FREE Board Certified Ophthalmologist Optics and Refraction Principles Questions and Answers Flashcards
6 cards from real Board Certified Ophthalmologist practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 FREE Board Certified Ophthalmologist Optics and Refraction Principles Questions and Answers flashcards as text
During retinoscopy at a working distance of 67 cm, a neutral reflex is observed with a +3.50 D lens in the phoropter. What is the patient's refractive error?
Answer: +2.00 D
Subtracting the working distance lens of +1.50 D (1/0.67 m) from the +3.50 D neutralizing lens gives a net refractive error of +2.00 D.
What optical principle explains why an IOL placed in the capsular bag requires less power than one placed in the anterior chamber?
Answer: Effective lens position closer to the retina requires less converging power
A lens positioned closer to the retina (more posterior) requires less converging power to focus light on the fovea compared to a more anterior position.
A myopic patient has a far point located 25 cm in front of the eye. What corrective lens power is needed at the spectacle plane (assuming negligible vertex distance)?
Answer: -4.00 D
The corrective lens must have a focal length equal to the far point distance, so power = -1/0.25 m = -4.00 D.
Which type of multifocal IOL design uses concentric rings of alternating distance and near focal zones to split incoming light?
Answer: Diffractive multifocal
Diffractive multifocal IOLs use concentric ring steps on the lens surface to create constructive interference patterns that split light between distance and near foci.
Jackson cross-cylinder testing refines which component of a patient's refractive correction?
Answer: Astigmatic axis and power
The Jackson cross-cylinder is specifically designed to refine both the axis and power of the cylindrical (astigmatic) correction during subjective refraction.
When light passes from a medium with refractive index 1.33 into a medium with refractive index 1.00, at an angle of incidence of 30 degrees, what is the approximate angle of refraction?
Answer: 41.7 degrees
Applying Snell's law: 1.33 × sin(30°) = 1.00 × sin(θ), giving sin(θ) = 0.665 and θ ≈ 41.7 degrees.