# Diopters

(Redirected from Diopter correction)

Diopter is a measure of the optical power P of a lens (or mirror) and is equal to the reciprocal of focal length in meters. The most common unit symbol for diopters is dpt, D, or m-1.

$P={\frac {1}{f}}=-{\frac {1}{d}}$ • In EM, we use the cm measurement to calculate diopters needed to correct refraction of the eye. If you can see 50 cm clearly, your diopters will be $-{\frac {1}{0.50}}=-2dpt$ .
• Lenses in series add their powers: if you're wearing -2 diopter contacts (adjusted for glasses strength) and put +1 diopter reading glasses over the contacts you're in effect wearing -1 diopters.
• There are some caveats such as vertex distance, since moving the lens away effectively gives you a weaker negative lens or stronger positive lens. There is also decentration, which induces prism when the lens is moved to the side. These effects become negligible for weaker lenses.
• According to the thin lens sign convention, negative focal power is diverging, and positive focal power is converging.
 0.00 to -0.75 dpt Probably don't need glasses -1.00 to -2.00 dpt Mild myopia, no differentials needed -2.00 to -5.00 dpt Moderate myopia, glasses always needed -5.00 to -10.00 dpt High myopia -10.00+ dpt Very high myopia. Field of view significantly reduced.

## Gap and ratio

Comparisons between two diopters is typically expressed using one of these terms:

• diopter gap (or diopter difference): absolute difference in diopters between the values of the two eyes
• diopter ratio: ratio of the diopters in one eye over the other one (right eye / left eye)

For example,the following correction:

OD: -1.5 SPH / -1.5 CYL
OS: -1.0 SPH / -2.0 CYL


can be expressed as a 0.5 dpt gap in both SPH and CYL, a 1.5 ratio in SPH and a 0.75 ratio in CYL:

|(-1.5 dpt) - (-1.0 dpt)| = 0.5 dpt
|(-1.5 dpt) - (-2.0 dpt)| = 0.5 dpt
(-1.5 dpt) / (-1.0 dpt) = 1.5
(-1.5 dpt) / (-2.0 dpt) = 0.75


Note that the term diopter ratio is often used interchangeably for diopter gap, for example when talking about reducing a correction while keeping the gap the same. This can also be expressed as a percentage difference between the two diopter values (e.g. the 0.5 dpt difference between the right and left eyes here is equivalent to 0.5 dpt / |-1.5 dpt| = 0.33 or 33%).

## Technical Details

This section is for the math-savvy people. It explains concepts in more detail, but knowledge of it is not strictly necessary to use the EM method.

### Thin Lens Equation

The focal length of a lens is given by the lensmaker's equation. By assuming the lens is much thinner than the radius of curvature, therefore assuming lens thickness is zero, we get a simplified version of the lensmaker's equation. We can do some further derivation, we arrive at the thin-lens equation:

${\frac {1}{d_{o}}}+{\frac {1}{d_{i}}}={\frac {1}{f}}$ According to the thin lens sign convention,

• di is positive if it is a real image on the opposite side of the lens as the object, and it is negative if it is a virtual image on the same side of the lens as the object.
• f is positive for converging lens and negative for diverging.

This is also sometimes presented in the Newtonian form:

$\left(d_{o}-f\right)\left(d_{i}-f\right)=f^{2}$ #### Examples

"Full correction" takes an object at infinity and produces a virtual image at your far point distance d:

${\frac {1}{f}}={\frac {1}{d_{o}}}+{\frac {1}{d_{i}}}={\frac {1}{\infty }}+{\frac {1}{-d}}=-{\frac {1}{d}}$ This is the resulting equation at the beginning of the article. It also explains why the focal power is increased for objects at closer distances: mainstream optometry calls this the "add" for presbyopia, although they typically use the minimum amount required for you to see at 40 cm with full distance correction using accommodation. For example, if you choose 80 cm as the working distance for your differentials (resulting in a +1.25 dpt "add"), and your blur horizon is 50 cm (resulting in -2 dpt), the formula is

${\frac {1}{f}}={\frac {1}{d_{o}}}+{\frac {1}{d_{i}}}={\frac {1}{80\ cm}}+{\frac {1}{-50\ cm}}=1.25\ dpt+\left(-2\ dpt\right)=-0.75\ dpt$ ### Cylinder

A cylindrical lens of power Fc has focal power F at angle θ from its axis:

$F=F_{cyl}(\sin \theta )^{2}$ #### Transposition

We can understand why there are two different ways to write a spherical and cylindrical lens combination, using the Pythagorean trigonometric identity and the complementary angle identity:

$(\sin \theta )^{2}+(\cos \theta )^{2}=1$ $\cos \theta =\sin {\left(90^{\circ }-\theta \right)}=-\sin {\left(\theta -90^{\circ }\right)}$ $F=F_{cyl}\left(1-(\cos \theta )^{2}\right)=F_{cyl}+\left(-F_{cyl}\right)(\sin {\left(\theta -90^{\circ }\right)})^{2}$ We can see that by adding the cylindrical power to the spherical power and inverting the cylinder and adding 90 degrees (same as subtracting 90 degrees, since the axis is in modulo 180 degrees) to the axis, we get an equivalent combination.

For example, -1 sph -1 cyl 1 axis is the same as -2 sph +1 cyl 91 axis.

In general, optometrists prefer to use a negative cylinder and ophthalmologists prefer to use a positive cylinder, but the two forms are equivalent.

#### Spherical Equivalent

By calculating the average value over all angles using an integral, the result is

$F_{avg}=\lim _{T\to \infty }{\frac {1}{2T}}\int _{-T}^{T}F_{cyl}(\sin {t})^{2}\,dt={\frac {1}{2}}F_{cyl}$ This is why the spherical equivalent has power equal to half of the cylinder's power.

Multiple lens, each with spherical and cylindrical components (not necessarily at the same axis) can be added to form one lens with a spherical and cylindrical component.

We can use the double angle formula to convert each cylindrical lens into a constant plus a cosine:

$\cos {2\theta }=1-2(\sin \theta )^{2}$ $F=F_{cyl}(\sin {\left(\theta +\phi \right)})^{2}=F_{cyl}\left({\frac {1-\cos {\left(2\theta +2\phi \right)}}{2}}\right)={\frac {1}{2}}F_{cyl}+{\frac {-F_{cyl}}{2}}\cos {\left(2\theta +2\phi \right)}$ The constant parts are added with the spherical components. The cosines can be added by converting them to phasors and adding the phasors together. The resulting phasor corresponds to one of two cylindrical lenses (see the section on Transposition), and its corresponding spherical component must be subtracted from the total spherical component.

### Decentration

Induced prism can be calculated using Prentice's rule. Similar to Vertex Distance, decentration is less of an issue for smaller power lenses.

The amount of prism P induced by decentration c of a lens of power f is

$P=cf$ 1 prism diopter displaces 1 cm for an object 1 m away. If c is in cm, and f is in diopters, then P is in prism diopters.

A prism with apex angle a and refractive index n results in angle of deviation of the light d, which is equal to P prism diopters:

$d=(n-1)a$ $P=100\tan {d}=100\tan((n-1)a)$ 