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  • (boing-boing sound)

  • (upbeat music)

  • - Hello, and I would like to welcome you

  • to the Laramy-K OpticianWorks Training Center,

  • where, in case you have not figured it out,

  • today we are going to talk about image jump.

  • Image jump is one of those topics

  • that you may very well find on an exam,

  • not something you're really gonna use

  • a whole lot in your day-to-day working life

  • as an optician.

  • But hey, it never hurts to work

  • another Prentice's formula problem.

  • What is image jump?

  • Well, this is image jump.

  • Super important, I'm gonna end up repeating myself,

  • this is a common, ordinary prescription.

  • We don't have anisometropia.

  • We don't have antimetropia.

  • We don't have super high powers.

  • We don't have anything special going on.

  • This is just a routine situation.

  • And I'm gonna talk about that a little bit more

  • in just a minute.

  • We have our human, we have our distance prescription lens,

  • and we have a great big bifocal lined segment.

  • Let's say it's a two, 2.50, quite strong.

  • Person is looking out into the world,

  • and they're looking out in the distance,

  • and the object appears just where it's supposed to be.

  • Everything's great.

  • They glance down, they converge, they look through

  • the segment, the little top of the segment there,

  • and suddenly that E appears to shift position.

  • It jumps.

  • That's image jump.

  • That's something people just get used to.

  • It's just a natural occurrence,

  • and I'll go into a little bit more detail

  • about that in just a second.

  • But that's the basic concept of image jump,

  • and it certainly is something

  • that you'd want to know happens,

  • but as to the rest of it ...

  • I think this is the 50th or roughly the 50th video

  • that we have produced, and there is a reason why

  • I have not addressed image jump before.

  • I think it's a lousy concept.

  • The videos are here to give you stuff

  • that you can really use in your day-to-day

  • life as a working optician.

  • Image jump is not really one of them.

  • Let me run through this and then I'll work

  • through a complete, full, straight top 28 example.

  • Especially for those of you that are working

  • through the complete program online,

  • the OpticianWorks program, you know

  • that I very rarely say easy or simple,

  • but image jump really is just

  • a very, very basic Prentice's formula problem.

  • P is equal to hcm times D,

  • which you should be familiar with.

  • Hcm in this case is relatively easy to determine.

  • D is simply your add power.

  • Really not a whole lot to this, all right?

  • We're not talking about any special case scenarios.

  • We're not talking about an application for slab-off

  • or something, we're talking about routine prescriptions.

  • Let's say Mr. Smith has been wearing a minus one

  • with a plus 1.50 add for the last three years.

  • Doc bumps him up to a two.

  • Comes in, puts on his new glasses,

  • "Oh wow, the print just jumped way up

  • "when I put my glasses on."

  • Well there's nothing you can do about that.

  • He's gonna wear those glass for a couple of days,

  • he's just gonna...

  • His brain's gonna overcome it.

  • He's just gonna be wearing his glasses.

  • So why are we making a big deal about image jump?

  • We're gonna go through all this stuff in a second

  • to figure out the amount that is created

  • as an error in the image jump in a straight top 28.

  • And you know what?

  • Knowing the amount doesn't help us at all.

  • Let's say we finally determine,

  • and I don't think that's the amount,

  • 2.3 base diopters total.

  • So what?

  • What I'm suggesting is perhaps it's time

  • to change the conversation.

  • If your state exam, practical exam, is forcing you

  • to do stuff with image jump, maybe after you pass

  • and get your wings, maybe it's time to say,

  • "Hey, maybe we should consider not doing it anymore."

  • If you look on books and online your gonna find all kinds

  • of stuff with old text, and ribbon segments,

  • and executives, and glass.

  • Glass doesn't even exist anymore in this kind of stuff.

  • Let's change the conversation, all right?

  • We've gotta start looking forward

  • instead of lookin' backwards.

  • Enough of my little mini rant for today on image jump.

  • I wanna wipe the board clean.

  • I will see you in a second

  • and we'll work through a complete problem.

  • All right, be careful I'm armed here today.

  • Let's work through a very basic image jump problem

  • with a straight top 28.

  • As is so often the case here, we're going to have to work

  • the white board two, maybe even three times today.

  • So, just because we worked through all of this

  • to figure this out, doesn't mean we're done.

  • What's kinda nice about this.

  • Let's just say you had a customer come in,

  • they had this new lens order, a -1.50 -50 x 135

  • with an add of 2.25, we know they're gonna be

  • in a straight top 28.

  • Their left is a -1.75 -0.75 x 140

  • add of 2.25, straight top 28.

  • Which is kinda nice is we don't actually need any of that

  • and I'll show you why that is in the next section.

  • We do need that and we know we're working

  • with a straight top 28.

  • We are working a Prentice's formula problem,

  • P is equal to hcm times D.

  • All we're gonna do right now is figure out hcm,

  • the distance moved.

  • Hcm is the distance from the segment edge

  • to the segment O.C.

  • And I use the word edge

  • because there are still round segments out there and stuff

  • but we're just gonna be concentrating

  • on that straight top 28.

  • From the edge to the O.C., to determine that

  • we can use a formula, the answer is 3.5

  • and it will always be 3.5,

  • hcm is the height of the segment

  • minus half the width of the segment.

  • So if you take a straight top 28

  • and you take your millimeter ruler

  • and you measure from the base

  • to the top of the segment, you get 17.5 millimeters.

  • I suggest you try it, go find a bifocal

  • and a millimeter ruler and do it.

  • Obviously a straight top 28 is 28 millimeters wide.

  • So, if I have my height of 17 1/2 and half of my 28 is 14,

  • if I subtract this to it I end up

  • with an hcm of 3.5 millimeters.

  • So I'm halfway there.

  • I just need to figure out D which is nothing more

  • than my add power, and I can solve this problem.

  • Gonna talk a little bit about why we didn't need this.

  • What direction image jump always occurs

  • and then we'll actually solve to determine

  • how much this would be creating.

  • We have solved for hcm which is 3.5 millimeters.

  • We have D which is 2.25, which is our add power.

  • The add portion of this, remember, is just

  • a super simple spherical

  • plus lens quite literally slapped

  • onto the front of that distance prescription,

  • whatever it might be.

  • That's why we can think about this as being a separate piece

  • that doesn't have to take into account

  • what that distance prescription is behind it.

  • Because it's a simple plus lens,

  • if we took it and we brought it out into here

  • and we took off this top,

  • we have our straight top bifocal segment here.

  • Remember when we looked at the height

  • of that straight top 28, we said it was 17.5,

  • half of 28 in a circle would be 14,

  • that extra 3 1/2 millimeters up

  • is really important in this...

  • working this through.

  • Because if it was at 14 you'd be at the O.C.,

  • but because we are above that,

  • we have that little extra piece,

  • the formula, or the concept, whatever you wanna call it,

  • for an image jump, assumes that we're actually looking

  • through this little portion, this part that's above

  • the line, the O.C. line which separates

  • the two prism shapes, so that we're actually looking

  • through this portion, so we are receiving base down prism.

  • That's where that comes from.

  • So your answers to your image jump problems

  • are always gonna be a base down answer, that's why.

  • So we took our hcm 3.5,

  • we multiplied it times our D

  • of 2.25, we end up with 7.87, we divided by 10

  • to convert our cm to millimeters,

  • we end up with 0.787 rounded

  • to 0.8 diopters base down

  • for our image jump.

  • That is how it would work through

  • a complete image jump formula.

  • (exhales) Sort of.

  • You can actually use five for almost all of this.

  • Slab-offs are actually calculated on five.

  • A lot of test questions and stuff

  • will just use five instead of 3.5

  • instead of working through that whole thing

  • that we just did.

  • Hey, what could I say.

  • I'm looking in the optical tutorial books,

  • they show it both ways, 3.5 and five

  • for that same segment.

  • Is it really gonna make any difference

  • on the long end here?

  • No, none whatsoever.

  • Another reason why I'm not that big a fan

  • of image jump, but...

  • I hope that made things a little bit clearer

  • for some of you that do struggle with it

  • or are facing an image jump problem.

  • For any of you that are forced to work some

  • of these questions, there are some default O.C. placements

  • or if your hcm in the formula.

  • For a round 22, you can use 11,

  • for a straight top 25, 28, or 35,

  • you can use the default of five.

  • For an executive, zero, a flat top 40 or 45, zero.

  • Or if anyone should ever give you

  • a progressive lens question like this,

  • why they would do that I don't know,

  • that too would be zero.

  • Thank you so much for watching.

  • If you enjoyed this and then found it at all useful,

  • hit the like button, leave me a comment,

  • always happy to see those.

  • And I will see you again next week.

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A2 初級 美國腔

如何計算影像跳轉 (How To Calculate Image Jump)

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    wei 發佈於 2021 年 01 月 14 日
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