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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: myopia

Let's help our friend, Suresh, to pick the right lens for his glasses. Created by Mahesh Shenoy.

## Want to join the conversation?

- so shouldnt a myopic person using concave lenses see farther thing smaller than they seem to a person with a normal eye because the images formed by concave lenses are diminished?? but this is not happening-a myopic person wearing proper lenses is seeing the same as a normal person.WHYY?(2 votes)
- I don't know about you, but I am myopic, and things do in fact look smaller when I put on my prescribed glasses.(2 votes)

- can you make a course on linear algebra? please(0 votes)

## Video transcript

Siraj cannot see things clearly which are farther than 200 centimeters maim the defect in his eye and prescribe a lens with suitable power so it's given that this is siliceous eye that that he cannot see anything farther than 200 centimeters so if there's any objects over here all the of to infinity he cannot see it so I would have just marked that with red over here so these are the regions where he cannot see clearly but if it's within to our centimeters we're just going to assume that it says it's not mentioned if it's within 20 centimeters he can see things so I'm gonna put the green over here and of course once you go very close you can't see it anyways even normal I can't see it so let's ignore that part so this is given to us so now we have to name the defect in his eye can you think about what that defect is just pause the video and think about what what this defect is well since he can see things which are near to him at close distances but he cannot see things which are far away the defect is short sightedness so let's write that down the defect over here is short sightedness he's nearsighted so short-sighted short-sighted or he is myopic so we say suresh has myopia or myopia any one of them all right the next thing we need to do is prescribe a lens with suitable power so we are like the doctor over here we have diagnosis problem we understood that he cannot see farther than 200 centimeters now we need to figure out exactly what power lens he should use and before I proceed how do we define power well quickly let's quickly recall power is defined as the reciprocal of the focal length this means you really need to figure out what the focal length over here is alright and you may think that you know okay what are we using a converging lens or a diverging lens I can't recall that well we don't have to worry about that at all you know what we will do we'll only think in terms of object and image and we can figure out the focal length so let's think about this you see Suraj cannot see farther than 200 centimeters so we usually call this as the far point far point all right so another way to say that is we could say Suraj has a far point of 200 centimeters that's the farthest he can see but what is the far point for a normal eye for an undetectable what's the far point well an unaffected eye can see as far as they want right you can see all the way to the moon or to the Sun now you can see the stars so far point for an undefeated eye is infinity which means to correct his vision we have to make sure that Suraj can also see all the way up to infinity you're getting that in other words in other words if our object is at infinity let me just write that down if we have an object which is at infinity then it's image our lens should bring its image to the far point that's what our lens should do then Suresh will be able to observe that object all right so from this we understand that if the object our lens should have a focal length in such a way if the object is at infinity then the image should be at 200 centimeters the image should be at 200 centimeters and yes what if you know the object is in the numerator distance we can figure out what the focal length is so I want you to pause the video and see if you can figure out what the focal length is given these two values and remember we have to use sign conventions so please use sign conventions all right let's do it we know our length is going to be over here so for sign convention we need to first choose our incident direction as positive so all the distances on this side will be positive and all the distances on this side would be negative so the object distance is negative infinity but negative infinity positive infinity sign matter the image distance is negative 200 centimeters and so if we use our lens formula now so let me just go down over here start using lens formula so well thence formula is 1 over F equals 1 over V minus 1 over U so if we substitute we will get 1 over minus 200 minus 1 or infinity and that's going to be well this is good this is zero one over infinity is zero so all we get is minus 1 over minus 200 which means the focal length is negative 200 centimeters that's the focal length notice we have solved the problem even without thinking about whether really quite a converging lens around I heard England's you don't even need that just thinking times some object an image now by looking at this focal length we can actually figure out whether it's a converging lens or diverging it's again pause the video see if you can understand whether it's a converging or diverging lens just look at the minus sign and try to think about it alright let's see since the focal length is minus 200 the principal focus is over here somewhere on this side the negative side somewhere over 200 centimeters is over here right so we understand that the principal focus is over here now which lens will have a principal focus on this side well let's see if you had a converging lens over here then if you had a ray of light that goes like this it would have gone converged like this and so the principal focus would have been positive so it's not a converging lens it said diverging lens so we are going to be using a diverging lens over here so that when you have parallel beam of light an array of light appears to come from here with this alright but what we are asked is the power so the power is just the reciprocal of the focal length and we need to keep this in SI unit so that's meter so let's do that 1 divided by minus 200 200 centimeter is 2 meter so minus 2 so the power is minus 0.5 meter inverse which we also call as diopters so power is minus 0.5 diopters so what we would do is we would give this suresh a prescription written the read writing that he requires a lens of negative 0.5 diopter power that negative sign itself is telling us it's a diverging lens one curious question we might have is what if the object were to come closer what happened then well if you look at the ray diagram of a concave lens then you see regardless of where the object is its image is always formed within the focal length and since our our lens has the focal length of 200 centimeters which is the far point regardless of where the object is the image will always be within the green region and Suresh will be able to see it now of course if the object itself had to come within the green region if the object itself we're to come somewhere over here then its image would be even closer you can see that in the diagram and then maybe it'll be closer than the near point and maybe soonish will not be able to see that because it's just too close well in that case he just has to remove his specs and that's why sometimes you may have seen that you know people who are you know short sighted there might be varying specs when they are looking at faraway things but men when they're reading a book maybe they will just remove those glasses and they will read it normally