22 Ophthalmic Optics Practice Questions & Answers
Every Ophthalmic Optics practice question from the Optician (ABO) Practice Test, with the correct answer and a short explanation.
Start practice test →1. A spectacle lens is made from a material with a refractive index of 1.60. What does that value tell you about light inside the material?
- A.Light slows to about 1/1.60 of its vacuum speed, so rays bend more at each surface.✓ Answer
- B.Light keeps its vacuum speed inside the lens, and only the curves bend the rays.
- C.Light speeds up to about 1.60 times its vacuum speed, so rays bend less.
- D.Light slows inside the lens, but the bending at each surface is unaffected.
Refractive index is n = c/v, the ratio of the speed of light in a vacuum to its speed in the material, so n = 1.60 means light travels at about 1/1.60 of its vacuum speed inside the lens. The greater the slowing, the more a ray changes direction at each surface, which is why higher-index lenses reach the same power with flatter curves.
Source: Brooks & Borish, System for Ophthalmic Dispensing, chapter on ophthalmic lens materials (refractive index defined as n = c/v)Report a problem with this question
2. A thin lens brings light from a distant object to a focus 25 cm behind the lens. What is the power of that lens?
- A.+4.00 D of plus power✓ Answer
- B.+2.50 D of plus power
- C.-4.00 D of minus power
- D.+25.00 D of plus power
Dioptric power is the reciprocal of the focal length expressed in metres, so F = 1/f = 1/0.25 m = +4.00 D. Because the focus lies behind the lens the focal length is positive, which makes the lens a converging (plus) lens.
Source: Definition of the dioptre: F = 1/f with f in metres (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic optics)Report a problem with this question
3. A +5.00 D lens and a -5.00 D lens are each held up to a distant scene. Which statement describes what each lens does to the light and the image formed?
- A.Both lenses converge the light, but the minus lens forms its real image much farther away.
- B.The plus lens diverges light into a virtual image; the minus lens converges it to a real image.
- C.The plus lens converges light to a real image; the minus lens diverges it to a virtual image.✓ Answer
- D.Both lenses diverge the light, and the plus lens forms its virtual image much closer in.
A plus lens adds convergence, so light from a distant object crosses behind the lens and forms a real, inverted image that can be caught on a screen. A minus lens adds divergence, so the rays never cross and the eye traces them back to an erect virtual image on the object side of the lens.
Source: Brooks & Borish, System for Ophthalmic Dispensing, image formation by converging and diverging lensesReport a problem with this question
4. Light travelling inside a lens material reaches the back surface at an angle of incidence larger than the critical angle for that material and air. What happens to that light?
- A.It emerges along the surface itself, bent to exactly ninety degrees.
- B.It emerges bent away from the normal, with only a faint reflection remaining.
- C.It splits evenly, half of it emerging and half reflecting back inside.
- D.It is reflected entirely back into the material, and none of it emerges.✓ Answer
Beyond the critical angle no refracted ray can exist, because the angle of refraction demanded by Snell's law would exceed 90 degrees, so all of the light is returned into the denser medium by total internal reflection. Exactly at the critical angle the refracted ray grazes along the surface at 90 degrees.
Source: Total internal reflection and the critical angle, standard geometrical optics (Brooks & Borish, System for Ophthalmic Dispensing)Report a problem with this question
5. A wearer of high-powered lenses in a material with an Abbe value of about 30 reports colored fringes around objects when looking through the edge of the lens, but not when looking straight ahead. What explains this?
- A.The high index shortens the focal length off-axis, so the colors focus at different depths.
- B.Dispersion happens only at the lens edge, where the surface is polished to a different curve.
- C.The anti-reflective coating splits white light into colors, most strongly near the edge.
- D.Dispersion is high in this low-Abbe material, and prism away from the optical center spreads the colors.✓ Answer
Abbe value measures how little a material disperses white light, so a value near 30 disperses strongly. Chromatic effects only become visible where the lens acts as a prism, which is away from the optical center, so the fringing appears in the periphery and grows with lens power and decentration.
Source: Abbe value and transverse chromatic aberration in ophthalmic lens materials (Brooks & Borish, System for Ophthalmic Dispensing)Report a problem with this question
6. The right lens of a finished pair is a +3.00 D sphere. The wearer's line of sight passes through a point 4 mm above the optical center. What prismatic effect is induced?
- A.1.20Δ base up
- B.1.20Δ base down✓ Answer
- C.0.12Δ base down
- D.12.00Δ base down
Prentice's rule gives the amount: prism in dioptres equals the power times the decentration in centimetres, so 3.00 x 0.4 cm = 1.20 prism dioptres. A plus lens behaves as two prisms base to base, so a point above the optical center lies in the upper prism, whose base points down toward the center.
Source: Prentice's rule, prism = F x d(cm), with base direction for plus lenses toward the optical center (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic prism)Report a problem with this question
7. A left lens is a -6.00 D sphere, and the wearer's line of sight passes 2.5 mm temporal to its optical center. What prismatic effect does the wearer experience?
- A.1.50Δ base in
- B.0.15Δ base out
- C.1.50Δ base out✓ Answer
- D.15.00Δ base out
By Prentice's rule the magnitude is 6.00 x 0.25 cm = 1.50 prism dioptres. A minus lens behaves as two prisms apex to apex, so the base at any viewing point lies away from the optical center; a temporal viewing point therefore gives base temporal, which is base out.
Source: Prentice's rule with base direction for minus lenses away from the optical center (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic prism)Report a problem with this question
8. Which statement defines the optical center of a spectacle lens?
- A.The point through which a ray passes undeviated, so no prismatic effect arises there.✓ Answer
- B.The geometric center of the finished lens shape, midway between the boxed edges.
- C.The point where the front and back curves are equal, so the power there is zero.
- D.The point of greatest thickness, where the two surface powers cancel each other.
The optical center is defined optically, not by shape or thickness: it is the single point at which the front and back surfaces are effectively parallel, so a ray through it emerges in the same direction it entered. Everywhere else the lens acts as a prism, which is why Prentice's rule measures decentration from this point.
Source: Definition of the optical center as the point of no prismatic deviation (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic prism)Report a problem with this question
9. A lens has a front surface power of +8.00 D and a back surface power of -5.50 D. Using the approximate thin-lens relationship, what is the power of the lens?
- A.-2.50 D of total power
- B.+2.50 D of total power✓ Answer
- C.+1.45 D of total power
- D.+13.50 D of total power
In the thin-lens approximation the power of a lens is simply the sum of its two surface powers, so (+8.00) + (-5.50) = +2.50 D. The exact value is slightly higher because centre thickness adds a small amount of plus, which is why laboratories compensate thickness on strong plus lenses.
Source: Nominal (thin-lens) lens power F = F1 + F2 (Brooks & Borish, System for Ophthalmic Dispensing, lens design)Report a problem with this question
10. The front surface of a lens made from a material of index 1.50 has a radius of curvature of 100 mm. What is the power of that surface?
- A.+0.50 D of surface power
- B.+10.00 D of surface power
- C.+5.00 D of surface power✓ Answer
- D.+15.00 D of surface power
Surface power is F = (n - 1)/r with the radius expressed in metres, so (1.50 - 1)/0.10 m = 0.50/0.10 = +5.00 D. The same radius in a higher-index material would give more power, which is the reason high-index lenses can be made with flatter surfaces.
Source: Surface power formula F = (n - 1)/r (Brooks & Borish, System for Ophthalmic Dispensing, lens surfaces)Report a problem with this question
11. A lens measure calibrated for an index of 1.53 reads +6.00 D on the front surface of a polycarbonate lens of index 1.586. What is the true power of that surface?
- A.+6.00 D, the dial reading is correct
- B.+6.63 D, higher than the dial reading✓ Answer
- C.+5.43 D, lower than the dial reading
- D.+9.00 D, well above the dial reading
A lens measure reads sagitta and converts it to power assuming an index of 1.53, so on any other material the dial must be corrected: true power = dial reading x (n_actual - 1)/(1.53 - 1) = 6.00 x 0.586/0.53 = +6.63 D. Because polycarbonate has a higher index than 1.53, the true surface power is greater than the dial shows.
Source: Lens measure calibration at n = 1.53 and the correction F_true = F_dial x (n - 1)/0.53 (Brooks & Borish, System for Ophthalmic Dispensing, instrumentation)Report a problem with this question
12. How do a spherical surface and a toric surface differ, and what does each correct?
- A.A toric surface has one curvature in every meridian; a spherical surface has two and gives cylinder.
- B.Both surfaces have a single curvature, and cylinder comes from decentering one against the other.
- C.Both surfaces have two principal curvatures, but the toric one sets them at oblique axes.
- D.A spherical surface has one curve in every meridian; a toric surface has two and gives cylinder.✓ Answer
A spherical surface has the same radius in every meridian, so it brings light to a single point focus and corrects only sphere. A toric surface has two principal curvatures 90 degrees apart, producing two focal lines, which is exactly what is needed to neutralise astigmatism.
Source: Spherical versus toric surfaces and the conoid of Sturm (Brooks & Borish, System for Ophthalmic Dispensing, lens surfaces and astigmatism)Report a problem with this question
13. A lens is made to -2.00 -1.00 x 180. Using F = sphere + cylinder x sin squared of the angle from the axis, what power lies in the 060 meridian?
- A.-2.25 D, a quarter of the cylinder
- B.-2.50 D, one half of the cylinder
- C.-2.75 D, three quarters applied✓ Answer
- D.-3.00 D, the full cylinder added
The 060 meridian lies 60 degrees from the 180 axis, and the sine of 60 degrees squared is 0.75, so three quarters of the cylinder is effective: -2.00 + (-1.00 x 0.75) = -2.75 D. This is the basis of the familiar rule that 30, 45 and 60 degrees from axis give 25, 50 and 75 percent of the cylinder.
Source: Power in an oblique meridian, F = Fsph + Fcyl sin^2(theta) (Brooks & Borish, System for Ophthalmic Dispensing, astigmatic lenses)Report a problem with this question
14. A prism displaces the image of a target by 3 cm when that target is 2 m away. What is the power of the prism?
- A.3.00 prism dioptres
- B.0.75 prism dioptres
- C.6.00 prism dioptres
- D.1.50 prism dioptres✓ Answer
One prism dioptre is defined as a displacement of 1 cm measured at 1 m, so the deviation must be scaled to a one-metre distance: 3 cm at 2 m is the same deviation as 1.5 cm at 1 m, giving 1.50 prism dioptres. Written as a formula, prism equals the displacement in centimetres divided by the distance in metres.
Source: Definition of the prism dioptre as 1 cm of displacement at 1 m (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic prism)Report a problem with this question
15. A single lens must be ground with 6.00 prism dioptres base in combined with 8.00 prism dioptres base up. What single resultant prism does this produce?
- A.10.00Δ, based up and in✓ Answer
- B.7.00Δ, based up and in
- C.14.00Δ, based up and in
- D.10.00Δ, based down and out
Horizontal and vertical prism are perpendicular vectors, so they combine by the Pythagorean theorem rather than by adding or subtracting: the square root of 6 squared plus 8 squared is 10.00 prism dioptres. The resultant base lies in the quadrant defined by the two components, here up and in.
Source: Resultant of perpendicular prism components by vector addition (Brooks & Borish, System for Ophthalmic Dispensing, ophthalmic prism)Report a problem with this question
16. Corrected-curve, or best-form, lens series choose the base curve for each power range mainly to control which aberration?
- A.Chromatic aberration, the color fringing seen through the periphery.
- B.Distortion, the change in image shape seen through the periphery.
- C.Spherical aberration, the blur from rays crossing at different points.
- D.Oblique astigmatism, the off-axis blur seen when looking through the periphery.✓ Answer
When the eye rotates behind a spectacle lens the visual axis strikes the surfaces obliquely, and that oblique incidence splits the focus into two focal lines, blurring off-axis vision. Best-form series pick the base curve that cancels this oblique astigmatism for each power, which is why a strong minus lens is supplied on a flatter base than a strong plus lens.
Source: Tscherning ellipse and corrected-curve (best-form) lens design for oblique astigmatism (Brooks & Borish, System for Ophthalmic Dispensing, lens design)Report a problem with this question
17. A wearer of high-minus lenses says straight door frames appear to bow outward near the edges of the field. Which aberration is this, and what causes it?
- A.Distortion, because prismatic power grows toward the periphery and alters the image shape.✓ Answer
- B.Chromatic aberration, because dispersion in the material grows toward the lens periphery.
- C.Spherical aberration, because peripheral zones of the lens focus light at a shorter distance.
- D.Oblique astigmatism, because peripheral rays strike the surface at a slant and split the focus.
Prismatic effect increases steadily with distance from the optical center, so magnification is not uniform across the lens and straight lines are reproduced as curves. A minus lens minifies more toward the edge and yields barrel distortion, while a plus lens magnifies more toward the edge and yields pincushion distortion.
Source: Distortion from non-uniform magnification with increasing prismatic effect (Brooks & Borish, System for Ophthalmic Dispensing, lens aberrations)Report a problem with this question
18. An eye is optically too long for the refracting power of its cornea and lens. With accommodation relaxed, where do rays from a distant object come to focus, and what corrects it?
- A.Behind the retina; a plus lens converges the light onto the retina.
- B.In front of the retina; a minus lens diverges the light onto the retina.✓ Answer
- C.Behind the retina; a minus lens diverges the light onto the retina.
- D.In front of the retina; a plus lens converges it onto the retina.
In myopia the eye has too much converging power for its axial length, so parallel light is brought to a focus in the vitreous ahead of the retina. A minus lens diverges the incoming light before it reaches the eye, moving the focus back onto the retina.
Source: Axial myopia and its correction with a diverging lens (Brooks & Borish, System for Ophthalmic Dispensing, refractive errors)Report a problem with this question
19. A presbyope with essentially no accommodation left wants to read at 40 cm. What add power places the near point at that distance?
- A.+4.00 D, supplying 1/0.25 m of focusing power for comfortable reading.
- B.+2.50 D, supplying the 1/0.40 m of focusing power now missing.✓ Answer
- C.+1.50 D, since half the demand at 0.40 m is met by the distance lens.
- D.+2.00 D, kept one step below the demand to preserve the range.
The dioptric demand of a working distance is the reciprocal of that distance in metres, so 40 cm calls for 1/0.40 = 2.50 D. When no accommodation remains, the add must supply the whole demand, and the resulting range of clear vision is centred on the plane the add focuses.
Source: Accommodative demand equals the reciprocal of the working distance in metres (Brooks & Borish, System for Ophthalmic Dispensing, presbyopia and the near addition)Report a problem with this question
20. A +8.00 D sphere was refracted at a vertex distance of 14 mm, but the chosen frame holds the lens at 10 mm. What power keeps the effective power at the eye unchanged?
- A.+7.75 D, a slightly weaker lens
- B.+8.25 D, a stronger lens✓ Answer
- C.+8.00 D, the very same power
- D.+9.00 D, a much stronger lens
Using F_new = F_old / (1 - d x F_old) with d = 0.004 m for the 4 mm the lens moves toward the eye, 8.00 / (1 - 0.032) = 8.26, which is ordered as +8.25 D. A plus lens loses effective power as it comes closer to the eye, so more plus must be ordered to hold the correction.
Source: Vertex distance compensation, F_new = F_old / (1 - d F_old) with d in metres (Brooks & Borish, System for Ophthalmic Dispensing, effective power)Report a problem with this question
21. Two finished lenses have the same power of -6.00 D and the same diameter, one in a material of index 1.50 and one of index 1.74. How do they compare?
- A.The 1.74 lens is thicker at the edge, since a higher index needs steeper curves.
- B.The two lenses have equal edge thickness, since power and diameter alone fix it.
- C.The 1.74 lens is thinner at the edge, since a higher index reaches power with flatter curves.✓ Answer
- D.The 1.74 lens is thinner at the edge and its Abbe value is higher, so less color fringing.
Surface power depends on (n - 1)/r, so a higher index produces the same dioptric power with a longer radius, that is a flatter surface, and a flatter surface needs less material at the edge of a minus lens. The trade-off is that raising the index generally lowers the Abbe value, so chromatic fringing increases rather than decreases.
Source: Relationship of refractive index to surface radius and lens thickness (Brooks & Borish, System for Ophthalmic Dispensing, high-index lens materials)Report a problem with this question
22. A +10.00 D sphere was refracted at 12 mm, but the dispensed frame holds it 4 mm farther from the eye. If the power is not recompensated, what does the wearer experience and what change is needed?
- A.Effectively too strong, so less plus power is needed to keep the correction.✓ Answer
- B.Effectively unchanged, since vertex distance matters only for minus lenses.
- C.Effectively too strong, so still more plus power is needed to sharpen it.
- D.Effectively too weak, so more plus power is needed to restore the distance correction.
Moving a plus lens away from the eye increases its effective power at the cornea, so the wearer is overcorrected and reports strain and a pulling sensation at distance. The remedy is to order less plus, which is the same rule applied in reverse when an aphakic or high plus lens is fitted closer to the eye.
Source: Effective power and vertex distance for plus lenses (Brooks & Borish, System for Ophthalmic Dispensing, effective power)Report a problem with this question
Practice questions based on the American Board of Opticianry Basic Certification exam content outline, standard ophthalmic dispensing references, and the federal rules that govern eyewear (the FDA impact-resistance requirement and the FTC Eyeglass and Contact Lens Rules). ABO and NCLE are marks of the American Board of Opticianry and National Contact Lens Examiners; this site is not affiliated with or endorsed by them. Opticianry is licensed in only some states and the requirements differ — confirm your own state board's practice act and the current exam format before testing. About the ABO exam →