Ophthalmic Optics and Principles Questions and Answers Flashcards
7 cards from real ABO NOCE Basic Opticianry practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Ophthalmic Optics and Principles Questions and Answers flashcards as text
Which type of lens aberration causes the image of a point source off the optical axis to appear as a comet-shaped blur?
Answer: Coma
Coma produces a comet-shaped blur for off-axis point sources because rays through different zones of the lens converge to different points.
The Abbe number (V-number) of a lens material is used to measure which optical property?
Answer: Chromatic dispersion
The Abbe number indicates the degree of chromatic dispersion; a higher Abbe number means less chromatic aberration.
A patient has 2Δ base-in prism OD and 2Δ base-in prism OS. What is the total prism effect?
Answer: 4Δ base-in
Base-in prisms in both eyes add together for horizontal vergence demand, totaling 4Δ base-in.
What is the purpose of a lenticular lens design in high-power prescriptions?
Answer: To reduce lens weight and thickness by limiting the optical zone
Lenticular designs place full prescription power only in a central optical zone, reducing overall lens weight and thickness for high-power corrections.
Which formula correctly describes Snell's Law of refraction?
Answer: n₁ × sin θ₁ = n₂ × sin θ₂
Snell's Law states n₁ sin θ₁ = n₂ sin θ₂, relating the indices of refraction and angles at a refracting interface.
A lens has a front surface power of +8.00 D and a back surface power of −4.50 D. Using the thick lens formula, the reduced distance (t/n) introduces a power correction. If t/n = 0.003 m, approximately how much additional power does this add?
Answer: +0.11 D
The thick lens correction ≈ (t/n) × F1 × F2 = 0.003 × 8.00 × 4.50 = 0.108 ≈ +0.11 D added power.
A patient's spectacle prescription is −5.00 DS. Using a vertex distance of 13 mm, the approximate contact lens power needed is:
Answer: -4.66 D
Effective power = F / (1 − d×F) = −5.00 / (1 + 0.013 × 5.00) = −5.00 / 1.065 ≈ −4.69 D, closest to −4.66 D.