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### Course: Optics (Essentials) - Class 12th > Unit 3

Lesson 3: How do curved surfaces change the path of light?# Solved example: Curved surface refraction

Let's solve a numerical on a curved surface refraction. Created by Mahesh Shenoy.

## Want to join the conversation?

- In the derivation, you brought about the equation by considering image distance as negative. So, shouldn't the negative sign not be put again while doing the question?(5 votes)
- Exactly what i was wondering. I mean he used those conventions already. Its like first you divided by n now u r multiplying by n(3 votes)

## Video transcript

we have a glass sphere which has a radius of 10 centimeters and it's made up of a material whose refractive index is 1.5 suppose we keep an object let's call it as o at a distance of 30 centimeters from the sphere we need to figure out where its image is going to be all right the first step to solving most of the problems in ray optics is to first make a drawing we already have that and since images are asked for us we can draw a bunch of rays of light so let's say we draw one ray of light from here it goes here another ray of light maybe goes over there this ray goes undeviated this ray will refract we know that much I don't know how much will it refract it's gonna refract maybe somewhat like this from this we see the rays of light are hitting a curved surface so this gives me the clue that we're dealing with curved surface refraction and there's only one formula that comes to my mind when it comes to curved surface refraction that's the curved surface refraction formula we have derived this formula in a previous video and our M is the refractive index of the medium that contains the refracted ray I M is the refractive index of the medium that contains the incident ray and UV and R are just the image distance object distance and the radius of curvature so let's see if we have all the data and we'll just plug it in and let's see what we get now wherever we are substituting we have to use signs and the sign convention we always choose is we take our pole as the origin so this is our origin and we choose the incident direction to be positive the right side is positive in this example right side is the incident direction so all the positions on the right side are going to be positive positions all the positions on the left side are going to be negative positions so let's apply this equation and see what we get so over here the refracted medium is 1.5 because this is where the refraction is happening so the refracted medium base 1.5 divided by the image distance which I don't know which I want to figure out - the incident medium incident medium is the one that contains the incident ray now nothing is mentioned so we're going to assume that to be one air or vacuum so that's just going to be one divided by the object distance well how far is the object our object is thirty centimeters so thirty but wait we have to be careful it's on the negative side so negative thirty that equals our M - I am so 1.5 minus 1 so that's 0.5 that is 0.5 divided by the radius of curvature radius is just 10 centimeters again we have to be very careful when we're dealing with the radius you see the center is over here isn't it this is the center of curvature so the center is on the positive side and that's why the radius becomes positive all right so whenever dealing with radius of curvature think about where the center is all right so we can solve this equation now pause the video for a while and just do the algebra and see what you get all right let's do this so we get 1.5 over V plus 1 over 30 is 0.5 over ten point five is 1/2 so this is 1/2 over 10 oh I took all this is 1 over 20 right 1 over 2 divided by 10 using 1 over 20 and from this we can now figure out let me let me do that over here so you get 1 point 5 over V equals I subtract 1 over 30 on both sides so you get 1 over 20 minus 1 over 30 that gives us you can take 60 as the common 3 minus 2 that's 1 over 60 which means V equals 60 times 0.5 1.5 so let me choose the same color so V equals 60 times 1.5 that's 90 centimeters it's centimeters because our object distance was in centimeters radius 1 centimeters so there we have that's the answer 90 centimeters so let's draw that 90 centimeters from there but remember our distance are always from the pole so positive 90 positive 90 would be on this side so it's going to be somewhere 90 centimeters from here so this is 20 because the radius is 10 so 90 would be somewhere over here maybe somewhere over here maybe so the ray of light is going to go and meet at this point these two rays are gonna meet at this point hmm do you think this can be the final answer I'm actually positive we didn't think about this for a while well you may have already guessed it these two rays of light before converging here are going to hit the second surface and as a result they will refract again which means this cannot be our final answer so what we have to do next is apply the same formula for the second surface so let's do that down over here I have all the things ready all we have to do now is apply that same formula because we still dealing with the curved surface right so apply that same formula now to the second surface so now this is going to be our new pole no longer this one I'm not care about this one this is the surface that we're interested again great idea to pause the video and see if we can substitute into this equation one more time all right let's do it refracted medium this time the ray of light is going to bend over here right somewhere somewhere like this I don't know exactly how it bends it doesn't matter but the bending is going to happen over here so this time the outside medium becomes the refracted medium so outside medium is just one so when we substitute you get one divided by the image distance we don't know the image this is not the final image we want to calculate with the final images - incident medium this time on this surface this is the incident ray I hope you can see that these are the incident Ray's on this surface we are going to completely neglect these incident rays now because that was for this surface all that matters now what's happening to this surface so the incident rays are now in this medium inside the glass so the incident medium now becomes one point five divided by the object distance there is the object now we might think that object is over here but we have to be careful you see that was the object for this surface what is the object for this surface well here's how I'd like to think about this I like to think that object is where the incident rays meet right initially these were the incident rays and wherever the incident rays met with each other was the object but now these are the incident rays so where are they meeting well if you trace them backwards they will not meet but if you trace them forward they meet over here so this this now becomes our new object this becomes the object for this surface now I know it sounds a little weird that the rays of light are appearing to converge at this point we don't have to worry too much about that just think of it this way wherever the incident rays meet that's where your object is in fact this is called as a virtual object because it's not really there it doesn't matter that's the object for us so where is the object what's the object distance well again we have to be careful is it 90 centimeters no when you talk about the object distance we have to always take it from the pole what's the distance from the pole well this distance is 90 centimeters this is 20 so our new object distance is going to be this distance and that will be 90 minus 20 that's going to be 70 centimeters and again we have to apply our sign conventions I'm sorry I should have done that before but let's do that if you apply sign conventions again incident Direction is positive incident Direction is positive and so everything to the right of the pole is positive so all these are positives and everything to the left of the pole is negative now this this is negative it's negative all right so our object is on the positive side I know this diagram has become a little shabby excuse me for that but now we are dealing with up so the object is going to be positive so we now have 70 over here plus 70 object distance that's going to be equal to this minus this this minus this so one point one minus one point five that's minus 0.5 divided by R now are think of it as the position of the center and the center now is on the negative side again we have to be very careful over here and so all now is negative hope that makes sense because when you consider from this pole the center now is on the negative side so you get minus 10 and if we solve this equation we will now get the final image so again pause the video and just try and do the algebra yourself all right let's do that so we get 1 over V equals I'll add 1.5 divided by 70 on both sides so again this right-hand side the part negative negative cancels and you have 1/2 half divided by 10 is 1 over 20 and we are plus 1.5 over 70 so if you take the LCM what's our LCM 70 and 20 LCM would be 140 so over here we multiplied by 7 here you multiply it by 2 so 2 times 1.5 is 3 so we get 10 over 140 and therefore our final answer our final image distance is going to be reciprocal 1/4 divided by 10 that's going to be 14 centimeters that is our final answer again here to check where that is it's 14 centimeters from this point on the positive side so that's going to be somewhere over here 14 centimeters 14 centimeters all right so the ray of light we know after a fraction after a fraction it's gonna go like this so this is where finally the two rays of light are going to meet and so this is the general way in which we can solve any problem where you have multiple reflections taking place you take the image of the first first surface as the object for the next surface again if there was another surface over here you just apply the same formula now this would be the object for that surface so that's the whole idea behind solving problems like this