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## Class 10 Physics (India)

### Course: Class 10 Physics (India) > Unit 2

Lesson 4: Defects of vision and correction numerical# Solved example: hypermetropia

Let's help our friend, Veena, to select the right lens for her eyes. Created by Mahesh Shenoy.

## Want to join the conversation?

- In here if the object is in bw (100/3)cm and 100cm, the convex lens used will form a real image on the right side. Then, how can Veena see that object?

Will she see an inverted image?(2 votes)- it's not the focal length of EYE lens. we're really deceiving the eye lens by showing the object's at 100cm away though it's just 25 cm away from the eye lens using an optinal convex lens.(4 votes)

- Why do you use the equation as 1/f = 1/o - 1/i instead of 1/f = 1/o + 1/i the sign of the answer makes a difference since that's how you determine type of lens used.(1 vote)
- We use it because 1/f = 1/o - 1/i is the lens formula.

1/f = 1/o + 1/i is the mirror formula.(2 votes)

## Video transcript

Vina cannot read when she holds books closer than hundred centimeters 100 centimeters from her eye this is the distance from her I name the defect and prescribe a lens of suitable power to correct that defect given that the normal eye in your point is 25 centimeters so let's see what's given to us oh if this is Vienna's eye then it's given that she cannot read when she holds books closer than her centimeters so if anything comes closer than hundred centimeters she cannot see it clearly that means if it's outside of hundred centimeters farther away than hundred centimeters she can see it and I'm gonna put a green over there saying that she can see any objects over here but if there's any object within hundred centimeters then she can't see it so we'll put a red for that so she can't see anything over here and I'm only going to go until 25 centimeters because beyond 20 or closer than 25 centimeters even a normal eye can't see it so we don't have to worry about this region it's only this region that veena is unable to see which a normal eye can but we know as I cannot do it so the first question is what is this defect well I want you to pause the video and think about what that defect is well since she can see things which are far away but she cannot see things which are close to her she is farsighted so the defect the defect over here is rather than over here far sightedness because she can see things which are far away so far sightedness or the biological term is hyper Metropia i will not write that's not me there so we got that that's the defect and now we need to prescribe a lens of suitable power to her so that she can correct that vision which means we're like the doctor over here and we have diagnosed her her issue and we need to now correct it so what lens should I put over here is the question and what should be the power of that at first we might be wondering well is it a converging lens or a diverging lens what should we use well we don't have to worry about that at all so let's think about this first of all how do we define power well in optics we define power P as the reciprocal of the focal and the reciprocal of the focal length so we need to figure out what is the focal length of the lens that we're gonna put over here but how do we figure this out well all we have to do is think in terms of objects and images and then we can figure this out think of it this way for a normal eye this is the near point all right so let me just call it as a misread this is the near point for a normal eye right because beyond this closer than this a normal eye can't see things clearly but for vena this is the near point because for her closer than her centimeter she can't see so this is a near point for rena so near point four I'm going to put a star over here saying that's a near point fariña so in order to correct her vision she should be able to see all the way up to 25 centimeters how do we do that well another way of saying that is that if there is any object right at 25 centimeters then its image must be formed at 100 centimeters does that make sense think about it if the object is at 25 centimeters and if the lens converts it and lens creates its image at hundred centimeters then she'll be able to see that so this is our object is the object object and this will be the image so we need a lens which does this so we know what the object distance is we know what the image distance is and so all we need to do now is the focal length of the lens needed so it'll be a great idea now to pause the video and see if you can solve this yourself all right let's do that so let's write that down object distance is given to us as 25 centimeters we also know the image distance the image distance is hundred centimeters but we need to be careful we have to use sign conventions since our lens is going to be over here in front of our eyes we're going to choose the incident direction as positive so this is going to be the incident direction and all that all the distances all the positions on the on the right side of the optics center is going to be positive this side would be negative so this will be negative object distance is negative because it's on this side so this is negative 25 centimeters even the image distance is negative because on this side in the opposite direction right so this would be minus 100 and so now we can go ahead and use the lens formula because since we know U and V we can calculate F so let's make some room and let's calculate this so the lens formula says 1 over F is 1 over V minus 1 over u and so that gives us 1 over V that's a negative 1 over 100 minus 1 over negative 25 well if you solve that let's see what we get you get 1 over negative 1 over 100 plus 1 over 25 take 100 as common you get minus 1 plus 4 that gives us 3 over 100 3 over 100 so so okay so the focal length is the reverse of that a reciprocal of that that's going to be hundred over 3 and we'll just keep it as it is so this is the focal length so this is so many centimeters and notice since we're getting a positive answer this means the principal focus is on this side so whatever the lens we're going to use its principal focus is on this side and so now we can figure out whether it's a convex lens or a concave lens think about it if the incident direction is if you have an incident ray in this direction which lens will give us the principal focus on this side well it should be convex because if it's a concave lens let me just quickly draw that then if the incident ray was somewhat like this you would have diverged away appearing to come from here so the principal focus would have been negative so it's a convex lens and we found out its focal length that's hundred over three centimeters but we require power and we can calculate that so power will be just the reciprocal of this so that's going to be 3 or 100 centimeters but if you put this in the SI unit because for power we always put things in the SI unit that's going to be 3 over 100 cm one meter one meter so three meter inverse r3 diopters so the prescription that we give is we say that she needs a lens which has a power of three adapters the positive power itself is telling us it's a convex lens so the beauty is just by looking at these signs we can figure out whether it's a convex or a concave lens we really don't have to remember that for farsightedness we have to use a convex lens it can be worked out directly