Class 10 Physics (India)
We will find the properties of the image formed by a convex mirror. We will do that using the mirror formula and apply sign conventions. Created by Mahesh Shenoy.
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- At6:25, Mahesh says that the magnification is 1.33 and therefore the image is going to be smaller than the object. But how is that possible? If the magnification is greater than 1, doesn't it mean that the image is bigger than the object?(1 vote)
a pencil of height two centimeters is kept upright five centimetres away from a convex mirror of radius of curvature 20 centimeters find the position height and nature of its image the first thing we'll do is make a drawing of this situation so it'd be a great idea to pause the video and see if you can first draw this whole situation yourself alright let's do that we have a convex mirror so let's say here is our convex mirror which has the radius of curvature of 20 centimeters so this distance let's write that over here this distance the radius is 20 centimeters in our mirror formula remember we need the focal length and one thing to remember is that the principle focus is right in between exactly in between the center of curvature and the pole and therefore if this is 20 our focal length must be 10 centimeters that's the first thing we can gather from this all right then we know that the pencil of height 2 centimeters kept 5 centimeters away from the convex mirror so the pencil must be kept somewhere over here it's height is 2 centimeters we'll put the height later and it's kept 5 centimeters away so this is 5 centimeters so we know the object distance we know the focal length we have to calculate the image distance or the position of the image so the first step is to write down the mirror formula remember the mirror formula 1 over F is 1 over V plus 1 over u but this formula is sign-sensitive so the next step would be to understand what are the signs and what we do that for that we're gonna treat this as a graph sheet we're gonna treat that this is on a graph sheet we're gonna call the pole as our origin and we'll choose the incident direction to be horizontal positive which is the incident direction to be positive and the incident direction is towards the left because the object is here the mirror is over here so the left side is positive therefore all the positions to the left side of the pole of the origin are positive positions and all the positions to the right side of the pole are negative positions as the first step now that we have this we can substitute over here and we can solve now again if we great idea to pause the video see if you can substitute and get the value of we alright let's do that so 1 over F F is 10 centimeters that's the position of the principal focus but that's positive so we're just gonna put positive over here 1 over V we don't know what the position of the image is we're gonna just keep it as it is plus 1 over u U is the position of the object all the objects on negative side so that position is going to be negative position so minus 5 and so all we have to do now is solve this we can write this minus sign over here because plus times minus is minus and now to calculate we will just add 1 or Phi on both sides that will give us 1 over V equals will that be equal to 1 over 10 plus 1 over 5 plus 1 over 5 we can solve this now if you take the common denominator we get 50 now we can we can take 10 as the common denominator L Thiam so you get 1 plus 2 1 plus 2 which is 3 let's just write that down over here and 1 over V is 3 over 10 therefore V you can figure that alright that over here itself V is 10 or 3 and there we have it that's our answer that's the image position now we need to interpret this result carefully what does it mean that we have a positive value for V the positive value is telling us that it's on this side of the mirror so our answer is going to be 1000 3s what about 3.3 3 right so that's going to be somewhere over here because it's positive we now understand this is the position of the image okay the next step would be to calculate the height of the image well the height of the object is given to us as two centimeters let me just write that down over here first this is 2 centimeters and therefore the height of the image we can now figure out using the magnification formula let's just make some room all right the magnification formula is the height of the image height of the image divided by the height of the object is going to be negative V native V divided by u so we can again substitute and I can great idea to pause the video and see if you can sub straight at it get the answer yourself alright let's do this now even for Heights remember the sign convention is above the principal axis the height is positive below the principal axis the height is negative all right so the height of the image I don't know so I'm just gonna keep it as it is height of the object or object is above the principal axis I don't care where the object is for the height all that matters is above the principal axis and therefore it's positive 2 centimeters so it's going to be positive that'll be equal to minus V well we already found what V is V is 10 over 3 divided by you or what's U U is negative 5 you know that right because U is the object position that's negative side so that's negative 5 all right so we just have to solve this let's see the negatives cancel 5 goes 1 times 5 goes to x so you get 2 over 3 and we can multiply the whole equation by 2 and so we'll get hatch I equals we multiplied by 2 you get 4 over 3 and that's our answer 4 over 3 again we need to interpret this carefully we got the height to be positive that means we already know the image or is over here that means that the height is going to be above the principal axis all right and notice the value is 4 over 3 what's 4 over 3 its 1.3 something right 1.33 that means it's smaller than the object and therefore we can pretty much now draw this it's going to be like this because it's positive and that's 4 over 3 and one last thing we need to know is the nature of this image well if it's fact it has to be that's the one thing we can remember so it's a virtual image if it was inverted it would have been real image so this is a virtual virtual image and so there we have it we have found everything that we wanted now of course you could stop over here but one last thing I usually love to do over here is draw a ray diagram and just make sure that everything is right I mean it's very easy to you know get some signs wrong and get some horrible answers over here so just to make sure we are on the right path we'll just quickly make a read wrong right a ray diagram so if you're into Ray diagrams over here one ray of light will shoot straight parallel to the principal axis and this will appear to diverge away from the principal focus so just a rough drawing will do and see what we get so that ray will like ray of light will go like this another ray of light we can shoot right at the pole oops I always like to shoot right at the pole and that ray of light after reflection will go somewhat like this and now these two rays well they are diverting away that means we'll never get a real image so it makes sense you are getting a virtual image that's correct and if you put a giant eye over here these two rays appear to be coming from somewhere over here so guess what we got everything right so we got it to be virtual you also found that the image has to be on this side so the ray diagram conforms that whatever we got is pretty much true so it's always great to find the answer and also then do a ray diagram to quickly check whether what have we gotten is right or wrong