ABO NOCE Basic Opticianry Ophthalmic Optics Questions and Answers 2 — Questions and Answers

Question 1: A lens has a power of +3.00 D. What is its focal length?

33.3 cm (Correct answer)
3.0 cm
0.33 cm
300 cm

Correct answer: 33.3 cm

Focal length (in meters) = 1 / Power. 1 / 3.00 = 0.333 m = 33.3 cm.

The relationship between focal length and lens power is f = 1/D, where f is in meters and D is in diopters. For a +3.00 D lens: f = 1/3.00 = 0.333 m = 33.3 cm. This means parallel rays of light converge to a focus 33.3 cm behind the lens. Understanding this relationship is essential for optical calculations on the ABO exam.

Question 2: Which of the following best describes the index of refraction?

The ratio of the speed of light in a vacuum to the speed of light in a medium (Correct answer)
The angle at which light bends at a surface
The measure of a lens's vergence power
The reciprocal of the focal length

Correct answer: The ratio of the speed of light in a vacuum to the speed of light in a medium

The index of refraction (n) = speed of light in vacuum / speed of light in medium. Higher index means slower light and greater bending.

The index of refraction (n) is a dimensionless number that describes how light propagates through a medium. It is calculated as n = c/v, where c is the speed of light in a vacuum (~3×10^8 m/s) and v is the speed of light in the medium. Glass with n=1.5 means light travels 1.5 times slower in glass than in vacuum. Higher refractive index materials bend light more and allow thinner lenses for high prescriptions.

Question 3: Snell's Law states that n1 sin θ1 = n2 sin θ2. If light travels from air (n=1.00) into glass (n=1.50) at an angle of incidence of 30°, what is the angle of refraction?

19.47° (Correct answer)
45.00°
30.00°
48.59°

Correct answer: 19.47°

1.00 × sin(30°) = 1.50 × sin(θ2). sin(θ2) = 0.5/1.5 = 0.333. θ2 = arcsin(0.333) ≈ 19.47°.

Snell's Law governs the bending of light at an interface between two media: n1 sin θ1 = n2 sin θ2. Here n1=1.00 (air), θ1=30°, n2=1.50 (glass). So 1.00 × sin(30°) = 1.50 × sin(θ2). sin(30°)=0.5, so 0.5 = 1.50 × sin(θ2), giving sin(θ2)=0.333 and θ2=19.47°. Light bends toward the normal when entering a denser medium. This principle underlies lens design.

Question 4: What is the vergence of light at a point 50 cm from a point source?

-2.00 D (Correct answer)
+2.00 D
-0.50 D
+0.50 D

Correct answer: -2.00 D

Vergence = 1/distance (in meters). Light from a source is diverging, so vergence = 1/(-0.50 m) = -2.00 D.

Vergence describes the degree of convergence or divergence of a wavefront. For diverging light from a point source at 50 cm (0.50 m), the vergence = 1/0.50 = 2.00 D, but since the light is diverging (spreading out from the source), the vergence is negative: -2.00 D. Converging light heading toward a focus has positive vergence. This concept is fundamental to the vergence/power/image calculations tested on the ABO exam.

Question 5: A concave mirror has a radius of curvature of 40 cm. What is its focal length?

20 cm (Correct answer)
40 cm
80 cm
10 cm

Correct answer: 20 cm

For a spherical mirror, focal length = radius of curvature / 2 = 40/2 = 20 cm.

For a spherical mirror (concave or convex), the focal length (f) equals half the radius of curvature (r): f = r/2. With r = 40 cm, f = 20 cm. The focal point of a concave mirror is the point where parallel rays converge after reflection. This is tested in opticianry as it relates to funduscopy and other ophthalmic instruments that use mirrors.

Question 6: Which type of lens corrects myopia (nearsightedness)?

Minus (concave) lens (Correct answer)
Plus (convex) lens
Plano lens
Prism lens

Correct answer: Minus (concave) lens

Myopia results from light focusing in front of the retina. A minus (diverging) lens moves the focal point back to the retina.

Myopia (nearsightedness) occurs when the eye has too much converging power, causing parallel rays to focus in front of the retina. A minus (concave, diverging) lens is prescribed because it diverges incoming light, effectively reducing the total dioptric power of the eye-lens system so that the focal point lands precisely on the retina. The amount of minus power prescribed equals the degree of myopia (e.g., -3.00 D corrects 3.00 D of myopia).

ABO NOCE Basic Opticianry Ophthalmic Optics Questions and Answers 2

A lens has a power of +3.00 D.
What is its focal length?