Biomedical Engineering Reference
In-Depth Information
Total lateral magnification
=
(
M
L
for first lens) (
M
L
for second lens)
Total lateral magnifi cation
=
(
+
0.57) (
−
1.71)
= −
0.97
×
Therefore, the final image size is
(
6.8 mm
This is the same image size we calculated previously.
−
0.97)(7.00 mm)
= −
Let's consider another example.
An object is located 33.33 cm to the left of a
+
10.00 D thin lens that is located 50.00 cm to the left of a
−
2.00 D thin lens.
Locate the image, and calculate the magnification.
We won't go through all the calculations, but you should confirm that they are
correct. Viewing Figure 5-4, we see that an object vergence of
−
3.00 D is incident
upon the first lens, resulting in an image vergence of
7.00 D. (The image vergence
is obtained by adding together the object vergence and lens power.) The image
is real (therefore inverted) and located 14.29 cm to the right of the first lens or
35.71 cm in front of the second lens.
The real image formed by the first lens serves as an object for the second lens.
This object emits diverging light rays that are incident upon the second lens with
an object vergence of
+
−
2.80 D. The image vergence of the rays leaving the second
lens is
4.80 D. The image formed by this lens, which is virtual, is located 20.83 cm
to its left.
−
-3.00 D
+10.00 D
-2.80 D
-2.00 D
-33.33 cm
50.00 cm
I
O
+14.29 cm
-35.71 cm
-20.83 cm
+7.00 D
-4.80 D
Figure 5-4.
The real image formed by the plus lens serves as an object for the minus
lens. A virtual image is formed by this optical system.
