關於近視度數與眼軸長的關係 - 眼鏡
![Rachel avatar](/img/girl.jpg)
By Rachel
at 2008-07-07T16:49
at 2008-07-07T16:49
Table of Contents
模型眼
/ |
空氣 / 眼內介質 |視
| |網
n=1 | n=1.336 |膜
\ |
\ |
軸長 22.78
屈
折
面
R = 5.73
========================================================================
運用標準模型眼公式
n/f = (n-1)/R .............(1)
在標準模型眼中..
眼軸長f為22.78..
介面曲率半徑R為5.73mm..
折射率n為1.336..
假設此人近視D度..佩戴近視D度隱形眼鏡可正常投影至視網膜上..
設(f+a)為近視D度的眼軸長..可知
n/(f+a) = (n-1)/R - D
當 a << f 的時候 ..用泰勒展開式取近似
n/(f+a) ≒ n/f - n*a/f^2 = (n-1)/R - D ............(2)
(1)代入(2)得
n*a/f^2 = D
a = (f^2)*D/n
近視100度的人..D為1.0代入得
a = (0.02278)^2*1/1.336
= 0.0003884 m
= 0.3884 mm
即每近視100度..眼軸長會增加0.3884mm
--
/ |
空氣 / 眼內介質 |視
| |網
n=1 | n=1.336 |膜
\ |
\ |
軸長 22.78
屈
折
面
R = 5.73
========================================================================
運用標準模型眼公式
n/f = (n-1)/R .............(1)
在標準模型眼中..
眼軸長f為22.78..
介面曲率半徑R為5.73mm..
折射率n為1.336..
假設此人近視D度..佩戴近視D度隱形眼鏡可正常投影至視網膜上..
設(f+a)為近視D度的眼軸長..可知
n/(f+a) = (n-1)/R - D
當 a << f 的時候 ..用泰勒展開式取近似
n/(f+a) ≒ n/f - n*a/f^2 = (n-1)/R - D ............(2)
(1)代入(2)得
n*a/f^2 = D
a = (f^2)*D/n
近視100度的人..D為1.0代入得
a = (0.02278)^2*1/1.336
= 0.0003884 m
= 0.3884 mm
即每近視100度..眼軸長會增加0.3884mm
--
Tags:
眼鏡
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