“Against motion” at 67 cm neutralized by −1.50 DS. True refraction is:
A. −1.50 DS
B. −3.00 DS ✅
C. Plano
D. +1.50 DS
Explanation: −1.50 − 1.50 (WD) = −3.00 DS myopia.
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Be respectful and professional Treat all members with respect, no harassment, insults, or discriminatory language. Debate ideas, not people.
Educational purpose only This group is for learning and discussion, not for giving personal medical advice or replacing a consultation with an eye care professional.
Case and patient privacy Remove all identifying details from clinical cases; no names, photos, or data that could identify a real patient.
Quality of MCQs and answers Write clear MCQs with one best answer, provide references or brief explanations when possible, and be open to corrections from others.
No spam or self‑promotion No advertising, links, or promotions unless explicitly allowed by the admin (courses, products, channels, etc.).
Language and clarity Use clear, understandable English, avoid excessive slang, and format questions/answers so they are easy to read.
Constructive feedback only When correcting an answer or MCQ, be polite, explain your reasoning, and focus on helping others learn.
Follow platform policies All content must comply with the website’s general terms and community standards.See less
More myopia pre-cyclo than post-cyclo indicates:
A. True high myopia
B. Accommodative spasm ✔️
C. Keratoconus
D. Instrument error
Explanation: Pre-cyclo myopia is pseudomyopia from accommodation, cycloplegia reveals true lower myopia.
Neutralize at 67 cm with +2.00 D “with” motion. True refraction is:
A. +3.50 D
B. +2.00 D
C. +0.50 D ✔️
D. Plano
Explanation: +2.00 − 1.50 (WD) = +0.50 D hyperopia.
You perform retinoscopy at 67 cm and find neutrality with +0.50 DS “with” motion. What is the most likely underlying refractive error?
A. +2.00 DS
B. +1.00 DS ✔️
C. Plano
D. −0.50 DS
Explanation: With “with” motion neutralized by +0.50 DS, subtract the working distance: +0.50 − 1.50 = −1.00 DS myopia. However, the question asks for the refraction that produced “with” motion at that point; the patient is actually −1.00 DS myopic, but when viewed at 67 cm shows +0.50 neutrality, so true refraction is −1.00 DS. (This accounts for the apparent +2.00 D error at the working distance.)
You perform static retinoscopy at 67 cm and reach neutrality with +4.00 DS in the trial frame. What is the patient’s approximate refractive error?
A. +2.50 DS ✔️
B. +3.50 DS
C. +2.00 DS
D. +4.00 DS
Explanation: At 67 cm, the working distance is +1.50 D. Subtract this from the neutralizing lens: +4.00 − 1.50 = +2.50 DS hyperopia.
During cover testing, a patient shows 10Δ exophoria at distance and 25Δ exophoria at near. The MOST likely diagnosis is:
A. Convergence insufficiency ✔️
B. Divergence excess
C. Basic exotropia
D. Accommodative spasm
Explanation: Convergence insufficiency is characterized by a remote near point of convergence and greater exodeviation at near than distance. The significant increase at near (15Δ differential) is pathognomonic, distinguishing it from basic exotropia (similar deviation) or divergence excess (greater at distance).
A unilateral, dilated, poorly reactive pupil in an otherwise healthy 25-year-old with normal accommodation is MOST suggestive of:
A. Third nerve palsy
B. Pharmacologic mydriasis
C. Adie’s tonic pupil ✔️
D. Argyll Robertson pupil
Explanation: Adie’s tonic pupil shows light-near dissociation (poor light response, better near response) due to ciliary ganglion damage. It’s typically unilateral, benign, and occurs in healthy young adults. Third nerve palsy would have ptosis and extraocular muscle involvement, while Horner’s syndrome causes miosis.
A patient with high hyperopia (+8.00 D) and esotropia shows reduced deviation with +3.00 D addition. This represents:
A. Partially accommodative esotropia
B. Non-accommodative esotropia
C. Divergence insufficiency
D. Sensory esotropia
Explanation: Partially accommodative esotropia has both accommodative and non-accommodative components. The +3.00 D addition reduces but doesn’t eliminate the deviation. Fully accommodative esotropia would be completely corrected, while non-accommodative would show no reduction.
A patient with a full-threshold 3+ nuclear sclerotic cataract experiences a −2.00 D myopic shift. The LEAST likely cause is:
A. Central lens index changes
B. Anterior lens displacement
C. Posterior lens displacement
D. Cortical hydration changes
Explanation: Nuclear sclerosis increases lens density and curvature, causing a myopic shift. Posterior lens displacement would create a hyperopic shift, not myopic. The nuclear changes include index of refraction increases, thickening, and anterior displacement.
In retinoscopy, a −2.00 D against motion is neutralized with a +2.00 D lens at 67 cm. The patient’s refractive error is:
A. −0.50 D
B. Plano
C. +0.50 D
D. −1.00 D
Explanation: Working distance compensation is required. At 67 cm, add +1.50 D to the neutralizing lens. Against motion means myopia, so the patient is −2.00 D (against motion) + +2.00 D (neutralizing) − +1.50 D (working distance) = +0.50 D hyperopic.