# Diopters

(Redirected from Diopters)

Diopter is a measure of the [ https://en.wikipedia.org/wiki/Optical_power optical power ] P of a lens (or mirror) and is equal to the reciprocal of the focal length in meters . The most common unit symbol for diopters is dpt, D, or m - 1 .

${\displaystyle P={\frac {1}{f}}=-{\frac {1}{d}}}$

• In EM , we use the cm measurement to calculate the diopters needed to correct refraction of the eye. If you can see clearly at 50cm, your diopters will be ${\displaystyle -{\frac {1}{0.50}}=-2dpt}$ .
• Serial lenses add their powers: if you wear - 2 diopter contact lenses ( adjusted for glasses strength ) and put on reading glasses +1 diopter on the lenses you actually wear - 1 diopter.
• There are a few caveats such as vertex distance because moving the lens further away effectively gives you a weaker negative lens or a stronger positive lens. There's also shift, which induces a prism when the lens is moved sideways. These effects become negligible for weaker lenses.
• According to the thin lens sign convention, the negative focal power is divergent and the positive focal power is convergent.
• A lens with a negative diopter sign compensates for myopia while a lens with a positive diopter sign compensates for hyperopia .
 - 0.00 to - 0.50 dpt Not really considered myopic, probably doesn't need glasses - - 0.50 to - 1.00 dpt Mild myopia, normalized sometimes unnecessary - - 1.00 to - 2.50 dpt Low myopia, possibly unnecessary differentials - -2.50 to - 3.00 dpt Low myopia, differentials probably needed - 3.00 to - 6.00 dpt Moderate myopia, glasses still needed - 6.00 to - 10.00 dpt High myopia - 10.00+ deposit Very high myopia. Significantly reduced field of vision.
 ==Gap and ratio==


Comparisons between two diopters are usually expressed using one of these terms:

*    diopter gap (or  diopter difference  ): absolute difference in diopters between two values
​​*    diopter ratio  : ratio of one diopter value to another (like right eye / left eye)

For example, consider the following correction:

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

It can be expressed as a difference of 0.5 dpt in SPH and CYL, a ratio of 1.5 in SPH and a ratio of 0.75 in CYL:

|( - 1.5 dpt) - ( - 1.0 dpt)| = 0,
- 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 [1]


    , for example when talking about reducing a correction while keeping the same  gap  . This can also be expressed as a  percentage difference   between the two diopter values ​​[2] (for example, the  0.5 dpt  difference between the right eye and the left eye is here equivalent to  0.5 dpt / | - 1.5 dpt | = 0.33  or 33%). The general recommendation is that the left -


diopter differenceshould be constant on all lenses used. However, some old EM papers show successful cases where the differentials are equalized but normalized with a deviation of 0.25 D.Cite error: Closing </ref> missing for <ref> tag

It is often useful to disambiguate what is being compared:
*    left - right deviation  : diopters left eye minus diopters right eye, without taking into account the axis

• In the example above , the left - right deviation is +0.5 SPH - 0.5 CYL.
• Axis is ignored and cylinder powers are subtracted without using goal combination calculations .
• diff - norm deviation  : differential diopter minus normalized diopter, without taking into account the axis
** For example, if the norm is - 2 SPH - 0.5 CYL and the differentials are - 0.75 SPH, the diff - standard deviation is 1.25 SPH 0.5 CYL or 1.5 SPH equivalent.

• The axis is ignored.
• This quantity is usually positive, because more positive sphere is needed for close-up than for distance vision .

## Technical Details

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


### Thin lens equation

The focal length of a lens is given by the lens manufacturer's equation. Assuming the lens is much thinner than the radius of curvature, so assuming the lens thickness is zero, we get a simplified version of the lens maker's equation. We can do a few more derivations, we arrive at the thin - lens equation: [3] $\displaystyle {1}\frac{1} d_o}+\frac{1}{d_i}=\frac{1}{f}$
According to the fineness lens sign convention,  * di is positive if it is a real image from the side opposite of the lens to 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 a converging lens and negative for a diverging lens. This is also sometimes presented in the Newtonian form:  $\displaystyle f\right)\left(d_i - f\right)=f^2$



#### Examples

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


${\displaystyle {\frac {1}{f}}={\frac {1}{d_{o}}}+{\frac {1}{d_{i}}}={\frac {1}{\infty }}+{\frac {1}{-d}}=-{\frac {1}{d}}}$

This is the resulting equation at the beginning of the article. This also explains why focal power is increased for objects at closer distances: traditional optometry calls this "addition" for presbyopia , although they generally use the minimum amount required for you to be able to see at 40cm with full distance correction using housing. For example, if you choose 80 cm as the working distance for your differentials  (resulting in an "addition" of +1.25 dpt), and your blur horizon is 50 cm (resulting in -2 dpt ), the formula is


< math>\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 [/itex]

### Cylinder

A cylindrical lens of focal power P cyl  has a power P at the angle θ of its axis: ${\displaystyle P=P_{cyl}(\sin \theta )^{2}}$ ==== Axis ==== L' axis is usually in degrees modulo 180. It is common for 0 to be written as 180 in some regions. ==== Transposition ==== We can understand why there are two different ways to write a combination of spherical and cylindrical lens, using the Pythagorean trigonometric identity and the complementary angle identity: ${\displaystyle (\sin \theta )^{2}+(\cos \theta )^{2}=1}$ $\displaystyle - 90^{\circ}\right)}$
$\displaystyle P_{cyl}\right)(\sin{\left(\theta - 90^{\circ}\right)})^2$


We can see that by adding the cylindrical power to the spherical power and inverting the cylinder and adding 90 degrees (like subtracting 90 degrees, since the axis is 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 shapes are equivalent.

#### Spherical equivalent

 Calculating the average value over all angles using an integral, the result [4] is


${\displaystyle P_{avg}=\lim _{T\to \infty }{\frac {1}{2T}}\int _{-T}^{T}P_{cyl}(\sin {t})^{2}\,dt={\frac {1}{2}}P_{cyl}}$

C' is why the spherical equivalent has power equal to half the power of the cylinder.


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

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

${\displaystyle \cos {2\theta }=1-2(\sin \theta )^{2}}$

$\displaystyle P = P_{cyl} (\sin{\left(\theta + \phi\right)})^2 = P_{cyl} \left( \frac{1 - \cos{\left(2\theta + 2\phi \ right)}}{2} \right) = \frac{1}{2} P_{cyl} + \frac{ - P_{cyl}}{2} \cos{\left(2\theta + 2\phi \right)}$

The constant parts are added to the spherical components. Cosines can be added by converting them to [ https://en.wikipedia.org/wiki/Phasor   phasors] and adding the phasors together. The resulting phasor corresponds to one of the two cylindrical lenses (see the section on transposition), and its corresponding spherical component must be subtracted from the total spherical component.


There are implementations of this at

*   http://opticampus.opti.vision/tools/cylinders.php

 === Decentration === The
Induced prism can be calculated using Prentice reign. Similar to Vertex Distance, shift is less of an issue for lower power lenses.


The amount of prism power P induced by the decentration c of a lens of power f is ${\displaystyle P=cf}$ 1 prism diopter displaces 1 cm for an object 1 m away. If c is in cm and f in diopters, then P is in prismatic diopters. A prism with vertex angle a and refractive index n gives an angle of light deflection d, which is equal to P diopters of the prism: ${\displaystyle d=(n-1)a}$ ${\displaystyle P=100\tan {d}=100\tan((n-1)a)}$

See Vertex distance#Calculation

 ==References==

1. see derivations athttps://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_III_ - _Optics_and_Modern_Physics_(OpenStax )/02%3A_Geometric_Optics_and_Image_Formation/2.05%3A_Thin_Lenses
2. just integrate one or two periods: https://www.wolframalpha.com/input /?i=average +de+%28sin+x%29%5E2+de+0+à+2+pi