Biomedical Engineering Reference
In-Depth Information
This erect and minified virtual image serves as an object for the second lens
(i.e., the
15.00 D lens). Light rays diverge from this image, which is located
10.57 cm (8.57
+
10.57 cm) to the left of the second lens. (You should
confirm this by examining the diagram.) At the second lens, these rays produce an
object vergence
of
+
2.00 cm
=
n
L
L
=
(100)(1.00)
−
L
=
= −
9.46 D
10.57 cm
This vergence is written above the second lens. To locate the image produced by
this lens, we use the vergence relationship as follows:
L
′
=
L
+
F
5.54 D
This value is written below the second lens.
Since the rays that emerge from the lens are converging—they have positive
vergence-the image is real. How far is it from the second lens?
L
′
=
−
9.46 D
+
(
+1
5.00 D)
= +
n
′
′ =
L
′
l
n
L
′
′
=
l
(100)(1.00)
+
′
=
l
= +
18.05 cm
5.54
The final image is located 18.05 cm to the right of the
15.00 D lens.
The lateral magnification produced by the second lens is
L
L
+
M
L
=
′
−
9.46 D
M
L
=
5.54 D
= −
1.71
×
+
Since the object for this lens (i.e., the image formed by the first lens) is
+
3.99 mm
in height, the final image's height is
6.8 mm
The (real) image produced by this optical system is inverted and minified relative to
the initial object (which is 7.00 mm in height).
Another way to determine the final image size is to calculate the lateral magni-
fication for the system as a whole and multiply this total magnification times the
initial object size. The lateral magnification for the system is the product of the two
lenses' magnifications:
(
−
1.71) (
+
3.99 mm)
= −
