Neutralizing Prism

[emloray]

I tried doing a search but was unsuccessful in getting my question answered. Thirty years in the business and my mind's gone blank!:o

I'm trying to neutralize an 11 D prism in a lens.
now, using a B & L Vertometer which reads up to 5D if I add a prism ring of 6D where should the target be readable? At the center ? Or at what resultant prism after putting the 11D lens in.
Or should I add a higher prism ring? Or does that matter?
Hope this is clear...
I'm confused...:idea:

Appreciate your thoughts......Thanks !

[YrahG]

I tried doing a search but was unsuccessful in getting my question answered. Thirty years in the business and my mind's gone blank!:o

I'm trying to neutralize an 11 D prism in a lens.
now, using a B & L Vertometer which reads up to 5D if I add a prism ring of 6D where should the target be readable? At the center ? Or at what resultant prism after putting the 11D lens in.
Or should I add a higher prism ring? Or does that matter?
Hope this is clear...
I'm confused...:idea:

Appreciate your thoughts......Thanks !

Assuming that you are putting the prism in the vertometer to offset the 11D you'll find the prism at the target at the 5D ring. If you had a 10D ring that would put your target 1D away from the center which will give you a better look at the mires.

[Luis davila]

Yes, but it depens the orinetation of your prism, you have to set against the base and it has to be in the center

[William Walker]

11 diopters of prism - Wow. I know it's out there, but I haven't seen anything like that in a long time. I know it doesn't answer your question, but out of curiosity, what's the Rx? I'm curious as to whether the 11D is part of a split or not, and at what power the patient has.

What would happen if the power was low enough, that the OC would be placed outside the area of the eyewire? When the patient looks through the lens, the Rx could be correct, but would there be a way to neutralize, since the OC would be edged out, and not able to be shifted to the center of the lensometer? Might be a silly question, but it just popped in my head.

[YrahG]

11 diopters of prism - Wow. I know it's out there, but I haven't seen anything like that in a long time. I know it doesn't answer your question, but out of curiosity, what's the Rx? I'm curious as to whether the 11D is part of a split or not, and at what power the patient has.

What would happen if the power was low enough, that the OC would be placed outside the area of the eyewire? When the patient looks through the lens, the Rx could be correct, but would there be a way to neutralize, since the OC would be edged out, and not able to be shifted to the center of the lensometer? Might be a silly question, but it just popped in my head.

The prism causes deviation of the mires (rays). The mires are real and will be deviated by the lens even if the mechanical center is edged off the lens.

Yes, but it depens the orinetation of your prism, you have to set against the base and it has to be in the center

I was hoping that by stateing "offset" the orientation was implied, but yes you are correct and more clear in your explanation thank you.

[William Walker]

That makes perfect sense - thanks. That's what happens when I don't think about a question longer than 30 seconds before I ask it. :cheers:

[YrahG]

That makes perfect sense - thanks. That's what happens when I don't think about a question longer than 30 seconds before I ask it. :cheers:

With a impressive list of accomplishments like your's I'm just glad I have something to offer in comparison.

[William Walker]

Thanks. Coming here to help others out to learn and grow is a much bigger accomplishment than adding letters to the end of your name. Thanks again!

[gunner05]

11 diopters of prism - Wow.

Had a patient about a month before I left wallyworld with 26 horizontal and 2 verticle, split. Had to get fresnels in for him. *shudders* Not quite sure how that panned out after the pressons didnt work--except that he got free side shields included with his lens when we ground in as much as we could.

[Darryl Meister]

but would there be a way to neutralize, since the OC would be edged out, and not able to be shifted to the center of the lensometer?

All good responses.

Since we are routinely told to "spot the optical center" by centering the focimeter target, this is a reasonable enough question. But you are not necessarily locating the optical center by centering the focimeter target.

When the focimeter target is centered, this simply means that you have reached a condition of zero prism. You could achieve this condition a number of ways:

1. By not placing any lens in the focimeter, which should be an obvious case,

2. By placing a Plano* lens (with no Rx prism) in the focimeter at any reasonable position,

3. By placing a powered lens in the focimeter with the optical center properly centered, or

4. By placing a lens with prism (surfaced or decentered) in the focimeter in conjunction with a neutralizing prism of equal magnitude and opposite direction.

So, just remember that the focimeter target is measuring prism, if any, not necessarily the location of the optical center.

* Technically, most Plano lenses will produce a very small amount of prism by decentration.

[jpways]

Instead of trying to measure it with a vertometer and trying to get the prisms lined up, wouldn't be easier to try and measure the displacement we would expect from that much prism?

The easiest way I can think to do this is tape a piece of letter size paper to the wall, it doesn't matter which way, because we're only taking about 4 in. Stand back 1 m and shine a penlight onto the piece of paper, centering on a pre-marked center dot (this is to take into account the light dispersement of your penlight). Now without moving the penlight bring the lens in front of the beam of light and see if you get the expected displacement. Again I'd pre-mark the target dot, probably with a few extra dots on either side to make sure you're getting within ANSI.

I know it probably is easier to get the prism rings lined up right, but for how often the average optician has to do this, by the time you get the ring exactly right, how much time do you expect that to take?

[YrahG]

Instead of trying to measure it with a vertometer and trying to get the prisms lined up, wouldn't be easier to try and measure the displacement we would expect from that much prism?

The easiest way I can think to do this is tape a piece of letter size paper to the wall, it doesn't matter which way, because we're only taking about 4 in. Stand back 1 m and shine a penlight onto the piece of paper, centering on a pre-marked center dot (this is to take into account the light dispersement of your penlight). Now without moving the penlight bring the lens in front of the beam of light and see if you get the expected displacement. Again I'd pre-mark the target dot, probably with a few extra dots on either side to make sure you're getting within ANSI.

I know it probably is easier to get the prism rings lined up right, but for how often the average optician has to do this, by the time you get the ring exactly right, how much time do you expect that to take?

1cm / 1m = 10mm / 1000mm = 1mm / 100mm = 1mm / 10cm

so if you were to hold a lens 10cm away from a PD ruler and match up your reference point of the lens with the "0" on the PD ruler then look through th lens for every 1mm displacement you would have 1 diopter of prism.

[Darryl Meister]

1cm / 1m = 100mm / 1000mm = 10mm / 100mm = 10mm / 1cm... so if you were to hold a lens 1cm away from a PD ruler and match up your reference point of the lens with the "0" on the PD ruler then look through th lens for every 10mm displacement you would have 1 diopter of prism

At one centimeter, the displacement would only be 0.1 mm (1 cm = 10 mm) for one prism diopter, which would be difficult to measure.

It would be more practical to place the PD ruler at 1 meter, which would make the centimeter scale equivalent to prism diopters (and the millimeters equivalent to tenth-diopter prism values).

Aiming a small laser pointer through the prism reference point of the lens would probably make this experiment pretty easy to conduct, although you would need something like a tangent screen to measure prism in both the horizontal and vertical directions.

[YrahG]

At one centimeter, the displacement would only be 0.1 mm (1 cm = 10 mm) for one prism diopter, which would be difficult to measure.

It would be more practical to place the PD ruler at 1 meter, which would make the centimeter scale equivalent to prism diopters (and the millimeters equivalent to tenth-diopter prism values).

Aiming a small laser pointer through the prism reference point of the lens would probably make this experiment pretty easy to conduct, although you would need something like a tangent screen to measure prism in both the horizontal and vertical directions.

Thank you a conversion error. Exactly what I was trying to describe, I believe that the prism ruler is called a tangent scale if memory serves me correct and is what I based this example off of.