(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.