Автор: PeterPAN
Дата: 17-01-08 12:53
Малко пояснения, не ми се превежда
Example: Step 1. Magnification
Let's pick EF50mm f/1.8 II. The table says the closest focus distance for this lens is 0.45 meters (= ~17.7 inches), and the maximam magnification is 0.15x. Using the above basic formula:
x / 50 = 0.15
Solving for x, the derived maximum built-in lens extension is: x = 50 * 0.15 = 7.5mm
If you add Canon's 12mm extension tube to this lens, you will achieve a maximum total extension of 7.5 + 12 = 19.5mm
Therefore, using a 12mm extension tube with EF50 1.8 II lens, you can obtain a maximum magnification of:
19.5 / 50 = 0.39
a magnification of 0.39x [Filling the frame with an object of actual size 58.2mm x 38.7mm]
Example: Step 2. Minimum focusing distance (a rough approximation)
[NOTE: The following was corrected and updated Nov. 7, '03]
To continue with the example and figure out new minimum focusing distance (MFD) from the new magnification, we need the help of a few additional formulas (found in any general physics book):
1) M = b/a
2) 1/a + 1/b = 1/f
3) MFD = a + b
where
M: Magnification
a: lens-to-subject distance
b: lens-to-film/sensor distance
f: focal length
[NOTE: for a and b above, the position of lens being referred to is that of a theoretical equivalent lens of single element, NOT of the actual camera lens which usually have multiple elements.]
So, given the new magnifcation of 0.39x and focal length of 50mm:
M = 0.39 = b/a
1/a + 1/b = 1/50
Solving for a and b, you get:
a=178.2mm
b=69.5mm
MFD = a + b = 247.7mm
That is to say, using EF50mm f/1.8 II lens with a 12mm extension tube, the minimum focus distance should be reduced from 0.45 meters to ~0.25 meters (17.7 inches to ~9.75 inches)
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