Biomedical Engineering Reference
In-Depth Information
- 3.86 D
+ 14.89 D
H
H
′
- 25.92 cm
- 25.00 cm
+ 0.92 cm
- 0.46 cm
O
I
+ 8.61 cm
+ 9.07 cm
+ 11.03 D
Figure 6-4.
When locating the image formed by an equivalent lens, the object distance
is measured from the primary principal plane and the image distance from the second-
ary principal plane. Object vergence and equivalent lens power are given, respectively,
at the top of the equivalent lens and image vergence at the bottom.
Figure 6-4 shows the locations of the equivalent lens's primary and secondary
principal planes. The object is located 25.00 cm to the left of the lens.
When we use
the vergence equation to solve this problem, should we use an object distance of
25.00 cm?
The answer is “no” because when we create an equivalent lens, object (and image)
distances are measured from the principal planes, not the lens surfaces. Thus, the
object distance is
−
25.92 cm
(Fig. 6-4). The object vergence is calculated the same
as it would be for a thin lens situated in air
−
n
l
L
=
(100)(1.00)
L
=
−
25.92 cm
L
= −
3.86 D
We can now use the vergence relationship to locate the image
L
′
=
L
+
F
e
L
′
= −
3.86 D
+
14.89 D
11.03 D
The positive vergence tells us that the image is real. As with a thin lens in air, the
image distance is calculated as
n
L
′
= +
′
′
L
=
′
l
