If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:10:36

Video transcript

we have a thin convex lenss and the radius of curvature of both the surfaces are 25 centimeters and let's say the medium the material is made up of a medium which has a refractive index of 1.5 we need to figure out we need to figure out the focal length of this lens when it is kept in two cases air in the second case in oil of refractive index 2.0 all right let's do this so we've been given the radii of curvature and the refractive index of the material and we need to calculate the focal length so what expression connects the focal length and these things the only thing that comes to my mind is the lens makers formula that's what connects the focal length and the properties of a lens right so remember what lens makers formula is we derived this in a previous video 1 over F where F is the focal length equals n 2 and 2 is the refractive index of the material of the lens divided by N 1 which is the refractive index of the surrounding medium minus 1 times 1 over R 1 minus 1 over r2 where r1 and r2 are the radii of curvature of the two surfaces so let's apply this formula and see what the focal length is so let's do this so let's look at case 1 so 1 over the focal length let's call it focal length this f1 for the first case whatever focal length 2 yet that equals n 2 over N 1 well n 2 is the refractive index of this medium so that's going to be 1 point 5 divided by N 1 which is the refractoriness of the surrounding medium which is air so for us refractive index of the air is one pretty much 1 minus 1 times 1 over R 1 minus 1 over R 2 now R 1 R 2 R the radius of curvature of the two surfaces and both are 25 centimeters so is it let's see is it just 1 or 25 minus 1 over 25 we have to be careful remember whenever you're using 4 when you're substituting into a formula please use sign conventions sign conventions are extremely important the sign that we use over here are Cartesian sign conventions and the convention goes like this we are always let's first draw a principle axis we are always going to choose the incident direction let's call this as the optics center okay if you're always going to choose the incident direction as positive but the big question is which is the incident direction over here well you know what you can choose whichever direction you want to be incident over here so let's just assume the right side to be incident so let's say this is our incident direction then all the positions to the right of the optic centre is going to be positive we're going to treat this as the origin and all the positions to the left of the optic center will be negative positions negative positions and because the incident direction is to the right so you can imagine there's an object somewhere over here the rays of light are going to hit this surface first and so we'll call that as our surface 1 and then the ray of light will hit this surface so that will be surface number 2 so you could have chosen the left side as also positive that's also I mean you would have chosen the left side as incident then this would be surface number 1 and this would be surface number 2 so you're free to choose whichever direction you want as incident but accordingly your surface one and surface two will change so now that we have this let's see which radii are going to be positive and which really are going to be negative so remember these are spherical lenses so they are parts of a sphere so if you just draw this imaginary spherical surfaces that they are part of can you see that the first surface it will be a sphere like this which has the center on this side since the center is on the positive side the radius of curvature of the first surface is going to be positive so R 1 is going to be positive 25 that's correct minus R 2 1 over R 2 now notice when you consider the second surface its center of curvature its center is over here which is on the negative side as a result its radius is going to be negative 25 centimeters therefore it's going to be minus 1 over negative 25 so let's put that over here negative 25 that's the only thing we need to be careful about over here and so now we can go ahead and solve this one point 5 divided by 1 is just 1 point 5 minus 1 is 0.5 times 1 or 25 plus 1 or 25 the negative negative becomes positive so you get 2 over 25 and 0.5 times 2 is just 1 so that gives us 1 over 25 again this is 1 over F 1 and therefore the focal length in the first case turns out to be 25 centimeters and there we have it that is the focal length in the first case also notice that the focal length we are getting is a positive value what does that mean well since the focal length is positive it means the principal focus relies on the right side because that's where the positive side is so our principal focus let's write that down is going to be somewhere over here at a distance of 25 centimeters or from here which in turn means that if we have incident beam which is like this incident rays of light is like this then these rays will get converged at this point and that's what we would expect right this is a convex line so we expect it to converge at some distance and there we have it all right now let's go to case 2 in the second case our length is kept in an oil of refractive index 2.0 which means the surrounding medium of our lens now is going to be oil with the refractive index of 2.0 so it would be a great idea to pause the video over here and see if you can try this yourself all right let's see well if you think about it notice we're still going to use the same formula lens makers formula what's going to change well adding 2 is going to remain the same the material has not change and one is now 2.0 because the outside medium is now this era this oil the radius of curvature is going to remain exactly the same whether you put it in oil or whether you put it in air the radius of curvature is not going to change and so if you use the same sign conventions everything will remain the same so the only thing that's changing is this number this number is going to become 2.0 so again if you have not tried this before great idea to again pause the video and see if you can try this now and do the algebra so our new focal length let's call it as f2 now this will be 2.0 so this is going to become 2 and everything else will remain the same which means this remains the same let's see what happens over here this is going to change now isn't it so what's going to happen over here well 1.5 over 2 minus 1 we can take the common denominator that's going to be 1.5 minus 2 divided by 2 let's get rid of this okay so what we get now is 1 over f2 equals 1 point 5 minus 2 is minus 0.5 this 2 cancels so divided by 25 and 0.5 is 1/2 so it's going to be minus 1/2 or 25 that's going to be minus 1 over 50 which means if you take more space over here which means f2 let me write that down somewhere over here so let's say certain f2 is minus 50 centimeters and there we have it that's our result but here's the question what does it mean that our focal length is now negative again great idea to pause the video and think about this if you haven't already all right so negative just means one thing the principal focus has to be on the negative side that means the principal focus is over here and this distance now is 50 centimeters but what what does that mean but what that means is if you have incident rays let's draw in trees with purple now let's say if you have insulin raised like this then after refraction will the rays of light get focused here well that's that can't happen we know that but what that but what can happen is that the rays of light would diverge away and they appear diverge from this point that is what this means that's the only thing that this can mean in other words this negative focal length is telling us that our lens is now diverging lenss and this might sound weird to us because it's a convex lens how can convex lens act as a diverting lens well it turns out as you can see from the formula as if you can see from this it turns out but it can behave like a diverging lens when you put it inside a different medium and in fact if you look carefully the reason it's acting like diverging is because the medium outside has now become denser than the medium which is inside and as a result of rays of light are going from denser to rarer medium and so they're going to refract in the opposite of they're gonna refract Ave from the normal and if you actually draw the normals you'll see that this is happening great exercise to try and convince this yourself by just drawing ray diagrams properly but this is what's happening so from this example we can learn one very important thing that even if you have a converging lens in the air that same lens can become diverging if you put it inside some medium which has a higher refractive index than the material of this lens and so this is how we can use lens makers formula in different cases to figure out what the focal length is