The equivalent of a lens +1.00 +2.50 X 90 is:
a. +1.50D
b. +2.25D
c. +2.50D
d. +3.50D
our answer would be +3.50D however the answer key is saying the answer should +2.25...
Please help me
HELP again another question
The equivalent of a lens +1.00 +2.50 X 90 is:
a. +1.50D
b. +2.25D
c. +2.50D
d. +3.50D
our answer would be +3.50D however the answer key is saying the answer should +2.25...
Please help me
Last edited by kristiekyle; 11-11-2014 at 01:51 PM.
Kristie J Meyer
The person who says it cannot be done should not interrupt the person who is doing it. ~~Chinese Proverb
The spherical equivalent would be +2.25D. +3.50 would be the total power in the horizontal meridian. You find the spherical equivalent by adding half of the cylinder power to the sphere power and ignore the axis.Originally Posted by kristiekyle
Kristie-
What's the equivalent if the power is +1.00 +1.00 x 45
a +2.00
b +1.00
c +1.50
d none of the above
+1.50 !!! Sorry Uncle Fester the anticipation was killing me
As above answer but background info on what I think your asking.
By the sound of it your wanting to locate to circle of least confusion in a toric prescription.
The location of the circle of least confusion is equal to the spherical equivalent of the prescription:
Spherical Equivalent = Sphere + Cylinder / 2
For example, given a prescription of +1.00 DS +2.50 DC 90, the spherical equivalent (and the location of the circle of least confusion) is given by +1.00 + (+2.50) / 2 = +2.25 D
Does it work very well on PTs with an oblique axis?
Hi Bok,
This notation might be confusing; some might read this as adding the sph to the cyl and dividing that value by 2.
Brackets might work... Spherical Equivalent = Sphere + [Cylinder / 2]
or probably just cozy up the forward slash and the divisor...
Spherical Equivalent = Sphere + Cylinder/2
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
Or for this simple mind...
Ignore the axis
Divide the cylinder by 2
Add that to the sphere power
We are ignoring the axis! See Sibirer's post above. And my post. And Bok's post. The axis is not relevant when determining the sphere equivalent.
Last edited by Robert Martellaro; 11-12-2014 at 04:27 PM. Reason: damn spell checker!
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
Yes I know! This is for math dummies like me who don't absorb formula's the way you do.
Cheers for the correction, quickly typed that one whilst at work, didn't double check. lesson for next time :)
It's a necessary evil. If you look closely, you'll see it's not my strong suit. Now Harry C., wow!Yes I know! This is for math dummies like me who don't absorb formula's the way you do.
When I type something from my smart phone, it almost always comes out looking like an alien language!
Science is a way of trying not to fool yourself. - Richard P. Feynman
Experience is the hardest teacher. She gives the test before the lesson.
Thank you Robert, I have always enjoyed your posts your command of optics is way above mastery of the subject. One of the few on the board that I consider vital to read. Even when I know the answer sometimes your explanation adds knowledge.
Einstein said if you can't explain it simply then you don't understand it as well as you thought (not verbatim).
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