Optics related math

Here's a page with maths related to diopters and glasses.

You don't really need to know any of this stuff to improve your eyesight, but it's good to know for deeper understanding

The thin lens equation

${\frac {1}{f}}=({\frac {1}{s}})+({\frac {1}{s'}})$ where

• $f$ = focal length of lens
• $s$ = distance to object
• $s'$ = distance to image

Infinity

The term object at infinity is often used. When $s=\infty$ is substituted into the thin lens equation, that term vanishes, so that the focal length is then just the image location. For any sufficiently large image distance, the contribution from the reciprocal becomes negligible.

Virtual image

A converging lens (such as a magnifying glass) behaves like a "typical" lens - the incoming light is brought to a focus on the opposite side of the lens.

A diverging lens behaves differently - the light rays spreading from the source object are refracted outwards so that they are diverging even faster. They are not brought to a focus in any intuitive sense. Instead, the light behaves as if it was coming from a closer object. This is termed a virtual image - it lies between the source object and the lens. The thin lens equation still works as long as you use negative numbers to describe both the (virtual) image location and the focal length.

The corrective lens for myopia is a diverging lens. It works by forming a virtual image of objects far away, and it is that virtual image that the near-sighted eye is able to focus on.

Diopters are inverse meters

Intuitively, the more powerful a lens is, the more rapidly it can bring incoming light to a focus. So the power is defined as the inverse of the focal length.

See Also Diopters

See Also cm Measurement

Remember that 100cm = 1m.

$D={\frac {1}{meters}}$

conversely

$meters={\frac {1}{D}}$

Point of refraction

See also Refraction

Visual acuity equation

$({\frac {font\ height}{distance\ to\ sign}})({\frac {180}{pi}})\times 60=arcminutes=a$

Note: 5Arcminutes = 20/20

Set up proportion: ${\frac {a}{({\frac {20}{x}})}}={\frac {5}{({\frac {20}{20}})}}$

Visual acuity (mm/metres)

${\frac {font\ height(mm)}{distance\ to\ sign(m)}}\times 13.75=denominator\times of{\frac {20}{x}}$

Visual acuity (in/feet)

${\frac {font\ height(in)}{distance\ to\ sign(ft.)}}\times 1146=denominator\times of{\frac {20}{x}}$

With text that we are familiar with, the brain may clear up that text more than our vision would actually allow.

Average axial length accomodation/rate of change

$typical\ emmetropic\ eye=25mm=25,000\ microns$

$change\ in\ axial\ length\ of\ 1mm=3D$

If someone with typical eyes wanted to adapt say 20/20 to .25 less normalized within 3-4 months would need to decrease axial length 0.083mm about 0.92microns/day - 0.69microns/day average Credit: Mark Podowski

Converting from Glasses to Contact Lens Prescription or vice-versa

• Vertex distance
• Vertex distance formula (also for astigmatism)