Biomedical Engineering Reference
In-Depth Information
Suppose this patient's vision deteriorates so that in 2 years the smallest print that
can be read at 40.0 cm is 5 M. How can we help this patient?
Because of this deterioration, the patient now requires angular magnification of
5
to resolve 1 M print. He could hold the correspondence five times closer to his
eyes—at a distance of 8.0 cm (i.e., 40.0 cm/5
×
8.0 cm). This is the equivalent view-
ing distance. The patient, however, is not likely to be comfortable holding reading
material so close to his eyes. Moreover, the accommodative demand is 12.5 D (i.e.,
100/8.0 cm
=
12.5 D); it is difficult to sustain this amount of accommodation for
very long, even for young patients.
How can a plus lens be used to obtain this same angular magnification? As we've
learned, the patient can use a plus lens whose focal length is equal to the equivalent
viewing distance. In this case, where the equivalent viewing distance is 8.0 cm,
the power of the plus lens—the equivalent viewing power—is
=
12.5 D. As long as
the reading material is held at the focal point of the lens, the print will subtend the
same angle at the eye regardless of the distance from the eye to the plus lens.
4
+
Let's look at another example.
A 70-year-old patient with fully corrected myopia
has age-related macular degeneration and can barely see 5 M print at a dis-
tance of 40.0 cm when looking through his bifocal add. If our goal is to allow
the patient to read 2 M print, what power handheld magnifying lens should we
prescribe? How far should the lens be held from the reading material? How far
should the lens be held from the patient's eye? Should the patient look through the
magnifying lens with his distance correction or bifocal add?
At first glance, this may appear to be a rather confusing problem. If we keep our
wits about us and approach it one step at a time, it's not that complicated. The
patient can read 5 M print when it is located at the reference distance of 40.0 cm. In
order for the patient to read 2 M print, an angular magnification of 2.5
×
is required
(i.e., 5 M/2 M
=
2.5
×
). The equivalent viewing distance is 2.5
×
closer than the ref-
erence distance, which makes it 16.0 cm (i.e., 40 cm/2.5
=
16.0 cm). Therefore, the
plus lens must have a power of
+
6.25 D (i.e., 100/16.0 cm
=
6.25 D). (The equiva-
lent viewing power is
6.25 D.)
As long as the print is at the focal point of the plus lens, the patient can hold
the lens at whatever distance from the eye he finds most comfortable. The angle
subtended at the eye and the ability to see detail remains the same, even though
the field of view becomes smaller as the lens is positioned farther from the eye.
Since the light rays that emerge from the plus lens are parallel to each other, they
+
4. A slight variation on this approach is to use the formula
M
θ
(reference distance)(
F
)
, where
M
θ
is the
desired angular magnification and
F
is the power of the plus lens that will provide this magnification
when the reading material is held at the focal point of the lens. The
reference distance
is in meters. In the
current example,
M
θ
is 5
=
×
and the
reference distance
is 0.4 m. Solving for
F
gives a power of
+
12.5 D.
