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?
Telescopes and Optical Instruments
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
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