Solved example: magnifying power of compound microscope

we have a compound microscope whose objective focal length is 5 millimeters eyepiece focal length is 2 and 1/2 centimeters a sample is kept at 6 millimeters from the objective find the magnifying power of this microscope if the final image is formed at infinity let's quickly draw our compound microscope it consists of two lenses the objective lens is over here via the principle of the objective the goal of the objective is to create a large magnified image and as a result we usually keep the sample very close to the principal focus but outside the principal focus and we can see that the objective has a 5 millimeter friends focal length but it's kept at 6 millimeters a little bit outside the principal focus what this does is that this produces a large magnified image which here was here and now we can further magnify this by using a magnifying glass or another convex lens and this now acts like an object for this next convex lens that we're going to use so here's our magnifying glass under convex lens and notice that since we want the final image to be formed at infinity it this means that the rays of light falling on our eyes have to be parallel to each other and that can only happen if this object and this image it's the image of the first lens which is the object for the second lens is right at the principal focus because we've seen that only when you have objects that principal focus the refracted rays are parallel to each other so this is the setup that we have over here and all we have to figure out now is what is the magnifying power of this now we've seen in the previous video we've talked all about this in in great detail in the previous video and we've seen that the magnifying power of a compound microscope is just the magnifying the magnification produced by the objective this is the linear magnification produced by the objective multiplied by the magnification produced by the eyepiece now if you're not familiar with this or you need more clarity it would be a great idea to go back and watch that video and then come back over here let's see how we can solve this to figure out the magnification of the produced by the objective we just need to figure out what is the ratio of this image height to the object height and guess what we can do that because the object distance is given to us you see we know the object distance this is given to us as six millimeters we know the focal length of the objective this is the size of the objective okay so we know the focal length so we can calculate the image distance and so from that we can use the magnification formula and figure this out so this is something we can do by just using lens formula how do we figure out the eyepiece magnification well the eyepiece is just a simple microscope so we can directly use the magnification of a simple microscope and solve this so every great idea to pause this video and see if you can try this yourself first all right let's do this let's start with figuring out the magnification produced by the objective alright so first do the objective part so here we'll first try to figure out what the image distance is and then we can use the magnification formula so for that we're going to use the lens formula lens formula is 1 over F I don't want to write it down because you know we don't have much space but 1 over F equals 1 over V minus 1 or u so that's just directly substitute 1 over F what's F here for the objective F is 5 millimeters so let's put that in 5 millimeters now we have to be very careful with our sign conventions the incident direction is always positive therefore all that all that all the positions to the right of this optic center is positive and our focal length our principal focus is this one because the rays of light are going through over here and so our focal length also becomes positive and that becomes plus 5 millimeters so we're gonna keep on everything in millimeters okay so 1 over F equals 1 over V which we don't know so just keep it as 1 over V minus 1 over u minus 1 over u will U is the object distance well notice it's on this side so that's negative so that's negative 6 and this negative times negative makes it positive so this will end up becoming positive so from this we can figure out one over V is so just have to subtract 1 or 6 on both sides so we get 1 or V as 1 over 5 minus 1 or 6 minus 1 over 6 and that gives us that gives us we can take LCM as our common denominator 30 this is multiplied by 6 this is multiplied by 5 so you get 1 over V as 6 minus 5 over 30 that means V well let's just make some more space over here okay so what's V from this from this we can say V is 30 by 1 so 30 millimeters that's our image distance so in our diagram this distance from here all the way to here that is 30 millimeters or about 3 centimeters all right now we can go for the magnification formula so the magnification of the objective that's what we want right there over here magnificient of the objective is the height of the image divided by the height of the object but it's also same as V over you lens formula in the lens formula we've seen that's the same as V that is 30 millimeters will keep things in millimeters 30 millimeters divided by you while you is minus 6 that's over here minus 6 so that gives us minus 5 minus 5 let's hit minus 5 as our magnification which means the height of the image is 5 times more than the object and the minus sign is just telling us it's an inverted image we don't have to worry too much about the minus sign we just need to know the number the value is what we're interested in so we got this this is the first part next we need to figure out the magnification produced by the eyepiece well that's the magnification of the simple microscope and we've already seen before in previous videos that the magnification of the simple microscope which is our eyepiece over here is just the ratio of the near point distance divided by the focal length of the eyepiece or the simple microscope right now the focal length of our simple microscope is given to us let's just see what was that it's given to us as so here 2.5 centimeters that's given to us which means this distance this distance is given to us as 2.5 centimeters and D near point well that's usually taken as 25 centimeters it'll be dimension in the problem but if it's not mentioned we'll take it as 25 centimeters so we know that as well so that's 25 centimeters divided by 2.5 centimeters 2.5 centimeters and that's 10 that is 10 because you know this cancels so you get 10 and so we found the magnificient produced by the eyepiece as well and so the total magnification produced by this compound microscope is going to be the product of this and make sense right I mean notice the first this gets magnified five times and then that gets further magnified ten times so the 12 magnification will be the product right so five times ten that's going to be 50 usual right it is 50 X or 50 times like this sometimes they could also ask you what is the distance between the objective lens and and the eyepiece now you can see from the diagram we can clearly see what that distance is it is 3 centimeters plus 2.5 centimeters so if there was asked what is the distance between the 2 lenses that's about 5 and 1/2 centimeters in our example