 And so a plus lens, I kind of think that like this where you're putting basically two triangles bottom to bottom. And I like to think of that way because then if we I know that if I'm moving 10 millimeters down from the center, I'm going to have base up prison. If I'm talking about the right lens. I'm talking about the left lens I'm going to be adding base down prison. Based on Prentice's rule, patients reading one centimeter below the optic center. And so we're getting 1.75 prison doctors from the right eye, minus 2.25 from the left. So prismatic effect is for isn't that for space up in the right eye, right. Any questions on that one I think that was one that I sent out so we'll be at a chance like that already. Any questions. I guess I had a question, and maybe it's intuitive, but we have a base up prism in the right eye base down prism in the left eye. And so why are we adding the prism, as opposed to subtracting the prism. Good question so if you have opposite prisms in the two eyes. I guess if we had. Can you guys see my cursor moving around at all on here. Yeah. Okay, so if this, if this I was a plus 2.25, then we're getting the same direction of prism in both eyes. Does that make sense. There's the difference between the two at that point would only be a half diopter, which isn't going to induce very much prism. Because they're opposite signs. This left eye is bending the light up this right eye is bending the light down. And so it's almost like you're getting double the difference because each eye is doing a different thing. It's the dip it's, it's the induced amount of prism it's not the total amount of prism, the induced amount of prism. Yeah, and so that's that's one reason that you know somebody who the prison the prescriptions are pretty similar to each other you wouldn't expect there to be induced prism, but the more they have or especially if they have like a plus in the vertical and one eye and a minus in the other are you really have to pay attention to induced prism. Yeah. Now frankly, I in my mind I just think okay if there's a difference greater than about three diopters between the two eyes. This is something that I'll probably have to pay attention to less than that most patients don't complain about it. But definitely like if they are complaining and you look at the prescription and they are anisomatropic. That could be a reason so. Yeah, does that make sense. Yeah, yeah that does. Thank you. I included the ad. So, what do you do with that. Good question so the ad basically cancels it cancels each other out. So, you can get into some, some differences if you if you're talking about, I just don't know if I want to go there. But basically if you, if you include the ad, that would then make this left eye a plus a quarter. And then you'd add 2.5 plus 1.75 to this. And it would that the difference between the two eyes would still end up being the same. And the ad power is the same in both eyes just ignore it basically because it cancels each other out when it comes to parenthesis rule. Sorry, I just had a quick question as well. I'm Jordan on the intern. So if you're, I think where I got tripped up on this is that if the astigmatic axis the steep axis is that 180. I think I like ignored the sill because if they're looking like in the vertical meridian down isn't that the flat axis for the astigmatism so I just took the like sphere at like face value for each prison. Got it. So, yeah, that's why a power cross like this is something that's really helpful I think. I know in one of the video segments they talked about the power cross but I can't remember which part that was but so power cross basically just means you and you may have drawn one of these out but so for a power cross. So you take the first number and put it at whatever axis is in the second number so in this case, you put the plus one at 180. You add the two together and put it 90 degrees away. You take a plot you put this plus one on horizontal at the astigmatism. So plus 1.75 on the vertical. When it comes to parentheses rule. We're really only paying attention to the vertical especially what it says it's looking at 10 millimeters below almost every problem that I've seen is if it's somebody looking down in the glasses. And so you really. You do want to put it on a power cross and maybe just draw really quick these triangles. Just to help you visualize is the image getting pushed down. We have based up prison or based down prison. But yeah, I always put full powers on there just so I don't get tripped up by that. Yeah, and if you, you know, if you're thinking retinoscopy and I do have some retinoscopy problems in this presentation. You know, then you're talking about okay the beams falling across like this and yes I'm measuring the horizontal meridian, or now I flip my bean up my beam like this and I'm moving it up and down. And I'm measuring the vertical meridian. I wouldn't think of it like that in these problems I would just, if you see the prescription the glasses prescription I would just put it on a power cross really quick. If the ad powers are the same I would just ignore the ad. And then draw that power cross for both eyes draw your little triangles and within. At the same time you should be able to take the difference between the two and come up with an answer. Perfect. But yeah, we'll do, we'll do a few more of these problems as well as practice rule because they're, they're high yield on the test. So I hear if you and if you include the ad. And the right eye in the vertical meridian you're still eight but then doesn't it flip the prism in the left eye because you turn into plus a quarter in the vertical meridian. Is that true or no. It would, it would flip that prism but you're still basically taking the difference between the two. So, even though that is now technically base up it's much less base up than the right. That right eye is much higher so even though. So, yeah, like basically plus 4.25 in the right eye, and you have plus a quarter in the left 4.25 minus 0.25 still comes out to be for diopters. And then you're one centimeter below. So even though now that left eye would technically be base up, you have much less base up in the left that you do in the right and the difference still ends up being for diopters. That makes sense. Yeah, sounds good. Sounds good. Okay, we'll do it. We'll do it. What's that. I was going to say like, like the, the bifocal ad has to cancel out right because it doesn't even specify if their bifocals are progressives. So like you don't even know, like at what ramp that it's increasing power so it can't actually calculate the true prism from the bifocal or progressive ad right so it has to cancel out should not given that information. Correct you'd have to know if it's the seg the seg height. Different things like that. But once again if the if the bifocal power is the same. Really you're just dealing with the vertical power of the distance prescription, you can just kind of ignore that so. Yeah, unless the seg heights were different if the ad powers were at different levels and you might have to take that into account. That's getting pretty complex for like a problem that's only a minute long so. We'll get some more practice of these things because I think they're high yield they're, they're really helpful so. Okay, let me try to switch back to the who will go on to the second one. Nice. JCC and the lead here. I'm not even sure why it keeps points but probably some brownie points at the end I guess so. Okay, so question number two, can you guys see this. Yep. Okay, can you guys see the actual presentation view now. Okay, so very good job on this one. This is talking about a telescope, plus 20 lens minus 40 objective. One of those equations on that equation list today, but it should have been sent out by lane. It's this mag question. If you have a plus eyepiece lens and in minus objective lens just in your head if you have opposite science. That's a Galilean telescope. And this one's a little different because basically they're holding it backwards. Anyway, based on the equation, we have 20 over 40 which is a half point five X. So, the difference between their focal lengths ends up being 2.5 centimeters. So very good job I think there's only one. I can't even see who didn't get that right just. Oh that was that was me sorry I flipped my, my eyepiece and objectives in the. Yeah, my dad. No you're good. Okay, we'll do some more of these, these telescope questions as well. And, and sometimes drawing these out can be helpful to but I was going to keep the eyepiece and objective. In your head as far as that equation should be pretty good so. Okay, any other questions on this one. Go back to the food, especially as a rack up some points on this prerequisite. Okay, question number three, everybody pulling through on this one. What are you based on how quickly those were answered I think a lot of you did that one earlier. Let's go to the presentation. There was one more slide on the telescopes from before this is just to show where the focal point is in between the two lenses in the telescope. It's an astronomical. The focal point is inside so those are going to be longer. It's a Galilean focal point is an eyepiece behind the ocular. So those will be shorter telescopes. Here's question number three. So this e at the plane of the minus three lens. I think just drawing these ones out is the best way to do this to picture it in your mind and make sure that you're keeping your equation straight. Everybody got that one right any questions on question number three. Take that as a no. Here's the equation issues V equals you plus D. I think Jennifer you and I you learn this is V equals you plus P but very good. Yes, we're going to question number four. I'm loving these names you guys. Here's question for is just me or cannot select the right answer. Well, some of these. So there's a B or C. D and E. Some of these. Who would only give me four options that I could put in there so. Yes, I had to do some for anyone so if there's ever a question with five answers, one of the answer choices on who will have two options in there I didn't know how else to do it so. Awesome you guys all got that right. Again, so when you guys are figuring out question number four. Did you guys figure out the answer based on equations or did you just draw it and look at the image that was formed. I'm just curious how you guys came to that answer. I was told all minus lenses created upright virtual images and it's clearly smaller because it's closer than the object. Nice. Did anybody draw the chart in question number three. And then just use that to see the image look like. I was calculating the mag but I couldn't figure out where to get the sign that shows that it's upright. So let me share back to the presentation. So here's the drawing. And so, like I said for me whenever I do these problems, I draw it out, just to keep my signs correct. But also, on O caps you don't have an unlimited amount of time so if it's something that you would you look at it you can figure it out pretty quickly that drawing it no words but. So if you actually kind of drew out these lines doesn't have to be perfect but if you drew them out. And you'd see that the image is minified it's shorter. The image form is upright and virtual because it's behind the lens. So, the magnification use this quote this. This equation down here. So I'm sorry you were asking about the sign is that right. Yeah, so I used instead of the distances in my magnification I was using the virgin says, and kept winding up with a negative. A negative value on that on Magso like. So, I was using M equals negative you over V and had negative values for you and V. So, it was coming out to a negative magnification. Got it. I have to just look at the equation sheet again sir and plan through these back here. So this this equation here but then the negative you over V. Yeah, maybe, maybe that's the equation is not supposed to have a negative in front of it. I don't know. I will have to look. I, the equation that I don't want to use is that one that I put in in that equation list. I mean I do remember that negative you over V. But I will have to look back in time at my notes. No noise. Graduates and cobwebs on that one so I will put a star next to that. Okay. Yeah, based on, like I said, for me like if you've drawn the image already from the question before. Then you can just draw you can kind of tell what the image looks like and that's an easy way without even having to do any equations. The things that you guys talked about already though, generally work pretty well just to have in your mind as far as a bright virtual. So, okay. So you guys are doing is there is there a situation we didn't counter where my lens created either a real or inverted image, or can we just know that as a rule for our test. Good question so yes, a minus lens will create that, but it all depends on what they're asking the question in relation to So, you know because in this there's to a two lens system there could be multiple lenses. If they were saying, you know, in relation to the. No, I think I think if it's a minus lens of pretty much always would be a virtual upright image. But you just have to be pay attention to what they're saying in relation to because sometimes they might say something I just remember like, and maybe they were just complex questions back in the day, but they might say something like, you know, if from the perspective of the object. What what is this image, you know, look like, and they might just say like it's smaller and upright, but still be virtual so I think that I think pretty much keep that in your mind that it would be upright and virtual from a minus lens. I think keep in mind is just thinking about the incoming virgins to the lens. So if you have a lens system where the incoming virgins to the minus lens is like a plus five, and then that lens is a plus two. Then the outgoing virgins is still going to be a plus and so it's going to be a, you know that will when it holds so I think that's important to keep in mind. Gosh, I think that does make sense. Thank you Jennifer. Once again, we're drawing it out might be helpful, but yeah. It's a single end system then it would always be the case if the lights coming from the left, but the very the incoming virgins definitely matters. Okay. More practice problems. So please keep in mind I believe this is still the case where you have 480 minutes for 460 questions. That's a little over a minute per question. Some questions you'll fly through some of these optics one just take more time. But if you find that you're getting stuck on something just, you know, come back to it but no calculator so kind of do some today. We already have some scratch paper and pencil. So this is the next question let me put it on the food so you guys submit your answer. JCC prisms on a roll here this is good stuff. Okay. All right. So it looks like a couple of you got fooled by the working distance. Let me see. All right. Who is JCC. That's me. Nice. Optics master. Jennifer, can you explain why the answer is C and not. I think the other one, excuse me, why the answer is the. So you have to take into account your working distance when you're doing retinoscopy you as the observer are only 50 centimeters away, you're not at infinity. So you have to take one divided by point five to get two diopters of, you know, kind of induced virgins because you're an observer closer. So whatever answer you get like the plus 22 you just have to take two away from it. So you end up with a 20 and a 25 instead of a 22 and 27. Nice. So can you guys see the presentation now. Are you guys still seeing the cahoot or are you seeing the presentation now. We're seeing the presentation. Okay. So this is exactly Jennifer saying so. We're doing retinoscopy working distance matters a lot. And so it should tell you a working distance and you have to account for that urgency. So like you said one over one over point five is a minus two. And so, if you put your, your retinoscopy powers on a power cross, take that virgins into account will end up with a plus 20 and plus 25. But if you forget this virgins, it can definitely throw you off. You forget that working distance. And I am and I am very impressed you I you all got the powers on the right lines, which with retinoscopy can be a little bit tricky sometimes. Any questions on on this problem. Going once points. Okay. Great job on that one. Let's go back to the compute everybody rocking it on that one. spherical equivalent is something that thankfully is pretty simple. I don't know how high yield that is but it's good to have, have that skill down clinically and potentially on the test as well. You guys all got that one right basically. Here is the equation. The sphere plus half the sale. And I guess the only thing that I would pay attention to be close attention to using is if it's plus sale or minus sale. Because that would determine if you're either adding or subtracting from the sphere power. Do you guys all did really well enough. So, oh, there's a sneak peek at the next question. Any questions on spherical equivalent that's that's pretty simple stuff. Any questions there. Any questions on the next question. Hey, some Jordy. Talking about that one. So optically empty just just made a push there who is optically empty. It's me. Allie. Is that right. Yeah. Awesome. So, Ali, can you explain how you got that answer and how you got it so quickly. With prisms. They work off like 100 centimeters being how much they're going to displace. So, with this question. They, um, I think it was like 30 or 33. So I knew at 100 centimeters, it'd be six. So then 30, it'd be about two. Yeah. That's a good presentation here. Exactly. So we're at 33 centimeters, which is about a third of a meter, right. So if it's a six prison after you take a third of that, which should be about two centimeters. Great job. So you can draw this one out as well. With optics drawings are usually helpful. They just take a minute to draw. But yeah, if you have that 100 centimeter number in your head. You can just figure out how what length. Or how far from the prison they're talking about. Do a quick ratio and apply that to the prison. I think there's one person that didn't get this one right. Are there any questions. On prison. I just put the wrong ones. There we go. All right. Okay. And another thing to keep in mind. I'm going to go back to prison. Maybe applicable later. We'll see. For angles under 45 degrees each degree. Equals approximately two prison diopters or two prison doctors per degree for angles under 45. The big one I think they're going to be testing on is, you know, is to find this deviation of one centimeter at one. Love it when everybody gets it right here. So this basically explains. So far point in your point. So for an uncorrected mile. Our point of 40 centimeters. The power is just one over negative point four meters 40 centimeters. It's up to minus two and a half. Great job. The next question is also related to this. This minus two and a half. So don't forget that number yet, but. Any questions on how to determine. The refractive error for a mile or a. Hyper over any questions on that. Okay, let's do the next question. Like I said, it's related to question number three. Yes. Okay. So one for num or for answer D. And then five for minus one and a half. So what are the questions making the remove whose four eyes. I am. Yes. So hard. So, Ariana, can you. Can you tell us how you got that answer? Yeah. The patients. Regular far point. Told us that they have. Normally minus two. And then they would have to accommodate. So how do you know if you're supposed to like add them together or take the difference there? Well, you know that he's a mile. And so he has. Power in their eyes. And then to get to this closer point, they would need. Minus four. And that's calculated from the 25 centimeters. So the difference in power. Is what they have to accommodate. So how do you know if you're supposed to like add them together? And so he has this. Near point like he's near sighted. And then this is now a closer near point. They used to get to. Awesome. Yeah, absolutely. So once again, if you ever can't keep that in your mind, you could draw something out really quick, but. Yeah, it's basically just taking that difference. And you end up with one and a half diopters. Of accommodation. Yeah. It's much, much easier if you put it into the powers and diopters to do a problem like this when you're accounting for equation. Excuse me, accounting for accommodation. So yeah. And they're, do they do questions on O caps that are kind of sequential. Are there. I don't think so. Not necessarily. Okay. So they could easily come. Yeah. Again, I feel like sometimes they had a two part question. Maybe like two questions are tied together. Got it. So they could put something where these are either separate or together easily. I mean, I just split it up just so you can do that, but. Yeah. Anyway. Putting it into the power and diopters is very helpful for some of this. Any, any questions on. Accounting for this accommodation right here. Next question. True or false. So glad you all got that right. Okay. So these next questions are tied together. Even if they don't have this type of question on. Oh cap where there's multiple questions in a row together. I think that it's actually very, very valuable. Practice to do this. So. So this is question number five. And you guys see it on the presentation there. Back in presentation. Does anybody need more time on question five. It's harder without being able to tell when you guys return. Okay, I'm going to move on to question number six. Here's question number six. Okay. So we have five more seconds for this one. Okay. Question number seven. Anybody need more time for question seven. Here's a question eight. Is 1.5 the distance between the two lenses. Yes, it is. Great question. Yes, it is. Anybody know that need more time on this one. Let's go question nine. The last one in the series. This is where it'd be great to be in person. Okay. I'll be COVID. Five more seconds. So. Are you on here? Yeah. I'm going to talk through Ariana's. Okay. Perfect. Can you tell me what you've got for question number five. Okay. So I got B. So you is minus one. One over four is point two five. And that's positive. So it's 25 seven years to the right of the lens. My wonderful brother. That's one of my favorite pictures. If you mind just kind of put that on. Don't show them that in the recording. Okay. So you ask question number five. The 25 seven years to the right of the lens. Awesome. And you just use that B equals you plus D. Equation for that. Okay. So the big thing here is just make sure you're keeping your science straight. If the object is to the left of the lens, which is the standard. Then. That's going to be minus one. Added to a plus five gives you that plus four. Plus is going to be to the right of the lens. So 25 centimeters to the right. Excellent job. All right. Any questions on this problem that anybody. How many questions on this. If you have any questions, feel free to ask Cole. He's got this one down. All right. Tyler, are you available? Yeah. Okay. So for question six. Can you tell us what you, what your answer was and how you got that answer? So I think the answer was. Sorry, B. It is real because it's a converging lens. So light actually passes through that point. It is if you draw the central ray from the tip of the object through the center of the five that are thin lens and beyond. So if you do the magnification equation, which states the image distance divided by the object distance, it should also be smaller. Awesome. So let me just go to the next slide here. Yeah. So. I just did a quick drawing as well. And you can also buy that see that it's inverted real to the right of the lens and reduced it is smaller. But absolutely just keeping those things coming in your mind. Okay. It's a converging lens. And then you can do the magnification equation. Jennifer, do you usually draw the rays or do you just. Do the equation and I don't know what to expect. Just curious. I like to think about the distances. So for, you know, this example, I forget how much it was, maybe 35 centimeters, but I'll do positive 35 over negative 100. So then I know it's minified because it's less than one. And it's minus. So it'll be inverted in real. Got it. That's great. That's great. I asked Jennifer because she's, you know, fresh out of optics boards and all that fun stuff. So yeah, there's a couple ways to kind of keep it in your mind. That's a great one. Just using the distances. So, okay, any questions on question six. Lydia. Are you on here? I am on yes. Great. Can you pull up question number seven here. You tell us what you got for question seven. Yes, so for question number seven. I said that the size of one second. The size of the intermediate object. So it is smaller. I actually thought that it was one half of the size, but based on the drawing and the previous question, I believe that the answer should be a force of the size. It is one fourth of the size. So. Yeah. So equals you plus T is a good way to kind of determine how. Or what the size is. So once again, like if you draw it. If you just, if there were these series of what is that four or five questions together and you did a drawing initially. And just drew where the image was, then a lot of these questions could be answered just by looking at that image. But it does take a little bit of time. So just as long as you're going to be able to keep that kind of image in your mind. Then you could just answer with that. You can also just use the distances, like what Jennifer was saying. The distance of 25 centimeters versus one meter. That's one fourth of the distance. So we expect one for the size. Yes, or you can just use the equals you plus the equation. I think the picture is very helpful the drawing. Yeah, pretty helpful. And so it goes over in the video. How to do these drawings. And also, so that grand rounds on Wednesday. I had not seen any of those resources on the Academy website. I'm going to try to incorporate some of those into my pre-work for next time. It sounds like some of you, each UI twos have been able to utilize that resource. But when I was browsing through it looked like it had some really good information on telescopes and I think one of them was a tracing. But basically, you, if you have your, your, your object and you draw a line straight through there, if it's converging lens, you always, you always draw it straight through where the center of the lens meets the horizon. And just you can continue to continue that one on for infinity. If it's a converging lens, you would. Yeah, I guess you do kind of need to know the distance away from where the lens meets the horizon. So it would be converging and just draw it. Anyway, it goes over on there. I don't want to take too much time doing that, but if you, if you learn how to ray trace quickly, it can be super helpful for questions like. Any questions on this one. Okay, for question number eight. Harry on it. Can you walk us through question number eight. Yes. Okay. Question number eight. The, the new you for the new image suffrage and for objects. I first placed the object 25 centimeters to the right of the plus five lens, which puts it at 1.25 meters to the left of the minus two lens. And then my new, that's my new object. So my new object versions is. The negative of the inverse of 1.25. I made that a fraction, which is like negative four fits. And then I added in the. Our lens, which is minus two. So now it's minus two added to minus four fifths. That's negative 14 fifths. And then our new distance is the inverse of that, which is five fourteenths. That's about a third. And it's negative. So it's to the left of our new lens. So about a third of a meter to the left, which is D. Yeah. Very good. So, I mean, you basically just walked us through exactly what I have on here. You have to make sure that your object starting out is in the correct position. 25 centimeters to the right of the first one. Take that difference. And then it's just a matter of keeping your sign straight. And that equals you plus the equation. And exactly. And sometimes you can get caught up on exact numbers. So things like. The dimensions or rounding can be very helpful. Are, you know, you, you're exactly right. It ends up being about a third. So close to 33 centimeters to the left. And that while there's not an exact 33. To the left option. There's a 36, which is pretty darn close. And so that's what you go to. Yeah. So that series of questions, I guess we have one more. So let's do question nine and then we'll. We'll talk about how helpful you since our, but any questions on question number eight for now before we move on. Okay. Let's go. Tyler. Are you on here? Yeah, I'm here. Okay, can you walk us through what you got for number nine? Um, so. Yeah, I mean, and if you don't know, that's totally. No, I think. I'm a so given that it's a negative two diaper lens. My assumption is that it's going to be a. Virtual image, which gives C or D. And virtual images tend to be. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Okay. Well, thankfully. You were right. Um, once again, just coming. From just like a, just a sheer rate tracing. If we started out with our initial. Our first image ends up being right here. Okay. And so if you had drawn this out or just, you know, and then you drew your primary rate through the center of that second lens. And then you knew that the new image was to the left. You could tell just by this image, if you just put about where, where it was on here on the drawing. It's going to be reduced. It's going to be virtual. And it's going to be inverted. It's going to be inverted. It's going to be even just drawing like one single rate can be pretty helpful if you know where the object image or ending up. But yes, this is a case where since it is a minus lens that it is inverted. But yeah, I, I find for me personally drawing it out. It's very. Great job. Okay. So I think if you were able to do those questions. That is awesome. I mean, if you got all this right. I think that's going to cover a lot of material as far as images. I mean, a lot of high yield stuff. And especially using that equals you plus the equation. That's what most of this is just keeping your sign straight. I don't know that ever do anything this complex in a row. Okay. So that's questions five through nine. Any questions on any of those before we move back to the boot. Okay. So I wanted to. Quick question. I think I remember. At somewhere along the lines, somebody saying that. A. Hyperropic. Well, a. Negative lens. So like a negative two diapter lens will only produce virtual images. But a plus diapter lens can produce virtual or real images depending on. Where the object is. So that was one thing that Jennifer was talking about earlier. Yeah. For both a plus or a minus lens, the virgins coming in. Matters. As far as the type of object that will be produced. Okay. And so. If. You know, if this image was inside the primary focal point of the lens, that will determine. If it's a real or virtual image that is produced. And if it's to the right and the left. I. Cannot imagine that they would do something like that on. Okay. Because it would take like those problems on tests that we did in optics would take five to 10 minutes each to calculate sometimes putting our clean, like having to draw it all out. I don't know. Those of you who have taken O cap. Have you run into a scenario like that? Not that it's all about the test, but I'm just curious. Has there been a scenario where. Where they had put an image inside the primary focal points that you were dealing with. Something different, I guess. I feel like. In my memory, they could. The optics could be anything like there could be a straightforward question, but there are also questions where you. Really, you have to do some work on it. So. Yeah. It didn't surprise me. So even if it was inside the primary focal length, if you keep the V equals you plus D equations straight. You should be able to determine how far to the right or to the left. The image is produced. And if it's to the right of that lens, it's going to be a real image if it's to the left of the virtual. And really that's what it all comes down to is. Where is that image being produced to the right or to the left of the lens. So I think if you, you know, if you had to draw it out, that's where it gets complicated. If it's inside that primary focal point. But if you just keep your equations straight. And then you should be able to determine if that image is to the right or to the left of the lens. Does that make sense? Hopefully that does like that vehicles you plus deal with these lens equations. I mean, if you only knew that equation, you could come up with the answers for almost. Okay. Let's, let's keep cruising. We're going to go back to who can you guys see the commute screen now. Yes. Okay. So here's the next one. Double points. Okay. I don't have a double points, but. I had about 10 seconds. Nice. This is why it was double points here. So this one was obviously a tough one. Kind of through a curve ball there. Let's see. Optically empty. Nice. Solid work. Hey, Ali. Was that a guess? Nice. Okay. Does anybody know what equation we would use here? I know. I think you still do. You plus D equals B, but you have to. First calculate the power of the mirror. Which is a two over R. I think. Two. So the power of the mirrors. One day after. And so I think the object distance was 33 centimeters. So the last one is. Minus two. So your object distance is then. 50 centimeters. So your image just. Sorry. Your image distance is 50. So it's 33 divided by. Or 50 divided by 33, right? Let me share my screen. Okay. So exactly right. So. Power of the mirror is two over R two over the radius. So you have to calculate. So we have to calculate that power first. Two over two is one. Our object. This plays 33 centimeters to the left. Minus three diopters. And then the, so did any, was everybody able to get this minus two diopters for. The image. Did anybody get to that point. Or did the radius just throw everybody off? I kind of got my time to make stuff. That's where. Where I lost, but yeah. I think it makes sense to think of a. The mirror as a lens. Which is, which in my mind, when I think of it as a lens, then I think of the U plus the virgin's formula. But if I didn't think of that, then I probably wouldn't have gotten to the virgin's formula. I don't. I didn't fully. I still really understand how you know, it's like a plus versus the minus 1.5 X. That's what I. That's what I'm still kind of stuck on. Got it. So it. It all comes back to this equals U plus the equation. And this is once again, we're drawing it out can be helpful just to keep. Objects and images. So. Exactly right starting out so you have to figure out the power of the mirror once you get that power of the mirror. That ends up being a plus one, then you can plug in that be equal to plus the equation. The equals so you are object distance is minus. 33 centimeters because it's to the left of that original lens. So that would give me that you is a minus three. Be. We take minus three plus the one from the lens we get minus two so both are B and are you are negative. Our transverse magnification equation is transverse mag equals you over. And so you would plug the minus three. Over the minus two. And then you get a plus one point five X. So in this case, it's all just based on that B equals you plus the equation. So if the signs get mixed up, it'd be super easy to get thrown off on it. It's a plus or a minus. But since you is a negative. The negative three. And B is a negative, a negative two from our equation. The negatives cancels each other out. You have a plus one point five X. Does that make any sense? The, the signs make sense in that, but what is it? Do you know what it means like the real world? Is that signifying like inverted or upright? If it's plus or minus. Jennifer, that would mean it's an upright image, right? She's on. Yes. So the plus signifies that there's an upright. I believe so. Yes. And magnified by one point five times. And then the negative would mean it was inferred. Okay. Cool. Thank you. Yeah. No bro. So that was a tough one. Obviously that's where the double points came from. So. Solid work on guessing that one correctly. Yeah. Keeping the sign straight. Makes a huge difference, but. Taking a second just to draw it. Draw it out. Okay. Let's go back to the boot. I'm a pretty nice person in real life, but this question does not. Make me a nice person. All right. Three seconds left. So. Let's see who got that one right. Okay. Did anybody actually try to do the math on that one? I tried, but full disclosure didn't get done. I just guessed what I thought might be about, right? Yeah. I hope that. That's all that work. Did anybody, I mean, were those all just guesses or did anybody have. Like some thought processes to what, how they came up with their answer. Let's see. Arianne, I think you got that one, right? How did you come up with that answer? All right. Well, like Jennifer. My calculations were starting, but then had to quickly click one before finishing. I was starting to use. Like a lens maker formula with the. Like the power is proportional to the index of refactions, but did not get far and writing everything down. So how did you come up with the, an educated guess on that? It was asking for a corneal power. Which is close to 40 in a real eye. So I went with the answer close to 40. Awesome. So let me, is that kind of what everybody did? Was there any other clue that led you to. To answer D for anybody. You can also use the thick lens equation. Let me just share this really quick. So. You can actually come to this answer using all these equations. I'd be crazy to try to do that when you only have a minute and seven seconds per question on average. So basically the thing to pay attention to, if you get a question like this, there are good ways to make educated guesses on this. Like Arianna said, like a corneal power is close to 45, right? But that's in air. Okay. So the, the kind of key words if there is such a thing as assume index of refraction of air equals one. And it gives you the cornea and the aqueous. So if it's asking in air, it's going to be probably pretty close to 45 because that's what the average corneal power is. But it could give you this in water. Okay, it could say that somebody's opening their eyes under water and that changes things drastically. So as kind of a rule of thumb. If it's in air, it will be close to 45 diopters. If it's in water, it will be close to minus a half diopter. Like the lens maker equation and stuff is, and the thick lens equation, both very useful. And, and frankly, I think, you know, you could calculate those out. But something like this, if you just, if you can just keep that in your mind, like, I mean, I use this in contact lenses and in cornea, I use this in all the times that the average corneas around 45. So that's something that's good to have in mind clinically as well. But if you just keep those two rules kind of in your, in the back of your head and then pay attention to if the index or a fraction is in air versus water, that should be able to help quite a bit. So that was kind of a mean one. I didn't actually expect anybody to get that question right in 90 seconds. Do those rules. Make sense to people. Any questions or in terms of that. Okay. All right, let's keep going to the next good. So kind of a split on that one. All right, so let's see. Dr. petty my team over here, we just clicked a little too fast. That was our bad. No, you're good. Let's see. Cole, can you tell us what your answer was? Well, I was the one who got it wrong. I changed the sign too early. But basically you add the plus still to the sphere. And then switch the sign as well as change the axis by 90 degrees. So we did that we just did it incorrectly. All right, so let's go here. I would normally do a seventh inning stretch, but we got to get through all these questions here so. Yeah. So to switch from plus still to minus still form exactly what you said, you add the seal and sphere together just make sure you're paying attention to the signs, flip the sign of the sale and then change the axis by 90 degrees. We come up with the question. I'm curious, did anybody like mess up on that? Without just accidentally hitting the wrong button. I get any questions on how to switch from plus still to minus so. And vice versa. I got something that could easily show up and also very helpful in real life. You know, if you're switching back and forth between contacts and glasses. Nice job on that one. Here's the next question. Okay. I'm going to give you a little bit of a split again on that one. So let's see how it came up. Remind me who prison is. I don't know who that one is. That's me. Abigail. Okay, can you tell us how you came up with that answer? So. I knew that. Each prison. Diopter is equal to like half of the degrees when it's less than 45. So I picked the closest to 20, right? So the closest to 10. Exactly. Yeah. So, okay. I think that was something that I mentioned on an earlier slide about prisons. The exact number is 0.572 degrees. Times the amount of prison. Or gives you 0.572 gives you the amount of degrees. So it's, you know, it equals 11.4. That's about 10. If you just, it, like I said, sorry, let me back up. If it's less than 45. Then it's about half of the amount of prison. So that's a pretty good rule for a most prism. Certainly that we're prescribing is going to be much less than 45. Did anybody accidentally get that wrong? Or just not remember that? Look like maybe there are a couple of them. Just keep that in your head. Degrees half the prism. As long as we're under 45. Next question. Starting right now. This is a telescope question. That's rock that one. Can you tell us how you came up with an answer? So I, I'm not participating in the Kahoo chat. I have to be honest because I was late because I was. Oh, I'm so sorry. But it is because it's the, like the way the. How should I explain it the way the. Lenses are aligned. It's the Galilean. And then can we go back to the numbers of the question, please. Sure. Let me pull up. And if you do want to hop in, hop on the pin and pattern around the bottom, if you want to hop on there at some point, but let me go back to the question. Okay, so here's the question here. Yep. So. If you're subtracting the focal length of it. From the, from the two objective lenses. That will tell you that it's the. Galilean lens. And it's like, yeah, you're subtracting 10. Like minus 10 plus four. Let me put this up for you. So, yeah, so one thing to keep in mind, as far as like Galilean versus astronomical. If they're two plus lenses, you know, it's astronomical, which is going to be longer. The focal length is going to be in between. If it's Galilean. Then you're going to have a plus lens and a minus lens. I guess that's how you know that it's Galilean plus objective and a minus ocular. And then the focal point will be outside. So it's going to be a shorter telescope. So one thing to keep in mind, if you're given like the powers to find a distance, you'll, you'll want to take the inverse of the power. To find the focal length of that. So one over four. Plus one over a minus 10. You add those together. So 25 centimeters. Plus a minus 10 centimeters equals 15 centimeters. And yeah, so. There are definitely some telescope equations up there, but if you can just keep that in your head like somehow like astronomical is two plus Galilean is a plus and a minus. And the astronomical one is longer. If it's longer, that means that the focal point is inside or in between the two lenses. You'll be able to, like, that will help you to answer a lot of the telescope questions, just knowing that. If we just look at here, we got a plus four and a minus 10. Okay, well it's a plus and a minus. I know it's a Galilean that eliminates C and D. And then you can go ahead and, and plug in those powers. You take the difference and then that's about 15 centimeters. So, thank you. Yeah, you're very welcome. Yeah. Anybody else have any questions on this one? I think this is something very typical of what you might end up seeing. So you've been kind of those just general rules. Any other questions on this? Okay. Because the next one. Okay. I'll move that all over the place here. Telescope ones are tricky. So I think, so let me, let me pull up my, let's see. JCC man. Okay. So, for angular magnification. Server that just go. The answer is C minus five X. This gets back into the question of the minus versus plus on magnification. What does that mean. So the equation is for the telescope is magnification equals a negative power of the octave over the power of the objective. The reason that the negative and the plus matters. So a negative magnification will give an inverted image. So it's still magnifying it when you use an astronomical telescope. It's still magnifying and it will still give you a larger image. But it will be inverted. In other words, everything that you look at will be upside down. It's a different formula than the formula that's like. Diopter of the eyepiece over diopter of the object of the, of the, like, objective, I guess, is equal to magnification. I mean, it's for all the way back up. I think I may have confused ocular and objective or eyepiece or something. You may have just seen the answers. I think in one of my lectures I did before before we had the reverse classroom. I had some pictures of what the different telescopes look like. And a, and a Galilean telescope, those will usually just be straight like if you think of what a pirate would look at, you know, look out on on a ship or something like that just like a straight telescope. An astronomical telescope usually has a band in it. They're not straight. So a lot of the telescopes you just see like somebody at a football game it might have like a band in it. That's because they have to have. They have to have mirrors in there to fix the image so it's not inverted when you look at it. So an astronomical telescope. There's just kind of some rules with them for an astronomical telescope would be greater than 4x. You'll have a plus power ocular or both lenses will be plus power, and the magnification will be negative or inverted. Yeah, I was just looking at that equation for the magnification. The one that is on the handout that I gave you guys. And it doesn't have that minus there for some reason. I don't know when I was looking at this, this problem online I definitely have that one so I'm going to have to look at that equation again. Jennifer do you remember if the equation has a minus in it. It does. The equations minus the power of the I piece over the power of the objective. That's my thought but for some reason the one that is on the paper which I just, I got from the vocab book. Because the image, they're talking about its image distance and object distance. And the image is a minus because it's coming from the left, whereas the object is a plus, if you're considering it in that way. There's a separate equation that's the telescope magnification equation that's the second to last one that doesn't seem to be wrong. Yeah, yeah, because it should definitely have that minus in there. I've never heard the telescope magnification. Because that's how you know that the astronomical hasn't heard it magnification so I will fix that on your equation sheet. The whole stars. Yeah, we only have two minutes let's do a few more questions if we can. I'm flies when having so much. I thought I'd fixed all the issues with last year. Okay, here's number 14 be glad you guys are nailing. I'm sure so. Okay, I think it's all about that one right. I'm just going to show you this page really quick so present after one centimeter at one meter distance we have half a meter. So we need to double that. Object size to get a prism. Yeah, you guys all got that right. Any questions on that one. They will keep moving through to get through as many more of these as we can. Do you see the good boot again. Yeah, we can. Okay. So this is that same equation that we were just looking at that and equals the power of the ocular over. I'm sorry the minus power of the ocular over the power of the objective. So, how are you able to tell astronomical versus gala way, Jennifer. We know it's astronomical because both powers are positive. Right. And to Jordan. Were you able to come up with, or how are you able to come up with the magnification on that. It's just the telescope magnification equation. So it's just the negative power of the piece over the objective power. Great. Yeah, you guys are really well on that one type race. All right. Okay, next question. You might get through. Fantastic. All right. There's kind of the images here. Ariana, can you tell us how you got that answer in the right eye. So first of all, we're looking below that. So we're interested in vertical in the right eye. The cell. Is actually acting at 180 so we can ignore the cell there. And then we multiply one centimeter from the 10 millimeters times minus three and get minus three and then in the left eye. So minus one and then the plus one that's acting vertically cancel out. So, left eye is zero. And then the minus three, like you drew there is, it's a diverging lens, it's a base down so three base down on the right eye, nothing in the left. Awesome. Yeah. So, like, draw the picture or do you kind of just do those in your head. I'm just curious. What's most effective for you guys because everybody got that right. Did anybody draw a picture. I draw a quick picture, just two quick triangles. I draw two. May I ask a question just to see if I understand this correctly. If the patient was not looking telling 10 millimeters below the optical center but 10 millimeters above the optical center. The only thing that would change is that it would not be a base down but it would be a three day after base up. Correct. Yes, and that's where drawing the picture is helpful to see which direction the prison base is going. So, drawing this little question, or excuse me, these little triangles will show which direction the prism is now are going to do great job saying okay that you know they're looking one centimeter below. So we know that we only have to worry about the vertical. If this is saying that they're looking 10 millimeters to the right, then we'd be worrying about these horizontal numbers K which would end up being different. And in this case would end up being a different answer. Just pay attention to how far and which direction. The patient is looking all the ones that I've shown here have been vertical and looking down, but they could technically ask any direction and and amount I would hope that they would do 10 millimeters one centimeter or two centimeters something to keep it simple for you. But they could do you know, half a centimeter or something that would you just have to plug that into your equation here. I have a question on that though they were looking up it would end up being base up prison. Any other questions on this. Okay, I think that time for our last question to the right or to the left I don't think I would get to that answer. So how would the equation differ if this patient would look 10 millimeters to the right. Like, how would we describe that with the basis, it would be. If it's looking to the, like it would be the base out. But how would the present day of just be good question so, so let's say that the patient was looking to the right. Okay. So that now we have to like think about the patient's perspective so that would actually move what we're looking at over to the left on this can you guys see my cursor there. Yes, yes. Okay. So, in this case 10 centimeters we would multiply that by the minus two. So two minus two times one centimeter. We'd have two prism diopters from this lens. We would have one person diopter from this lens because minus one times one centimeter gives us one. And in this case it'd be still be a base out. If you drew the picture you'd have triangles like this. Hopefully you can kind of tell what I'm doing from a cursor there. But it would be base out in both cases and then we just take the difference, a two person doctor minus one person doctor means that they would have one person diopter base out in the right eye induced by that lens system. Wonderful. Thank you. Yeah. Yeah, great question. Let's do our last one really quick. The great double points. Really just want to hash these ones is apparently pretty heavy. Oh, can you guys see it so I can you guys still see the cahoot on there. Yes. Okay, perfect. All right. Got that. So, see, see could be either, or I guess it was C or D so which one did you guys come up with. See, see. Great. Let's see. Ali, could you just walk us through this one really quick. Sorry, I missed the beginning of the question so I was still doing my math, but it is similar but the thing is that you have to flip it to, because you gave it to us and negative. So, right. Correct. Yeah. So you have to flip that, and then it's similar to what we did before so in making your power crosses so it'd be like the right I had minus three at the 90 minus four at 180. The left was minus seven at 90 minus six at 180. The right was minus three minus seven to do the multiplication with the 10 millimeters below. I think I can't see the problem. No, you, you got it totally right so. Hey, you're exactly right to you put it on a power cross. Now you had mentioned, you know, putting this in plus still form first. You don't necessarily have to do that. I know that that's kind of how we think in ophthalmology. Yeah, so, so it's completely fine to do that. But on a, you know, putting it on a power cross like this like you just put the first number or whatever axis this up this tells you so minus three at 90. And then you just add these two together and put it 90 degrees apart so you could just go straight to the power cross. And at the same time, you know, like, if you're more comfortable working and plus so just put that in plus certainly quick, and then put it on there, but you could technically save yourself a step there, but absolutely did the rest right. We have a minus seven in the vertical on the left eye minus three vertical on the right. Take the difference or you know, together basically and you get minus four diopters were 10 millimeters below. One centimeter. So we have four base down in the left eye. Great job. Any questions on that one. Back to our who here. Let's see who took the podium here. Keplerian King. Who's the Keplerian King. I don't even know who that was. That's cool. Cool. Nice. Okay. Well done you guys. Yeah, optics. There is a ton of potential things to cover if you watch those videos. I hope you paid attention he kind of has like the top 10 list of hot ticket items on the test of high yield material. And I think in his lectures he does a really good job covering those top 10s and a few other things as well. I tried to gear, you know this towards most of the top 10 items and a few other things mixed in. But I know we're kind of at a time does anybody have any questions. This is the last couple minutes here. Before we end. Okay. Thanks Dr petty. Hey, you guys are so welcome.