Difference between revisions of Vertex distance

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The '''vertex distance''' is the distance between the surface of your eye and the center back of your lens. As the lens moves further from the eye, the perceived strength of your lenses is altered. This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes.

The '''vertex distance''' is the distance between the surface of your eye and the center back of your lens. As the lens moves further from the eye, the perceived strength of your lenses is altered.

* When a plus lens is moved away from the eye, it is ''perceived'' as being stronger

* When a minus lens is moved away from the eye, it is ''perceived'' as being weaker

This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes: to achieve a given correction for short-sightedness, the lenses of glasses needs to be stronger than the lenses of contact lenses.

==Practical Guidelines==

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:If you switch from glasses to contact lenses at high myopia or vice-versa you definitely will need to account for vertex distance before buying new corrections.

:If you calculate your own correction based on blur distance from your eye, but intend to buy glasses, you will need to adjust for the vertex distance.

* <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>

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<math>D_C=\frac{1}{\frac{1}{D}-x}</math>,

where <math>D_C</math> is the ~~corrected (~~perceived) diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.

where <math>D_C</math> is the perceived diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. (More generally, <math>x</math> is the distance the lens is being moved from its original position.) It is important to note here that this equation is sensitive to minus signs of your diopter strength and the direction of movement.

Example for a vertex distance of 15mm (=0.015m):

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In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses. It can be seen that vertex distance '''~~increases~~''' the strength of [[plus Lenses]] and '''~~decreases~~''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].

In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses, for distant objects. It can be seen that vertex distance appears to '''increase''' the strength of [[plus Lenses]] and '''decrease''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].

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where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>

Conceptually, the focus ~~length~~ is ~~reduced~~ (~~power~~ is ~~increased~~) ~~because it has moved closer~~ to the source of the ~~image~~.

Conceptually, the location of the [[Optics_related_math#The_thin_lens_equation|image]] is changed because the lens has moved relative to the eye. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.

* for a minus lens, moving the virtual image further from the eye is equivalent to using a weaker minus lens

* for a plus lens, moving the (real) image towards the eye (so that it is not as far beyond the retina) is equivalent to using a stronger plus lens.

Note that the equivalence relationship holds ''only'' for distant source objects: the behaviour of glasses and contact lenses will be different for all closer objects, and the near point in particular will be different.

==References==

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-The '''vertex distance''' is the distance between the surface of your eye and the center back of your lens.  As the lens moves further from the eye, the perceived strength of your lenses is altered.  This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes.
+The '''vertex distance''' is the distance between the surface of your eye and the center back of your lens.  As the lens moves further from the eye, the perceived strength of your lenses is altered.
+* When a plus lens is moved away from the eye, it is ''perceived'' as being stronger
+* When a minus lens is moved away from the eye, it is ''perceived'' as being weaker
+This is particularly important to know about when switching between contacts and eyeglasses, and for very high myopes: to achieve a given correction for short-sightedness, the lenses of glasses needs to be stronger than the lenses of contact lenses.
 ==Practical Guidelines==
@@ Line 9: / Line 13: @@
 :If you switch from glasses to contact lenses at high myopia or vice-versa you definitely will need to account for vertex distance before buying new corrections.
+:If you calculate your own correction based on blur distance from your eye, but intend to buy glasses, you will need to adjust for the vertex distance.
 * <div> '''I wear contact lenses as [[normalized]] and put [[plus lenses]] over them as my [[differentials]]. Do I need to worry about vertex distance?'''</div>
@@ Line 26: / Line 32: @@
 <math>D_C=\frac{1}{\frac{1}{D}-x}</math>,
-where <math>D_C</math> is the corrected (perceived) diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. It is important to note here that this equation is sensitive to minus signs of your diopter strength.
+where <math>D_C</math> is the perceived diopter number, <math>D</math> is the diopter strength of your lenses and <math>x</math> is the vertex distance in meters. (More generally, <math>x</math> is the distance the lens is being moved from its original position.) It is important to note here that this equation is sensitive to minus signs of your diopter strength and the direction of movement.
 Example for a vertex distance of 15mm (=0.015m):
@@ Line 34: / Line 40: @@
 <math>-4.0 dpt: D_C=\frac{1}{\frac{1}{-4.0}-0.015}=-3.774 dpt</math>
-In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses. It can be seen that vertex distance '''increases''' the strength of [[plus Lenses]] and '''decreases''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].
+In the above example the -4.0 dpt glasses yield the same level of correction as -3.75 dpt contact lenses, for distant objects. It can be seen that vertex distance appears to '''increase''' the strength of [[plus Lenses]] and '''decrease''' the strength of [[minus lenses]]. The effect is noticeable above 4.0 dpt and is mostly negligible for [[low myopia]].
@@ Line 42: / Line 48: @@
 where <math>f_C = \frac{1}{D_C}</math> and <math>f = \frac{1}{D}</math>
-Conceptually, the focus length is reduced (power is increased) because it has moved closer to the source of the image.
+Conceptually, the location of the [[Optics_related_math#The_thin_lens_equation|image]] is changed because the lens has moved relative to the eye. For a distant source object, the image is formed at the focus point of the lens, and so simply moves with the lens.
+* for a minus lens, moving the virtual image further from the eye is equivalent to using a weaker minus lens
+* for a plus lens, moving the (real) image towards the eye (so that it is not as far beyond the retina) is equivalent to using a stronger plus lens.
+Note that the equivalence relationship holds ''only'' for distant source objects: the behaviour of glasses and contact lenses will be different for all closer objects, and the near point in particular will be different.
 ==References==