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Opthalmoscope - retinal image formation

  1. Mar 8, 2005 #1
    I was just wondering if somebody could check to make sure my logic is correct for this lab exercise. The setup is basically this: using an ophthalmoscope, a doctor can peer into the pupil of a patient to view the retina and innards of the eye. The opthalmoscope is a device with a built-in light and changable converging or diverging lens, which can be adjusted to bring the image of the retina into better focus.

    We were told to predict what kind of lens in the opthalmoscope would be necessary to bring the retina into focus under the following conditions:

    Doctor's eye: Emmetropic (normal), Myopic (near-sighted), or Hyperopic (far-sighted)
    Patient's eye: Same as above

    For all combinations.

    Essentially, my thinking is simply that any time either the patient's eye or doctor's eye is myopic, a diverging lens will be needed, and any time one or the other eye is hyperopic, a converging lens will be needed. In the case that one is myopic and one is hyperopic, it depends on which eye has the stronger defect.

    That look correct? I'm quite comfortable dealing with defects with just one eye, but for some reason I'm making it difficult on myself when dealing with two (as in, light -> eye 1 -> reflected light -> eye 2).
     
  2. jcsd
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