 A plus, oh, 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 it that way because then if I know that if I'm moving 10 millimeters down from the center, then I'm going to have base up prism. If I'm talking about the right lens, I'm talking about the left lens, I'm going to be adding base down prism based on Prentice's rule, patient is reading one centimeter below the optic center. And so we're getting 1.75 prism doctors from the right eye getting minus 2.25 from the left. A little prismatic effect is four prism doctors base up from the right eye or right hyper. Any questions on that one? That was one that I sent out so we'll be at a chance like that already. Any questions? Good. I guess I had a question and maybe it's intuitive but we have a base up prism in the right eye and a 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 eye was a plus 2.25, then we're getting the same direction of prism in both eyes. Does that make sense? 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. So it's the induced amount of prism. It's not the total amount of prism. The induced amount of prism. That makes sense. Yeah. And so that's one reason that somebody who the prescriptions are pretty similar to each other, you wouldn't expect there to be induced prism, but the more anti-symmetropia they have, or especially if they have like a plus in the vertical on one eye and a minus in the other eye, you really have to pay attention to induced prism. Now, frankly, 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 anti-symmetropic, that could be a reason. Yeah, does that make sense? Yeah. Yeah, that does. Thank you. Yeah, no worries. I included the ad. So what do you do with that? Good question. So the ad basically cancels each other out. So you can get into some differences if you're talking about, I don't know if I want to go there. But basically, 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 the difference between the two eyes would still end up being the same. So if 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 at 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? I just took the sphere at 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. 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 this second number. So in this case, you put the plus one at 180, you add the two together and put it 90 degrees away from there. And so you take a plot, you put this plus one on a horizontal, add the astigmatism, so plus 1.75 on the vertical. When it comes to parenthesis rule, we're really only paying attention to the vertical, especially if 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 base up prison or base down prison. Yeah, I always put both powers on there just so I don't get tripped up by that. Okay. 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 beam up my beam like this and I'm moving it up and down. And I'm measuring the vertical meridian. And if I do 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 within a pretty quick amount of 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 their high yield on the test. So I hear. If you include the ad, then in the right eye, in the vertical meridian you're still faith, 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 eye. That right eye is much higher. So even though, so you have like basically plus 4.25 in the right eye, then you have plus a quarter in the left 4.25 minus 0.25 still comes out to be four diopters. And then you're one centimeter below. So even though, yes, even though now that left eye would technically be base up, you have much less base up in the left eye than you do in the right. And the difference still ends up being four diopters. Does that make sense? Yeah, sounds good. Sounds good. Okay. We'll do it. We'll do it. What's that? I was just 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 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 give them 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, the. The vertical power of the, 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, then you might have to take that into account. That's getting pretty complex for. Like a problem that's only a minute long. So. Yeah, we'll get some more practice of usings because I think they're high yield. They're really helpful. So. Okay. Let me try to switch back to the cahoot. We'll go on to the second one. Nice. JCC and the lead here. I'm not 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? Yeah. Okay. So very good job on this one. This is talking about a telescope. Plus 20 lens minus 40 objective. So if you have a plus eyepiece lens and a minus objective lens, just in your head, if you have opposite signs, 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. 40, which is a half point five X. And 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, which is. Well, 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 sometimes drawing these out can be helpful too, but. As long as you can 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 pull them through on this one. I think a lot of you based on how quickly those were answered. I think a lot of you did that one. Earlier. Let's go to the. 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 a telescope. So if it's an astronomical. The focal point is inside. So those are going to be longer. It's a Galilean focal point is. The eyepiece behind the ocular. So those will be shorter telescopes. Okay, 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 is used to equals you plus D. I think Jennifer you and I you learn that says equals you plus P. But. Okay. Very good. So going to question number four. Kepler. I am loving these names. You guys. Here's question for. Is it just me or can I not select the right answer? Well, some of the. 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 finet one. So. If there's ever a question with five answers. One of the answers. We'll have two options in there. I didn't know how else to do it. Awesome. You guys all got that right. And once 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. The chart in the, in question number three. And then just use that to see the, 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 can look at it and you think you can figure it out pretty quickly about drawing it. No words, but. So if you actually kind of drew out these lines, it 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 formed is. The image formed 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 virgins says. And I kept winding up with a negative. A negative value on that. On max or 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. But. Have to. Just look at the equation sheet again. So I'm flying 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. The equation that I don't really 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, at my notes. No, no worries. So I will put a star next to that. Okay. But yeah, based on. Like I said, for me, like if you've drawn the image already from the question before. You can just draw it, 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. If there's 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 paying attention to what they're saying in relation to. Because sometimes they might say something, I just remember like, you know, they might say something like, you know, if from the perspective of the object. What, what does this image, you know, look like? And they might just say like it's smaller and upright, but it's still be virtual. So I think they, I think you can pretty much keep that. In your mind that it would be upright and virtual. From a minus lens. I think so. 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 role when it holds. So I think that's important to keep in mind. Gosh, I think that. That does make sense. Thank you, Jennifer. Yeah. Once again, we're drawing it out. Might be helpful, but. Yeah. If it's a single lens 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 just have these, 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 objects when 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 try to do some today. Hope you already have some scratch paper and pencil. Ready. So this is the next question. Let me put it on the food. So you guys can submit your answer. JCC. Prisms on a roll here. This is good stuff. Okay. I'll. So it looks like a couple of you got fooled by the. Working distance. So. Let me see if I can tell. 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. Sure. 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 0.5 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 Cougoud or are you seeing the presentation now? We're seeing the presentation. Okay. So this is exactly what Jennifer is saying. So. When you are 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. That's one over 0.5 is. A minus two. And so if you put your. Your retinoscopy powers on a power cross. Take that versions into account. You will end up with a. Plus 20 and plus 25. Very good job. But if you forget this virgins, it can definitely throw you off. You forget that working distance. But I am very, I am very impressed. 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. Twice. Okay. Great job on that one. Let's go back to the boot. Everybody rocking it on that one. Spherical equivalent is something that. Thankfully is pretty simple. I don't know how high yield it 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. It's a sphere plus half of the silk. And I guess the only thing you need to do is. If it's plus silk or minus silk. Is that would determine if you're either adding or subtracting from the sphere of power. You guys all did really well on that. 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? 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. Alley. Is that right? Yeah. Yeah. Okay. Okay. Okay. Okay. Any questions on spherical equivalent? That's that's pretty simple stuff. Any questions there? Next question. Hey, some Jordy. Talking about that one. So optically empty. Just, just made a push there. Yeah. Awesome. So Alley, can you explain how you got that answer? And how you got it so quickly? Um, With prisms. They work off. Like a hundred 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 a hundred centimeters, it'd be six. So then. At 30, it'd be about two. Yeah. Go back to the presentation here. Exactly. So we're at 33 centimeters, which is about a third of a meter, right? So if it's a six prism after you take a third of that, which would be about two centimeters. Great job. So you can draw this one out as well. Um, with optics drawings are usually helpful. They just take a minute to draw, but. Yeah, if you have that. A hundred centimeter number in your head. Then 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. Um, 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. With prisons. 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. Um, So 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 meter. Love it when everybody gets it right here. So this basically explains. Our point in your point. So for an uncorrected my oh. Our point of 40 centimeters. Um, the power is just one over. Negative. And that's going to be a point for me. 40 centimeters. Is 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, um, Any questions on how to determine. There are a factor there for a mile. Or a. Hyper. Okay. Let's do the next question. Like I said, it's related to question number three. Yes. Okay. So one for them or for answer D. And then five for minus one and a half. Four eyes is making to remove whose four eyes. I am. Yes. Got it. So hard. Can you, um, can you tell us how you got that answer? Yeah. The patients. Regular far point. Um, told us that they have. Normally minus two. Oh, minus 2.5. Power in their eyes. And then to get to this closer point, they would need. To get to this. Uh, minus four. And that's calculated from the 25 centimeters. So the difference in power. Uh, is what they have to accommodate. So how do you know if you're supposed to like add them together or take the difference there? Um, well. You know that he's a mile. And so he has this. And he's a mile closer and you're pointing to us to get to. 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. Yeah. And putting it into diopter form. It's much, much easier if you put it into the powers and diopters and put it into the powers and diopter form. And so that's kind of what they're doing. They're doing 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. Say that again. I feel like so as they had a two part question. Maybe like two questions are tied together. Yeah. 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. Okay. I don't know. I don't know. I don't think they're 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. O 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, five more seconds for this one. Okay, question number seven. Anybody need more time for question seven. Hey, here's a question eight. Is 1.5 the distance between the two lenses. Yes, it is. Great question. Yes, it is. Hey, anybody now that need more time on this one. Yeah, let's go question nine. The last one in the series. This is where it'd be great to be in person. So I can see if you guys are done. I'll be covered. Right. Five more seconds. So. Are you on here? Yeah. I'm going to talk through. Okay. Perfect. You tell me what you got for question number five. Okay, so I got B. So you is minus one. Plus D is plus five. So then the equals plus four. 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. Don't show them that in the recording. Okay. So you ask question number five, the 25 semesters to the right of the lens. Awesome. And you just use that B equals you plus D. Equation for that. 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. That's going to be minus one. Add it 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. That's not the numbers or anybody. How many questions on this. If you have any questions, feel free to ask. I think that's cool. He's got this one down. All right. Tyler, are you available? Yeah. Okay, so for question six. You tell us what you, what your answer was and how you got that answer. So I think the answer was. Yeah. 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 draw a process of elimination, I got reduced, but 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 then you can do the magnification equation. By 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 to mind. Okay. It's a. And then you can do the magnification equation. Jennifer, do you usually draw the rays or do you just. I'm 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 the 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. I asked Jennifer because she's, you know, she's a little bit of a rush 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. All right. Can you send a full question number seven here. 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 initially 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 fourth 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. Once again, like if you drive. 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. Yeah. 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. EG Y twos have been able to utilize that resource, but when I was browsing through it looked like it had some, some pretty good information on telescopes. And I think one of them was a tracing. But basically. You, if you have your, your to be 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 the converging lens you would. Yeah, I guess you do kind of need to know the distance away from the lens will be converging and just draw. 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. Ariana, could you walk us through question number eight? Yes. Okay. Question number eight. We were calculating the. The new you for the new image. So, for agents and for objects. Okay. First place 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 first of 1.25. I made that a fraction, which is like negative four fifths. 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 same direction, 25 centimeters to the right of the first lens. Take that difference. And then it's just a matter of keeping your sign straight in the B equals you plus the equation. And exactly, and sometimes you can get caught up on exact numbers. So things like fractions or rounding can be very helpful. 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 centimeters 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. Yeah, let's do question nine and then 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. If you did. Yeah, I mean, and if you don't know that's totally. No, I think. I'm 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. In for now. I said D. I don't remember how I got to be, but I said to you. Hey, well, thankfully. You were right. 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, sketch it in your own way. 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. So sometimes 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 virtual. Yeah, I find for me personally drawing it out. It's very helpful. 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. I had a quick question. I think I remember. At somewhere along the lines, somebody saying that a hyperropic wall. 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. That 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. Those of you who have. Those of you who have taken. Those of you who have taken. Those of you who have taken. Those of you who have taken. Those of you who have taken. 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 point. So 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. 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 the 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 would be a real image. And if it's to the left, it would be a 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. Then you should be able to determine if that image is to the left or to the left of the lens. Does that make sense? Hopefully that does. Like that V equals you plus D 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, can you guys see the commute screen now. Yes. Okay. All right, so here's the next one. Double points. Okay. All right, 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. Does anybody know what equation we would use here? Oh, 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 diopter. And so I think the object distance was 33 centimeters. So minus three plus 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. The power of the mirror is two over R two over the radius. But you're right, we do have to calculate. So we have to calculate that power first. Two over two is one. Our object. This place 33 centimeters to the left. 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. Straight. 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. So that would be a plus one plus D equation. V equals, so you. Our 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. B. We take minus three plus the one from the lens. We get minus two. So both our B and our you are negative. So the magnification equation is transverse mag equals you over B. And so you would plug the minus three over the minus two to get a plus 1.5 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 if it's a plus or a minus, but since you is a negative. And B is a negative, a negative two from our equation. The negatives cancels each other out. Yeah, but plus 1.5 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? Is she still on? Yes. So the plus signifies that there's an upright. I believe so. Yes. And magnified by 1.5 times. And then the negative would mean it was inverted. Okay, cool. Thank you. Yeah. No problem. 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 good. 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. Did anybody actually try to do the math on that one? I tried, but full disclosure didn't get done. I just guess what I thought might be about, right? 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 Arianna, 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. 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 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 would be crazy to try to do that when you only have a minute and seven seconds per question on average. But 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 or a fraction of air equals one. And then 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. It could give you the idea 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. 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, all the times that the average cornea is 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 of interaction 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, but those rules make sense to people. Any questions or concerns of that? Okay. All right, let's keep going to the next good news. So kind of a split on that one. All right, so let's see. Thank you. Thank you. Thank you. Thank you. Thank you. 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 you add the minus still to the sphere. And then you add the minus still to the sphere. And then you add the minus still to the sphere. And then you add the minus 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. And then you just flip the sign of the sill and then change the axis by 90 degrees. We come up with the. 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 still. And vice versa. I had something that could easily show up and also very helpful in real life. So I'm going to go for switching back and forth between contacts and glasses. Nice job on that one. Here's the next question. Okay. So a bit of a split again on that one. So let's see how we came up. Remind me who prism 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. I think it's 20 right? 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. And 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. In 45. Did anybody. Accidentally get that wrong or. Just not remember that. Look like maybe there are a couple of. Just keep that in your head. Degrees. Half the prism. So that's a pretty good rule for a most prism. Next question. Starting right now. This is a telescope question. Let's rock that one. Lydia, can you tell us how you came up with that 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. It's like the way the. How should I explain it? The way the. Lenses are aligned. It's the Galean. 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. Okay. 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. Galean. Lens. And it's like, yeah, you're subtracting 10. Like minus 10 plus four. And 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 that 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. Like if we just look at here, we had 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 in that type of 15 centimeters. So, thank you. 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. Okay, so you've been kind of those just general rules. Any other questions on this. Okay, it is the next one. Okay. We've got 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 5x. This gets back into the question of the minus versus plus on magnification. What does that mean. Okay, 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 is that 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. Is this 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. Let me scroll the way back up. I think I may have confused. Lock your objective or I piece or something. You may have just seen the answers. Yeah, so if you, 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 looked like. 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 a ship or something like that, just like a straight telescope. An astronomical telescope usually has a bend 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 abandoned it. So 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. These are just kind of some rules with them for an astronomical telescope will be greater than four X. 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 to you guys. And it doesn't have that minus there for some reason. But I know when I was looking at this, this problem online, it definitely had that line. 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. I thought, but for some reason, the one that is on the paper, which I just, I got from the. Okay. It's probably 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. Or is 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 does seem to be wrong. Yeah. Yeah, because it should definitely have that minus in there. For the telescope mag equation. Because that's how you know that the astronomical has an inverted magnification. So I will fix that on your equation sheet. Okay. Yeah, we only have two minutes. Let's do a few more questions. If we can. I thought I'd fixed all the issues with last year's. Okay. Here's number 14. You guys are nailing. Sure. Okay. I think you guys all got that one right. I'm just going to show you this page really quick. Okay. Okay. Okay. Okay. Okay. One centimeter at one meter distance. We had half a meter. So we need to double that. Object size to get our prism. Yeah, you guys all got that one right. Any questions on that one? Hey, we'll keep moving through to get through as many more of these as we can. Can you guys see the good boot again? Yeah, we can. Okay. So this is that same equation that we were just looking at. That's the minus. That's the minus, the power of the object. So. How are you able to tell astronomical versus Gallo. Jennifer. We know it's astronomical because both powers are positive. Right. And then to Jordan. 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 eyepiece over the objective power. Perfect. Great. Yeah. So you guys are really well on that one. Tight race, two through four there, I like it. Okay, next question we 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 the, so we're interested in vertical. In the right eye, the sill is actually acting at 180. So we can ignore the sill there. And then we multiply one centimeter from the 10 millimeters times minus three and get minus three. And then in the left eye, the 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 and nothing in the left. Awesome. Yeah. Did anybody 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 pictures. And then the quick triangles. Yeah, I draw two. May I ask a question just to see if I understand this correctly? If the patient was not looking 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 opto base up. Correct. Yes, and that's where drawing the picture is helpful to see which direction the prism base is going. And so drawing these little triangles will show which direction the prism is. Now, I got to do a great job saying, okay, 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, which would end up being different. And in this case, would end up being a different answer. So 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 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 half a centimeter or something that would, you just have to plug that into your equation here. Yeah, great question on that though. If they were looking up, it would end up being base up prism. Any other questions on this? Okay, I think we have 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 prism diopters be? Good question. 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 prism 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 with my cursor there, but it would be base out in both cases. And then we just take the difference, a two prism diopter minus one prism diopter means that they would have one prism 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. It's a great double points. Really just want to hash these ones. Because apparently they're pretty heavy over. Oh, can you guys see it still? All right, can you guys still see the kahoot on there? Yes. Okay, perfect. All right. Looks like most of you got that. So C could be either four or I guess it was C or D. So which one did you guys come up with? C. C? 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 in 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 would be like the right eye had minus three at the 90 and minus four at 180. The left was minus seven at 90, minus six at 180. And you use minus three, minus seven to do the multiplication with the 10 millimeters below. I think, I can't see the problem. Just mess it up. Oh, you got it totally right. So, hey, you're exactly right. So you put it on a power cross. Now you had mentioned, you know, putting this in plus silk form first, you don't necessarily have to do that. I know that that's kind of how we think in ophthalmology. My brain, not function. Yeah. 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 at whatever axis this tells you. So minus three at 90. And then you just add these two together and put it at 90 degrees apart. So you could just go straight to the power cross. At the same time, you know, like, if you're more comfortable working in plus silk, just put that in plus or really quick and then put it on there. But you could technically save yourself a step there. But absolutely, you 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, at this together, basically, and you get minus four diopters were 10 millimeters below. So that would be one centimeter. So we have four base down in the left eye. Great job. Any questions on that one? Go back to our Kahoo here. Let's see who took the podium here. The Keplerian King. Who's the Keplerian King? I don't even know who that was. That was Cole. Cole, 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 out of time. Does anybody have any questions? Just in the last couple of minutes here before we end.