general help

[mimicry]

I'm just starting in this field and the info at work is nil or not enough. One ? is through what meridian will light pass undeviated with this rx t 2.50 -2.50 x150 is it 60? another concerns vertex. a -5.00 ou has a vertex at 10mm then the question is what's the power at the corneal plane at 0 Would it decrease in power? What is the corneal plane? I know, I know, well someone has to ask the dumb questions. ed

[Pete Hanlin]

<FONT COLOR=#FF0000>"Through what meridian will light pass undeviated with this rx: +2.50 -2.50 x150?</FONT>

As you suggested, the axis in which this lens has no power will be 060. Consider a spherical +2.50 lens. Now put a cylinder through it (think of a Coke can running lengthwise across the back surface). Along the axis of the Coke can (the flat side running up the side of the can), there will be no change in power- so the lens will read +2.50. Across the can, however, the curvature of the cylinder will affect the power of the lens... so at 090 degrees away from the axis, you have the full -2.50 of cylinder.

<FONT COLOR=#FF0000>"A -5.00 ou has a vertex at 10mm then the question is what's the power at the corneal plane? Would it decrease in power?"</FONT>

First, figure out the focal length of the lens. 1 meter/-5.00 diopters gives you a focal length of -.2 meters.

Second, consider how the lens is moving. In this case, it is being moved .01 meter closer to the eye. Combine the amount of movement (.01) and the focal length (-.2 +.01= .199).

Finally, divide one meter by the new focal length, and the new power as a result of the change in position is -5.26. Therefore, if the patient needed -5.00 at 10mm, they would need only -4.74 at the cornea.

Note that a plus lens will decrease in power as it is moved closer to the eye. For example, a +5.00 lens would have a focal length of +.2 meters. Add the .01 meter that the lens is being moved for a new focal length of +.21. At the cornea the lens would have a power of +4.76 (so a lens at the cornea would need to be +5.24 to have the same effect).

Also, if you are figuring the opposite way (from the cornea to a certain vertex distance) remember to subtract the amount of movement from the original focal length (instead of adding).

Hope that made some sense. It's easier to demonstrate in person and on paper... In general, a minus lens gets weaker as it moves away from the eye and a plus lens gets stronger as it moves away from the eye (and vice versa).

Pete

[mimicry]

thank you for the info Pete. I tried to comprehend the vertex stuff in the other post but...I'm not there yet. thanks again. ed