Ophthalmic Lenses questions

[Ukht Zahra]

Hi everyone. This is my first post on this forum. I am due to start a course in Ophthalmic Dispensing in September. I will be using this thread to post questions. I am working through the book: The Principles of Ophthalmic Lenses by Mo Jalie. I have tackled quite a lot of questions, but some I need help with.

I will need time to re-read the chapter, but so far haven't found how to tackle this question.

The following thin lenses are placed together in close contact. Find the power of the thin lens which must be added to the combination to neutralise it.

+1.50 / +0.50 X 90
-1.75 / -0.75 X 135
+0.50/+0.50 X 180
+0.50 / -0.75 x 45

I don't need to complete answer, perhaps hints to help answer it. The answer at the back is -0.50DS.

[BigGuy]

It's been a long time since I was in school and worked with these theoretical problems but here goes:
Since the cylinder powers are the same but placed 90 degrees apart (90 and 180, 135 and 45) they will cancel each other out, so that leaves the sphere powers to add up. +1.50 -1.75 +0.50 +0.50 = +0.75 so I believe the resulting lens should be a +0.75 sphere
To neutralize it you would use a -0.75 sphere

[Uncle Fester]

I think you should draw optical crosses.

Transpose the + cyl to minus and adjust the 135 and 45 axis powers to 180 to see where the -.50 comes from.

[rbanting69]

I get the same thing -0.75 counting 1/8ths or 1/12ths

[Uncle Fester]

I must be misreading the problem. Not the first time!!!

[optical24/7]

Hi everyone. This is my first post on this forum. I am due to start a course in Ophthalmic Dispensing in September. I will be using this thread to post questions. I am working through the book: The Principles of Ophthalmic Lenses by Mo Jalie. I have tackled quite a lot of questions, but some I need help with.

I will need time to re-read the chapter, but so far haven't found how to tackle this question.

The following thin lenses are placed together in close contact. Find the power of the thin lens which must be added to the combination to neutralise it.

+1.50 / +0.50 X 90
-1.75 / -0.75 X 135
+0.50/+0.50 X 180
+0.50 / -0.75 x 45

I don't need to complete answer, perhaps hints to help answer it. The answer at the back is -0.50DS.

1st you need to find the spherical equivalent of each power which will be;

+1.75, -2.12, +.75, and +.12. Then add these together = +.50 To neutralize a +.50 takes a -.50. There's the answer.

[Uncle Fester]

1st you need to find the spherical equivalent of each power which will be;

+1.75, -2.12, +.75, and +.12. Then add these together = +.50 To neutralize a +.50 takes a -.50. There's the answer.

There ya go!

Now to clear the smoke from the room!!!

[Robert Martellaro]

I took the horizontal and vertical axes and transposed one of them to plus cylinder. It's easy to see that the cylinders cancel.

+1.50 +.50 x 90
+0.50 +.50 x 180

becomes

+2.00 -.50 x 180
+0.50 +.50 x 180

when combined, equals +2.50 DS

****************************

Same for the oblique axes.

+.50 -75 x 45
-1.75 -.75 x 135

becomes

+.50 -.75 x 45
-2.50 + .75 x 45

equals -2.00DS

+2.50 DS (-) -2.00 DS = +.50 DS, and is neutralized by -.50 DS.

For cylinders that don't cancel, which is typical, use a crossed cylinder formula.

http://www.opt.uab.edu/NBEOOptics/F%...s%20THEORY.htm

Or, D. Meister's crossed cylinder tool.

http://www.opticampus.com/tools/cylinders.php

[BigGuy]

I see the error of my ways and stand corrected.

[Uncle Fester]

I took the horizontal and vertical axes and transposed one of them to plus cylinder. It's easy to see that the cylinders cancel.

+1.50 +.50 x 90
+0.50 +.50 x 180

becomes

+2.00 -.50 x 180
+0.50 +.50 x 180

when combined, equals +2.50 DS

****************************

Same for the oblique axes.

+.50 -75 x 45
-1.75 -.75 x 135

becomes

+.50 -.75 x 45
-2.50 + .75 x 45

equals -2.00DS

+2.50 DS (-) -2.00 DS = +.50 DS, and is neutralized by -.50 DS.

For cylinders that don't cancel, which is typical, use a crossed cylinder formula.

http://www.opt.uab.edu/NBEOOptics/F%...s%20THEORY.htm

Or, D. Meister's crossed cylinder tool.

http://www.opticampus.com/tools/cylinders.php

This is what I kinda did in my head... ergo the smoke.

[MakeOptics]

Expect that in the course you won't be given a clean problem. Mo Jalies' book is going to use a method called astigmatic decomposition. Astigmatic decomposition is a method where the power is converted into 3 components which can be easily added together and then reconverted back into spero-cylindrical form. In the US the same method is used but steps are combined and the resulting 3 to 4 formulas can be used to directly add 2 lenses together. Astigmatic decomposition allows for a greater number of lenses to easily be added together simply.