Visual acuity and retinal image size
What is the size of the retinal image of an object seen by the eye? This issue is important to understand the factors limiting Visual acuity, and certain consequences of refractive surgery, particularly for the strong ametropia. The caclul of the size of the retinal image is also important for the problems related to the difference in size of the retinal image (aneisoconie). An excessive difference in the size of the retinal images between the eyes right and left of the same subject can induce an excessive disparity and disorders of binocular vision, with reduction of the perception of the relief)stereoscopy).
If the density, the size of the photoreceptors of the fovea varies from one eye to the other, it is obvious that for a same density of photoreceptors (the ability to sample), retinal resolution will be all the more important that the projected image will be great.
Size of the retinal image of an object at infinity
This problem can be generalized to n optical system, which we know the overall power (called D, diopters). This system forms in conditions paraxiales the image of an object at infinity in its plan focal image. The height of this image is you. Here's how to calculate:
The figure above allows to establish (by expressing these distances in meters, and the angles in radians)
= YOU ϴ' P 'F'
Using the paraxial (for small angles) equation, we can also ask:
n = no ϴ ϴ' (n and not are the indices of refraction of both objects and image)
So (n/n) = you ϴ P 'F'
Moreover, D = no / P 'F', so 'F' P = no/d:
We get to the end: = You n ϴ / D
Case of the eye:
The usual middle of observation of the eye is air, so n = 1.
The result is a simple formula: = YOU ϴ / D who sets the size of the retinal image of an object which underlies a ϴ angle is equal to the ratio of the angle with the equivalent power of the eye. We usually choose a rounded value of 60 D for the equivalent refractive power of the eye (horny + lens).
For an angle of one minute of arc (expressed in radians), and an eye power of 60 D, you get the size you of the retinal image is equal to 0.004850 mm is about 5 microns. Separate two points that form an angle of one minute of arc with the eye, to 3 photoreceptors, either a space between the center of these photoreceptors of 2.5 microns (two photoreceptors are stimulated, and separated by a photoreceptor not stimulated).
Some studies show that the minimum size of the cones in the fovea may be half this value (1.7 micron about), and this explains that the Visual acuity of resolution (resolving power) can reach 30 arc seconds (half a minute of arc, or about 20/10).
Effect of means of correction of optical defects on the size of the retinal image
In the previous example, we have considered an Emmetropic eye.
For a simplified ametropic eye, retinal image sizes before and after correction of the defective vision, which can be accomplished by a glass of glasses, one lens or photoablative surgery.
For an object at infinity, we can establish that the magnification (magnification or "magnification") or the (reduction) shrink to the size of the image (we call this factor GV) induced by a glass of bezel is linked to the report:
GV = image size after correction (net) / size of the image before correction (fuzzy)
This report is also the angles that underlie the objects seen with and without correction to the pupillary Center. Expressed in angular size, this report becomes:
GV = ϴ'/ ϴ
For example, a hyperopic eye, for which the image is formed behind the retina.
For this ametropic eye not corrected, using the formula giving the size of the image depending on the angle that is underpinned by the object, you get:
= You ϴ/D, where ϴ is the angle formed with the Visual axis for a ray passing through the pupil of entry. This image is blurred because it ametropic eye and not corrected.
With a corrector bezel convex glass, the eye is Emmetropic because glass focus parallel rays picked up the object in terms of convex glass (which is the point of the hyperopic eye punctum). "" Indeed, the ' action ' a glass correction for hyperopia (nearsightedness) is to focus the rays emitted by distant objects in terms of the point of the ametropic eye punctum. They form a virtual image that the ametropic eye 'sees' then clear, effortless (punctum point corresponds to the distance to which to lie a point to be imaged on the retina of an ametropic eye).
In the case of the hyperopic eye, this point punctum is a located at the back of the eye. the angle of the emerging ray of this 'virtual' image through the center of the pupil of entry is noted ϴ'. This angle is the object seen by the eye (this object is the virtual image formed by the corrector glass).
According to the geometry of the image, this angle can be calculated as follows:
Θ ' = you / (f - v) where v is the distance from the glass in the center of the pupil.
So, we get by replacing ϴ and ϴ' by the calculated expressions:
GV = you / (f - v) you D, or, after simplifications:
GV = 1 / (1 - vD)
This expression involves the distance of the corrector glass (v), and (D) power
For a myopic eye: D < 0, so GV < 1: the size of the image is reduced.
For a long-sighted eye: D > 0, so GV > 1: the size of the retinal image is increased.
These effects will be all the more marked that v is large.
Examples of modification of the retinal image of the glasses and practical consequences
A myopic eye is corrected by a glass of 3 diopters placed 12 mm from the eye, what is the percentage of reduction of the size of the retinal image?
Calculate the distance from the glass in the center of the pupil of entry of the simplified eye: it is situated about 3 mm from the top of the cornea, so v = 15 mm (= 0.015 m). Calculated GV = 95%, or a reduction of 5% of the retinal image in size. Such a difference can induce a difficulty to merge the images if the other eye is Emmetropic (flat glass). Conversely, the realization of a refractive surgery, issued in the plan (corneal)LASIK or PKR) allows (as in contact lenses) to "grow" the image focused on the retina, and a gain to the less academic in Visual acuity.
This gain is particularly interesting for the correction of high myopia in LASIK, or cataract surgery. The following illustration shows the effect of a correcting high myopia (-29 diopters in glasses) glass. The axial length of the eye is 33 mm.
The correction of the glass is cut within a central portion circular in order to thin the edges of the glass and improve the aesthetics of the scope mount. The edges of the glass have a power void.
A photograph taken at short distance (2 cm) with a smartphone allows you to compare the size of the image after refraction by the corrector glass-29 D. The calculation of the theoretical magnification glass-29 D is equal to 63% (assuming that the screen is "at Infinity"). This reduction corresponds roughly to that which is observed in the shot (in the Center). It is estimated that a correction of this myopia by surgery of the eye (replacement of the lens by an implant intended to emmétropiser or leave a low myopia) would allow a proportional gain in size of the image projected onto the retina, and therefore a significant gain of the best corrected Visual acuity.
A hyperopic eye is corrected by a glass of 6 diopters, placed 12 mm from the eye.
v = 15 mm, and GV = 1.10 is a 10% increase in the retinal image in size.
These effects of changing size of images are easy to see through glasses for an Emmetropic view image is "blurred" by the correction which introduced a defocus, but it is smaller through a concave glass correction of myopia, and bigger through a convex glass correction of hyperopia.
These percentages may seem small, but they can reach higher values for the strong ametropia as high myopia. For example, for a 16 D myopia, the correction in glasses induced a reduction of 20% of the size of the retinal image (relative to the size of the blurred image of the same eye uncorrected). Can be symmetrically as a correction in a nearest plan of pupillary Center (lens or corneal refractive surgery) can induce a "magnification" of the retinal image towards the clear image in glasses (myopia correction). The correction in the pupil plan is accomplished in case of cataract surgery, since replacing the lens by an implant).
A magnification of the retinal image from 15% to 20% is beneficial because it allows the retina to better sample the image formed: the expected in terms of Visual acuity gain is significant because the reduction of the minimum angle of resolution (MAR) is of the same order. This is even more beneficial because some studies show that strong short-sighted have lower than the emmetropes retinal sampling capability (more low density of photoreceptors at the level of the fovea). Thus, a strong myopic of 16 diopters with the best visual acuity corrected in glasses of 8/10 (MAR = 1.25 minute of arc) has a Visual acuity in lenses of 10/10 (MAR = 1 minute of arc). This difference is of course increased for even stronger myopia.
The correction of high myopia allows a better visual acuity when it is performed by a system similar to the eye (lens, corneal refractive surgery, intra ocular implant).
These changes in the size of the image come with other phenomena such as changes in the field of view apparent, which is more narrow in lens and glasses.
In contrast, the strong farsighted has a slightly better in correction glasses eyesight, since the image is 'blown up' on the retina. This is accompanied by a reduction of the apparent field of view.
Finally, one can show that if it was a technical correction to move forward or backward the retina to correct the ametropia (towards the cornea for short-sighted, and back the cornea to the hypermetropia), then the size of the retinal image would be the same once the eye corrected than he had previously with a glass of glasses (located about 16 mm from the plan focus of the eye).
The correction of theastigmatism imply a change in the size of the image that vary with azimuth. This explains why the straights observed through glass-rings (cylindrical) scorers seem to undergo a slight 'spin', a tilt shift.