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The Oblique Prisma
Author(s) -
G.H. Jonkers
Publication year - 1950
Publication title -
ophthalmologica
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.639
H-Index - 60
eISSN - 1423-0267
pISSN - 0030-3755
DOI - 10.1159/000300918
Subject(s) - oblique case , optometry , ophthalmology , medicine , philosophy , linguistics
The number of cases in which, both in the horizontal and vertical meridian, occurs a heterophoria, is not rare. Mostly it is sufficient to correct only the hyperphoria, but there are cases in which the correction of the hyperphoria appears to be insufficient, supposed an eventually existing error of refraction has been corrected already. Then it is necessary to correct the deviation in both meridians by means of rotating the prisma. Suppose, one finds in his examination with the Maddox tangent scale and rod at 5 m. a lefthyperphoria of 2 degrees and an exophoria of 5 degrees and it has appeared already that the correction of the hyperphoria gives insufficient relief to the patient, one must correct the existing exophoria too. We agree that we must totally correct a hyperphoria, but there is no “communis opinio” as to the extent in which we must correct an exoor esophoria. To clear this, I should like to refer to the results of my investigations with Verhoeff’s stereoscopic acuitymeter \ 100 normal men, examined at 5 m., appeared to have a hyperphoria of 0.5 degree average whereas the top of the distribution histiogram of the lateral heterophoria appeared to lie at approximately 1 degree esophoria. The normal values vary between 1.5 degree exophoria and 3.5 degree esophoria. When one wants to correct to the norm in the above-mentioned example, one will diminish the hyperphoria to about 0.5 degree or zero and the exophoria from 5 degrees to e.g. 1 degree. When one would correct only the hyperphoria, one would give a prisma of 2 diopters basis down to the left eye; when one wants to correct the mere exophoria, one would choose a prisma of 4 diopters basis nasal. Both are, in a simple way, to combine; the strength of the prisma and the deviation of the axis are calculated as follows: the strength of the prisma is determined by the length of the hypotenusa in the triangle COB: ]/~22 + 42 = y20 = 4.47, and the deviation is tg. a = 2/4 = ± 26 degrees. The 1 Ophthalmologica 118, 182 (1949). J o n k € γ s ‚ The Oblique Prisma 427

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