Biomedical Engineering Reference
In-Depth Information
Let's summarize a few key points on image formation by spherocylindrical lenses.
Assume that the major meridians of the lens are at 90 and 180 degrees and that
both have positive power.
• You can think of a point source (i.e., point object) as emitting light rays that
diverge vertically and horizontally. Vertically diverging object rays are focused
by the vertical meridian of the lens and horizontally diverging object rays are
focused by the horizontal meridian.
• For a point source, a horizontal line is formed where the vertical meridian is
in focus. The upper and lower edges of the line are in sharp focus because
they are formed by the in-focus vertical meridian, while the smeared horizontal
extent of the line is due the out-of-focus horizontal meridian.
• For a point source, a vertical line is formed where the horizontal meridian is in
focus. The left and right edges of the line are sharply focused by the in-focus
horizontal meridian, while the smeared vertical extent of the line is due to the
out-of-focus vertical meridian.
• The vertically diverging rays emitted by a horizontal line (object) are focused
by the vertical meridian as a horizontal line.
• The horizontally diverging rays emitted by a vertical line are focused by the
horizontal meridian as a vertical line.
This material can be difficult to conceptualize, so don't fret if you are finding
it confusing. Take a break—maybe sleep on it—and then reread the sections on
image formation for point and extended sources. It will click!
POWER IN AN OBLIQUE MERIDIAN
OF A CYLINDRICAL LENS
Consider the pl
030 lens in Figure 9-9. The power in the axis meridian
(30 degrees) is zero, and the power in the 120-degree meridian (the power meridian) is
−
−
5.00
×
5.00 D. What about the other meridians of the lens? Can we determine the power at,
say, 180 degrees? The answer is “yes,” we can calculate a power for this meridian, but it
may not be all that meaningful of a value. Let's first do the calculation and then I'll tell
you the problem with it. The formula to determine the power in the oblique meridian
of a cylindrical lens—sometimes called the
sine-squared formula
—is as follows:
F
=
(
F
cyl
)sin
2
θ
θ
where
F
is the power in the oblique meridian,
F
cyl
is the power in the power
meridian, and
θ
is the angle between the oblique meridian and the cylinder axis.
In the current example, the oblique meridian is 180 degrees. We are asked
to determine the power in this meridian. The power in the power meridian
θ
