1. ## accommodation through spectacles

Hi everyone!

A -3.00 myope will converge around 1PrismD less when looking through his spectacles , right?
(if PD for near is 0.3 cm less distance)

but how do we calculate his accommodation demand for 30 cm with the rx ?
doesn't it depend on his ac/c ratio?

thankssss

2. It is a bit complicated to determine the exact accommodation required for a given working distance at the plane of the eye, though I'm sure the formulas could be simplified (using a binomial expansion, for instance). You would first calculate the vergence of the spectacle lens at the plane of the eye, which is located at a distance from the back of the lens equal to the vertex distance.

$Ld = \frac{Fv}{1-dFv}$

where Fv is the back vertex power (in diopters), Ld is the image vergence (in diopters) for a distant object and d is the vertex distance, in meters (mm / 1000). Then you would calculate the vergence of the spectacle lens at the plane of the eye for a near object (located at a distance w):

$Ln = \frac{Fv-\frac{1}{w}}{1-d(Fv-\frac{1}{w})}$

where Ln is the image vergence (in diopters) for a near object at a working distance w from the lens, in meters (cm / 100).

The total accommodative demand at the plane of the eye is then given by the difference in vergence between the distance (Ld) and near (Ln) image vergences:

$Accommodation=Ld-Ln$

Note that hyperopes (with plus powers) must accommodate more than myopes (with minus powes) in a spectacle lens. Also, when the form of the lens is substantial, such as for high-plus lenses, additional calculations are needed because of the thickness of the lens.

Thanks D :)

4. ## Accommodative Demand

You can also simplify the entire issue. Accommodative Convergence/Accommodation ratios are not really the issue, despite being interrelated. There is a normal 6:1 ratio, which means that in a "normal" situation, a patient will accommodate 1D and converge 6 prism diopters. The pupil will also constrict (referred to as the synkinetic triad). What you are seeking is the accommodative demand at 30 cm, and can easily be measured theoretically by simply using the focal length formula, (D= 1/f or f= 1/D). The far point for a -3.00 myope is 33cm (-3.00= 1/f), so there will be a negative accommodative demand at 30 cm. With the Rx, then the demand at 30cm will be 3.33D (100/30).

I hope this is helpful, and is the answer you are seeking.

5. You can also simplify the entire issue.
I think he is asking about the change in accommodative demand produced by spectacle lenses. For Berno's problem, the prescription is -3.00 D, the working distance is 30 cm (0.3 m), and we'll assume a vertex distance of 13.5 mm (0.0135 m). Solving for the image vergences at both distance (Ld) and near (Ln) gives us:

$Ld=\frac{-3.00}{1-0.0135(-3.00)}$
$Ld=-2.88$

$Ln=\frac{-3.00-\frac{1}{0.3}}{1-0.0135(-3.00-\frac{1}{0.3})}$
$Ln=-5.83$

So, the demand on ocular accommodation (A), at the plane of the cornea, is:

$A=-2.88-(-5.83)$
$A=+2.95$

For comparison, note that the accommodative demand at the plane of the cornea with no spectacle lens in place is equal to:

$A=\frac{1}{0.3+0.0135}$
$A=+3.19$

Consequently, this myope has to accommodate almost a quarter-diopter less than an emmetrope with no eyeglasses. This is the main reason why myopes going from eyeglasses to contact lenses will need bifocals sooner.

Thanks D

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•