A patient presents with a refraction of -2.50 +0.75 x 90. He wants to be fit into a hard contact lens. His K readings are vertical 7.65 mm of radius (44.12 D @ 90) and 7.8 mm of radius (43.25 D @180). You chose to put a 7.65 mm radius of curvature contact lens on him. What should be the final spherical power of the hard contact lens?
Contact Lenses
No
U
D
Empirically fitting contact lenses is a right of passage for board exams. When fitting hard contact lenses, you should first change the refraction to minus cylinder format. The hard contact lens' tear film eliminates astigmatism, so you drop the cylinder part from the minus cylinder format. The remaining sphere is adjusted by whether or not the hard contact lens is flatter or steeper than the flattest K. In this case, the lens is steeper than K. For every 0.05 mm of radius, the sphere is adjusted 0.25 D. 0.15 difference = 0.75. Since it is steeper, you add minus 0.75 to -1.75 to get -2.50 sphere. There are a lot of subtleties with this question. First, note that higher mm of radius are actually flatter than lower mm or radius. This is evidence by the higher dioptric reading of 7.65 mm than 7.8 mm. Also note that you didn't need to know that 0.05 mm of radius changed the dioptric power 0.25 sphere because you fit the patient with a 44.12 diopter lens and the difference between this and the flat meridian (43.25) was about 0.75 diopters.
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