Difference between revisions of Vertex distance
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Formatted the fractions using LaTeX
(Formatted the fractions using LaTeX) |
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The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by: | The effect of vertex distance on the perceived diopter strength of your glasses can be expressed by: | ||
<math>D_C=1 | <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 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. | ||
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Example for a vertex distance of 15mm (=0.015m): | Example for a vertex distance of 15mm (=0.015m): | ||
<math>+4.0 dpt: D_C=1 | <math>+4.0 dpt: D_C=\frac{1}{\frac{1}{+4.0}-0.015}=+4.255 dpt</math> | ||
<math>-4.0 dpt: D_C=1 | <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. 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]]. | ||
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<math>f_C = f - x</math> | <math>f_C = f - x</math> | ||
where <math>f_C = 1 | 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 focus length is reduced (power is increased) because it has moved closer to the source of the image. | ||