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Questions pertaining to the body of knowledge of optometry can be posted here in this one thread.
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Hi folks
I've started a number of threads since the summer about questions relating to optometric pot-pourri (e.g. optyl/zyl, the UV thread, RMS error). I thank those who have responded - and in effort to clean up the threads a bit, I'll post (and invite others to post) questions pertaining to the body of knowledge of optometry here in this one thread.
My question today...
With regards to astigmatic lenses and the issue of calculating power in an oblique meridian (i.e. what is the power at 180 for a -2.00 -1.00 x 030 lens?), the equation I learned for power at 180 is:
F(alpha) = F(sphere) + F(cylinder)*sin(alpha)^2
I hope you can make out what I just wrote. The alpha/sphere/cylinder associated with the "Fs" in the equation are subscripts, whereas the alpha in the "sin" is the alpha value that is part of the equation. My question is what is the definition of alpha? I have,
alpha = angle between axis meridian and desired meridian.
Is it the acute angle between the two? Or is it the axis value subtract the desired meridian value? Or the desired meridian value subtract the axis value? In the last 2 cases, the value could be greater than 90 degrees, and can be negative (but I guess negative doesn't matter since the sine will be squared. The >90 might not matter either since sine is sinusoidal). So my question is, for desired power at 180, and axis 30, is alpha:
the acute angle = 30 degrees? (desired meridian drawn towards axis)
the acute angle = -30 degrees? (axis drawn towards desired meridian)
30-180 = -150 degrees?
180-30 = 150 degrees?
As I said, due to the squarring of the sine, and the sinusoidal nature of the sine function, they may all give the same answer (I think they do), but what is the definition of alpha? If you had to tell someone to calculate this, would you tell them to find the acute angle, or to simply subtract the desired meridian from the axis (or vice versa).
Thanks.
I think that's right. My head hurts. Please don't post **** like that again!
Hi folks
I've started a number of threads since the summer about questions relating to optometric pot-pourri (e.g. optyl/zyl, the UV thread, RMS error). I thank those who have responded - and in effort to clean up the threads a bit, I'll post (and invite others to post) questions pertaining to the body of knowledge of optometry here in this one thread.
My question today...
With regards to astigmatic lenses and the issue of calculating power in an oblique meridian (i.e. what is the power at 180 for a -2.00 -1.00 x 030 lens?), the equation I learned for power at 180 is:
F(alpha) = F(sphere) + F(cylinder)*sin(alpha)^2
I hope you can make out what I just wrote. The alpha/sphere/cylinder associated with the "Fs" in the equation are subscripts, whereas the alpha in the "sin" is the alpha value that is part of the equation. My question is what is the definition of alpha? I have,
alpha = angle between axis meridian and desired meridian.
Is it the acute angle between the two? Or is it the axis value subtract the desired meridian value? Or the desired meridian value subtract the axis value? In the last 2 cases, the value could be greater than 90 degrees, and can be negative (but I guess negative doesn't matter since the sine will be squared. The >90 might not matter either since sine is sinusoidal). So my question is, for desired power at 180, and axis 30, is alpha:
the acute angle = 30 degrees? (desired meridian drawn towards axis)
the acute angle = -30 degrees? (axis drawn towards desired meridian)
30-180 = -150 degrees?
180-30 = 150 degrees?
As I said, due to the squarring of the sine, and the sinusoidal nature of the sine function, they may all give the same answer (I think they do), but what is the definition of alpha? If you had to tell someone to calculate this, would you tell them to find the acute angle, or to simply subtract the desired meridian from the axis (or vice versa).
Thanks.
You can get an approximation of the power in the resulting meridian using a simple pythagorean equation.
I wasn't aware of this (but I'm not surprised that you can). I'm assuming the estimated power is the hypotenuse, and the two cylinders each contribute a sine and cosine component?
Hi folks
Back in the day when going through optom school, I clearly remember one of the bottom lines in ophthalmic optics was that polycarbonate (PC) had the best UV protection of all lenses, exceeding that of UV-treated CR39.
I've come across literature (advertising material of a specific lens company) recently showing treated CR39 has better UV protection than PC. Is this true? Is this what some of you have learned?
Hi folks.
Please confirm.
The item in the image is called a lens former. Correct?
Thanks in advance.
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I was wondering if anyone can point out what these bifocal styles look like:
-sovereign
-panoptik
-kryptok
-ultex
-ribbon
-perfection bifocal
Thanks.