An indirect ophthalmoscope condensing lens is held 10 cm from the nodal
point of a cyclopleged eye whose refractive error is +10.00. It forms a
real image of the retina 5 cm behind the lens (i.e. towards the
examiner). What is the power of the condensing lens?
Clinical Optics
Telescopes and Optical Instruments
No
U
A
Use the vergence formula to figure out the required power of the condensing lens:
U + P = V
(where "U" is the object vergence, which is always negative; "P" is the
power of the lens/mirror; and "V" is the vergence of light exiting the
lens/mirror to form an image. Remember: the object and image vergences
are the reciprocal of their distances from the lens/mirror
respectively)
In this problem, treat the "object" as the far point of the eye. Since
the eye has a refractive error of +10.00, you know its far point is 1/10
= 0.1 m = 10 cm behind the eye.
The total distance between this "object" (i.e. the far point) and the
condensing lens is therefore 10 cm + 10 cm = 20 cm. Therefore, the
object vergence is U = -1/0.20 = -5 diopters.
Since the image is 5 cm behind the condensing lens, then the image
vergence is V = 1/0.05 = 20 diopters. You know this vergence is
positive since a real image was formed
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