Vertex distance is the distance between the back surface of a corrective lens, i.e. glasses (spectacles) or contact lenses, and the front of the cornea. Increasing or decreasing the vertex distance changes the optical properties of the system, by moving the focal point forward or backward, effectively changing the power of the lens relative to the eye. Since most refractions (the measurement that determines the power of a corrective lens) are performed at a vertex distance of 12–14 mm, the power of the correction may need to be modified from the initial prescription so that light reaches the patient's eye with the same effective power that it did through the phoropter or trial frame. Vertex distance is important when converting between contact lens and glasses prescriptions and becomes significant if the glasses prescription is beyond ±4.00 diopters (often abbreviated D). The formula for vertex correction is F c = ( F − 1 − x ) − 1 {\displaystyle F_{c}=\left(F^{-1}-x\right)^{-1}} , where Fc is the power corrected for vertex distance, F is the original lens power, and x is the change in vertex distance in meters. The effect can also be noticed by moving the glasses further away from the eyes. For a short-sighted person, this weakens the effective strength of the lens, which may make it easier to read text up close. More plus power or less minus power as you move the glasses further away from the eye.
Derivation The vertex distance formula calculates what power lens (Fc) is needed to focus light on the same location if the lens has been moved by a distance x. To focus light to the same image location:
f c = f − x {\displaystyle f_{c}=f-x}
where fc is the corrected focal length for the new lens, f is the focal length of the original lens, and x is the distance that the lens was moved. The value for x can be positive or negative depending on the sign convention. Lens power in diopters is the mathematical inverse of focal length in meters.
F = 1 f ; F c = 1 f c {\displaystyle {\begin{aligned}F&={\frac {1}{f}};&F_{\text{c}}&={\frac {1}{f_{\text{c}}}}\end{aligned}}}
Substituting for lens power arrives at
1 F c = 1 F − x {\displaystyle {\frac {1}{F_{\text{c}}}}={\frac {1}{F}}-x}
After simplifying the final equation is found:
F F c = 1 − x F ⇒ F c = F 1 − x F = 1 1 F − x ⇒ F = 1 1 F c + x {\displaystyle {\begin{aligned}&&{\frac {F}{F_{\text{c}}}}&=1-xF\\&\Rightarrow &F_{\text{c}}&={\frac {F}{1-xF}}={\frac {1}{{\frac {1}{F}}-x}}\\&\Rightarrow &F&={\frac {1}{{\frac {1}{F_{\text{c}}}}+x}}\end{aligned}}}
Examples
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