 Let's also numerical on the eye defects. Here's the first one. A person needs a lens of focal length 18 centimeter to correct his distant vision. What is the power of the lens in diopters? Well, the first thing I see is the focal length is given to me. I just have to calculate the power. I know how to do that. Power is one over focal length. So if you just substitute, we will get one over 18. But remember, to calculate in diopters, I have to convert this into meters. How do you convert centimeters to meters? Well, you just divide by 100. So if I divide by 100, the 100 goes on top. So I get 100 divided by 18 meters. And if I just do this calculation, I'll get 5.5 meter inverse, which is diopters. Okay. But the question, but another thing we need to think about is this, should this be a converging lens or diverging lens? To do that, I look at this sentence. It's given that we have to correct his distant vision. What does this mean? Correct his distant vision. It means he can't see far away clearly. That's the word distant says that, is not unable to see it. What's the condition over here? If you can't see things far away, but you can see things nearby, it is near sightedness. So we know this condition is near sightedness. So how do we solve for it? How do we correct it? Think logically. I don't remember, just logically. Well, if you can't see things far away, that's the way I think about it. If you can't see things far away, it means your eyes are unable to relax. Your eyes require minimum power to see far away. Remember that. So your eyes are unable to reduce the power. And therefore this means your eyes need some help in reducing the power. So how do you reduce the power? How do you reduce the converging power? Hey, you added a diverging lens. So the way to correct the distant vision is by adding a diverging lens. So the power of my lens is negative 5.5 diopters because for diverging lens, the power is always negative. For converging lens, the power is always positive. So there you have it. Similarly, if the question was to correct his near vision, then it would mean he can't see things close by clearly. And if you can't see things close by clearly, that means you need help in increasing the power. In that case, you would have required a converging lens. That's how you can figure out whether you need a converging or a diverging lens logically. Let's look at the next question. We're given a myopic person has a far point of 60 centimeters. What is the nature and the power of the lens needed to correct the problem? Well, let's first try and draw a figure. So here's the person is myopic eye has a far point of 60 centimeters. What does that mean? That means he cannot see anything farther away than 60 centimeters. Remember, myopia means near sightedness. He can see things close by, but he cannot see things far away. So in this particular problem, this person can see anywhere between 25 to 60. You might be wondering why 25? Because remember, nobody can see. In all our problems, we'll always consider you cannot see below 25 centimeters. You cannot see closer than 25 centimeters anyways. The normal eye can see 25 centimeters all the way to infinity, but this myopic person can only see between 25 centimeters to 60 centimeters. So the question now is what is the nature and the power of the lens needed to correct the problem? So how do we figure this out? How do I figure out what is the power of the lens required because this time the focal length is not given, we have to calculate it ourselves. Here's how I like to think about it. See, what we need to do is make sure that whenever objects are beyond 60 centimeters, its image must come between 25 and 60 centimeters. That's how the lens should be. So if the object is farther and farther and farther away, its image should be somewhere in between here and here. And finally, if the object is at infinity, its image should be over here. If we can make sure of that, we are done. I repeat, if the object is at infinity, its image should be here. Then if the object comes anywhere closer, its image will come somewhere in between this. So the key to our problem is realizing that when the object is at infinity, the image should be at 60 centimeters. Why do I put a negative sign? Because remember, right side is positive, left side is negative. So now I know the object distance and the image distance. I can use the lens formula to figure out the focal length and from there I can use the power formula to figure out the power. So great idea to pause the video and see if you can solve the rest yourself. All right, let's do this. So we know our power equals one over F but the lens formula says one over F equals one over V minus one over U. Now we just substitute the values. So you get minus, sorry, one over minus 60 centimeters, minus one over infinity, one over infinity is zero. So that means power is just one over minus 60 centimeters. We'll convert this into diopters now. So we'll divide by 100 to convert this centimeters to meters. So I get 100 on the top and if I divide this, I get 1.67 meter inverse, which is diopters and the negative sign indicates that the power must be for a diverging lens. So notice automatically it tells me that I need a diverging lens and logically it makes sense. We've seen just now that if you are nearsighted, if you can't see things far away, then you need to have a problem in relaxing your eyes. You have a problem, you need help in reducing the power and therefore you need a diverging lens. There you go. Let's go to the next one, the last problem. Okay, we're given that a person has a near point of 30 centimeters this time. What is the nature and the power of the lens needed to correct the problem? Can you pause the video and try to draw and then see if you can solve it yourself? All right, let's do this. So this time the person has a near point of 30 centimeters. This means he has problem looking at things close by. He has no problem looking at things far away. And so he cannot see between 25 to 30 centimeters. Okay, and why do I say 25 to 30? Because remember, nobody can see, when normal I cannot see below 25 centimeters. So we will not consider below 25 centimeters. But normal eyes are able to see 25 all the way to infinity, this person can see from 30 to infinity. And so his problem lies between this region. So how do we solve for it? Well, the same trick. If there's any object over here, its image should come over here. And so the lens should be in such a way that if there's any object here, its image should come over here. That's the trick. All right, so if the object is anywhere here, its image should come anywhere here. This means if the object is right at 25 centimeters, its image should be at 30 centimeters. If we can do that, then if the object is somewhere here, its image will come somewhere here and everything will be solved. So the key point over here is if the object distance is at 25 and again minus because it's on the left side, then its image should be at 30 centimeters, again minus because it's on the left side of the eye. And from this we can figure out what the focal length is and we can figure out the power. Again, if you substitute into this equation, we is minus 30, you is minus 25. So I get one over minus 30, minus one over minus 25, these are in centimeters. And if I take the common denominator and simplify, I get five divided by 750 centimeters. Again, convert that to meters, so divide by 100. Then when it goes on top, I get 500 divided by 750. And if I do this, I get about 0.67 diopters. This time I get a positive sign that means I need a converging lens. And that makes sense. If you can't see things close by, it means you need help in increasing the power. And how do you increase the power? By adding more converging lens. So you need a converging lens of 0.67 diopters.