angle we want right angle triangles okay however you can look at this triangle over here can you see this triangle over here has the F or as one of its length and that triangle will be similar to this triangle over here so let me highlight those again we are focusing on these two triangles because they are similar to each other as we'll see and their side lengths are what we are interested in that's the whole idea okay so notice that this is right angle it should be already no this is also a right angle right this is also right angle and then notice these angles are equal to each other because they're vertically opposite which means the other angle will also be the same and therefore we can now say these two are similar triangles but before we proceed we might have question we were like wait wait wait this is a curve that's not a triangle this is not a straight line it's a curve so how can we treat it as a as a triangle well here's where we need to understand one approximation this is a spherical mirror which means it's a part of a sphere but look at where the center of curvature is and so if we redraw that complete sphere you can see a part of that entire I'm only drawing a circle but imagine it's a sphere that sphere is humongous compared to the part that we have chosen right and whenever you take a big sphere and you concentrate on a very tiny patch of that we can pretty much treat it to be a flat land or we can treat it to be a flat patch just like just like how earth is huge is a big sphere but when you look at a small patch of Earth it looks like flat so similarly we can assume that this mirror this curved part of the mirror is pretty much flat because it's a small part of that entire sphere alright so hopefully that justifies our claim that you know we can approximate this to be a straight line and therefore we can do the same thing that we did here we're going to exploit the properties of similar triangle their ratios are in the same proportion again I want you to pause the video and see if you can write that yourself alright let's see so if you take this side H I H I and divide by this side well what's this side equal to mm well if you look carefully that is nothing but H oh alright so that is hedge oh that will be equal to this side divided by this side well what is this side equal to hmm that's not something that we want right but we can convert it into something that we want notice this is equal to we minus F look at that carefully so that is we minus F divided by this side which is f you may be wondering if you're wondering why we're not taking the other side well that's not really because the hypotenuse is what we are not interested in right we are only interested in the sides which are over here and that's why we are not taking the hypotenuse all right and so this now forms our second equation because notice we now have two equations with two unknowns and so we can solve it alright so again it's a great time to pause the video and see if you can solve this yourself a give small clue whoa first try to connect vu and F only an equation containing vu and F can you can you try that again another clue look the left-hand side is the same which means the right-hand side must also be equal to each other just go ahead and try all right let's do that let's make some space for the algebra the geometry is pretty much done all we have to do now is algebra over here so since the left-hand side is equal we could say the right-hand sides are also equal to each other and therefore we could now say that V divided by U has to be equal to V minus F we minus F divided by F and strictly speaking we have gotten what we wanted this is the only unknown and U and F are known to us so you found out the relationship but you know this looks pretty ugly so we're gonna you know we're gonna try and make it a little bit pretty and so let's get rid of the denominator let's cross multiply what would we get we'll get V times F V times F equals if you multiply that over here you get UV u e- u F minus u times F in fact I've made it even more ugly so what you're trying to do is come up with an equation that makes it it becomes easier for us to remember because we don't want to derive this over and over again and do that one last step we'll do is we'll divide this whole thing by UV f and we'll see that we end up with a very simple equation now at least at a simple looking equation alright so if you divide the whole equation by UV F notice V and F cancel over here and so you end up all on the left hand side one or you that equals to note is UV cancels over here so you end up with one or F minus 1 or V over here and if you do one last manipulation to this equation we'll add 1 over V on both sides you will end up with the grand-grand equation that really wanting 1 over F equals 1 over F equals 1 over u plus 1 over V and there we have it given F and u we can now calculate the image distance from this and if we substitute that image distance in equation one we can now calculate what the height of the image is so there it is but one problem is that this is not a general formula so I don't want to box it as of now because we only derive this for a concave mirror and that - for real images what if the image is virtual what if you're dealing with the convex mirror are this will the same formula hold good maybe they will change right so in another video we'll see how to make a general formula which can be used for all the cases convex concave real virtual whatever you want we can use it so we'll do that in another video