Difference between revisions of Eyeballs
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We can use simple optics, the thin lens equation <math>\frac{1}{x} + \frac{1}{y} = \frac{1}{f}</math> and some very approximate numbers to give order-of-magnitude estimates of some of the quanties involved. | We can use simple optics, the thin lens equation <math>\frac{1}{x} + \frac{1}{y} = \frac{1}{f}</math> and some very approximate numbers to give order-of-magnitude estimates of some of the quanties involved. | ||
To estimate the focusing power of an [[emmetropic eye]], we might take the [[axial length]] as around 2.5cm. For [[distance vision]] (parallel incident light) that number is simply the [[focal length]] of the eye at rest, giving 40 [[Diopters]]. If we take the near point as about 25cm, that requires an additional 4 dpt of focusing power from the lens. | To estimate the focusing power of an [[emmetropic eye]], we might take the [[axial length]] as around 2.5cm. For [[distance vision]] (parallel incident light) that number is simply the [[focal length]] of the eye at rest, giving 40 [[Diopters]]. If we take the [[near point]] as about 25cm, that requires an additional 4 dpt of focusing power from the lens. | ||
If we now suppose that myopia is due entirely to elongation (ie the focusing power is unchanged), how much does the axial length need to increase to bring the [[blur horizon]] to 40cm ? With a 40 dpt lens and a source object at 40cm, the image would form 26.67mm from the lens, giving an estimate of elongation of 1.67mm or 6%. | If we now suppose that myopia is due entirely to elongation (ie the focusing power is unchanged), how much does the axial length need to increase to bring the [[blur horizon]] to 40cm ? With a 40 dpt lens and a source object at 40cm, the image would form 26.67mm from the lens, giving an estimate of elongation of 1.67mm or 6%. | ||