21 Prescription Analysis & Math Practice Questions & Answers

Every Prescription Analysis & Math practice question from the Optician (ABO) Practice Test, with the correct answer and a short explanation.

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1. A spectacle prescription reads OD +1.75 -0.75 x 180, OS +2.00 -0.50 x 010, Add +1.50 OU, with 2 prism diopters base down before the right eye. Which statement correctly describes what this prescription specifies?
- A.The 2 prism diopters base down must be divided into 1 prism diopter base down before each eye
- B.The cylinder of the left lens exerts its power at the 010 meridian
- C.The prism belongs only in front of the right eye, and each eye's near sphere is found by adding +1.50 to that eye's own distance sphere✓ Answer
- D.The +1.50 is the complete sphere power of the near portion for each eye
OD is the right eye and OS the left; OU means both. The Add is never a stand-alone power - it is added algebraically to that eye's distance sphere, so the right near sphere is +3.25 and the left is +3.50. Prism written for one eye stays on that eye unless the prescriber authorizes splitting it, and cylinder power is effective 90 degrees from its axis, so the left cylinder acts at 100, not 010.
Source: American Board of Opticianry Basic Certification exam content outline, prescription format and content; System for Ophthalmic Dispensing, reading the spectacle prescriptionReport a problem with this question
2. A prescription is written in plus-cylinder form as +1.50 +2.25 x 065. Written in minus-cylinder form, the same lens is:
- A.-0.75 -2.25 x 155
- B.+3.75 +2.25 x 155
- C.+3.75 -2.25 x 065
- D.+3.75 -2.25 x 155✓ Answer
Flat transposition has three steps in order: algebraically add sphere and cylinder (+1.50 plus +2.25 gives +3.75), reverse the cylinder sign (+2.25 becomes -2.25), and rotate the axis 90 degrees (065 plus 90 equals 155, because the original axis is 90 or less). The transposed form describes exactly the same lens in different notation.
Source: System for Ophthalmic Dispensing, transposition of sphero-cylindrical prescriptionsReport a problem with this question
3. A prescription is written in minus-cylinder form as -5.00 -1.75 x 110. Expressed in plus-cylinder form, it becomes:
- A.-3.25 +1.75 x 020
- B.-6.75 -1.75 x 020
- C.-6.75 +1.75 x 020✓ Answer
- D.-6.75 +1.75 x 200
Transposition runs the same way in both directions: the new sphere is -5.00 plus -1.75, which equals -6.75; the cylinder sign flips to +1.75; and because the original axis 110 is greater than 90, 90 is subtracted to give 020. An axis of 200 is impossible, since axis notation only runs from 001 through 180.
Source: System for Ophthalmic Dispensing, transposition and the axis notation conventionReport a problem with this question
4. A distance prescription for the right eye reads +2.00 -3.50 x 090. After determining the power present in each principal meridian, how should this eye's astigmatism be classified?
- A.Compound hyperopic astigmatism
- B.Compound myopic astigmatism
- C.Mixed astigmatism✓ Answer
- D.Simple hyperopic astigmatism
The sphere power acts along the axis meridian, so 090 carries +2.00, and the cylinder adds fully 90 degrees away, so 180 carries +2.00 plus -3.50, which equals -1.50. One meridian is plus and the other minus, meaning one focal line falls behind the retina and the other in front, which defines mixed astigmatism.
Source: American Board of Opticionry Basic Certification exam content outline, classification of refractive error from the prescription; Ophthalmic Lenses and Dispensing, types of astigmatismReport a problem with this question
5. A lens is found to have +1.00 D of power in the 030 meridian and -2.00 D of power in the 120 meridian. Written in minus-cylinder form, this lens is:
- A.-2.00 -3.00 x 120
- B.+1.00 -3.00 x 030✓ Answer
- C.-2.00 +3.00 x 030
- D.+1.00 -3.00 x 120
In minus-cylinder form the sphere is the most plus of the two meridian powers, so the sphere is +1.00 and the axis is written at the meridian where the sphere alone acts, namely 030. The cylinder is the difference reaching the other meridian: -2.00 minus +1.00 equals -3.00, which then acts at 120, ninety degrees from the axis.
Source: System for Ophthalmic Dispensing, the optical cross and principal meridian powersReport a problem with this question
6. A prescription reads -1.25 -2.00 x 180. What total power is present in the vertical (090) meridian of that lens?
- A.-2.25 D
- B.-1.25 D
- C.-2.00 D
- D.-3.25 D✓ Answer
The sphere power lies along the axis meridian and the full cylinder power is added at the meridian 90 degrees from the axis. With the axis at 180, the 090 meridian is exactly 90 degrees away, so it carries -1.25 plus -2.00, which equals -3.25 D. This is the power that must be used for any vertical prism or segment-height calculation on this lens.
Source: Ophthalmic Lenses and Dispensing, power in the principal meridiansReport a problem with this question
7. What is the spherical equivalent of the prescription -4.75 -2.50 x 015?
- A.-3.50 D
- B.-6.00 D✓ Answer
- C.-2.25 D
- D.-7.25 D
Spherical equivalent equals the sphere plus half the cylinder, carrying the signs through: half of -2.50 is -1.25, and -4.75 plus -1.25 equals -6.00 D. This value represents the dioptric midpoint between the two principal meridians, which is why it is used for base-curve selection and for approximating a sphero-cylinder with a single power.
Source: System for Ophthalmic Dispensing, spherical equivalentReport a problem with this question
8. A right lens is made to the prescription plano +2.00 x 180. Using the standard relationship in which none of the cylinder is effective at the axis, half of it is effective 45 degrees away and all of it is effective 90 degrees away, what total power is present in the 045 meridian?
- A.+1.00 D✓ Answer
- B.+2.00 D
- C.+0.67 D
- D.Plano
The 045 meridian lies 45 degrees away from the axis at 180, so exactly half of the +2.00 cylinder is in effect there, which is +1.00, and the plano sphere adds nothing. Power in an oblique meridian always builds from zero at the axis to the full cylinder power at 90 degrees from the axis.
Source: Ophthalmic Lenses and Dispensing, power in oblique meridiansReport a problem with this question
9. A wearer's distance prescription for the right eye is -3.75 -1.00 x 095 with an Add of +2.50. What single-vision near prescription should be ordered for that eye?
- A.-1.25 -1.00 x 095✓ Answer
- B.-1.25 +1.00 x 005
- C.-1.25 -1.00 x 005
- D.-6.25 -1.00 x 095
The Add is applied algebraically to the sphere only; the cylinder power and the axis are never changed. Adding +2.50 to -3.75 moves the sphere in the plus direction to -1.25, so less minus is the correct result. Subtracting instead of adding would give -6.25, which is the classic sign error.
Source: System for Ophthalmic Dispensing, deriving a near prescription from the distance prescription and AddReport a problem with this question
10. A wearer's left distance prescription is +1.25 -0.50 x 070 with an Add of +2.00. If a single-vision pair is ordered for computer use at half the Add power, what should the left lens be?
- A.+1.75 -0.50 x 070
- B.+2.25 -0.50 x 070✓ Answer
- C.+3.25 -0.50 x 070
- D.+2.25 -0.50 x 160
An intermediate single-vision lens is made by adding half the Add to the distance sphere, leaving the cylinder and axis untouched. Half of +2.00 is +1.00, and +1.25 plus +1.00 equals +2.25, so the lens is +2.25 -0.50 x 070. Using the full Add would over-plus the lens and pull the clear zone too close to the wearer.
Source: System for Ophthalmic Dispensing, intermediate and near single-vision prescriptionsReport a problem with this question
11. A right lens is a +4.00 D sphere, and the wearer's line of sight passes through a point 5 mm above the optical center of that lens. What prismatic effect does the wearer experience?
- A.0.80 prism diopters base down
- B.20.00 prism diopters base down
- C.2.00 prism diopters base up
- D.2.00 prism diopters base down✓ Answer
Prentice's rule gives the prism as the decentration in centimetres multiplied by the power, so 0.5 cm times 4.00 D equals 2.00 prism diopters. A plus lens behaves as two prisms joined base to base, so the base always lies toward the optical center as seen from the viewing point; looking above the optical center therefore produces base-down prism.
Source: Prentice's rule, standard ophthalmic optics; System for Ophthalmic Dispensing, induced prismatic effectReport a problem with this question
12. A left lens is a -5.00 D sphere and the wearer looks through a point 3 mm nasal to its optical center. What prismatic effect is induced?
- A.0.60 prism diopters base in
- B.1.50 prism diopters base out
- C.1.50 prism diopters base in✓ Answer
- D.15.00 prism diopters base in
Prentice's rule gives 0.3 cm times 5.00 D, which equals 1.50 prism diopters. A minus lens behaves as two prisms joined apex to apex, so the base lies away from the optical center toward the edge on the side being viewed through; a nasal viewing point therefore gives base-in prism, the opposite of what the same displacement would give in a plus lens.
Source: Prentice's rule, standard ophthalmic optics; Ophthalmic Lenses and Dispensing, prismatic effect of plus and minus lensesReport a problem with this question
13. A prescriber orders 1.50 prism diopters to be obtained by decentering a -2.50 D sphere. By how many millimetres must the optical center be displaced from the wearer's line of sight?
- A.15 mm
- B.6 mm✓ Answer
- C.3.75 mm
- D.1.7 mm
Prentice's rule rearranges to decentration in centimetres equals prism divided by power, so 1.50 divided by 2.50 gives 0.6 cm, which is 6 mm. Multiplying the two values instead of dividing produces 3.75, and dividing power by prism produces 1.7, both of which are the usual arithmetic traps.
Source: Prentice's rule, standard ophthalmic optics; System for Ophthalmic Dispensing, obtaining prescribed prism by decentrationReport a problem with this question
14. A finished pair of lenses is verified as 1.25 prism diopters base up in the right lens and 0.75 prism diopters base down in the left lens. What total vertical prismatic effect does the wearer experience?
- A.1.25 prism diopters
- B.1.00 prism diopters
- C.2.00 prism diopters✓ Answer
- D.0.50 prism diopters
Vertical prism in front of the two eyes compounds when the bases are opposite, because base up in one eye and base down in the other displace the two retinal images in the same direction relative to each other. The amounts are therefore added: 1.25 plus 0.75 equals 2.00 prism diopters. Only when both eyes carry the same vertical base do the amounts subtract.
Source: System for Ophthalmic Dispensing, compounding and cancelling prism between the two eyesReport a problem with this question
15. A single lens must be made carrying 4.00 prism diopters base in together with 3.00 prism diopters base up. What is the magnitude and general direction of the single resultant prism?
- A.5.00 prism diopters with its base down and out
- B.7.00 prism diopters with its base up and in
- C.5.00 prism diopters with its base up and in✓ Answer
- D.1.00 prism diopter with its base in
A horizontal and a vertical prism in the same lens act at right angles to each other, so they combine as the hypotenuse of a right triangle rather than by simple addition: the square root of 4 squared plus 3 squared is 5.00 prism diopters. The resultant base lies in the quadrant between the two component bases, so it is directed up and in.
Source: Ophthalmic Lenses and Dispensing, compounding and resolving prismReport a problem with this question
16. A frame has an eyesize (A measurement) of 52 mm and a distance between lenses of 18 mm, and the wearer's distance pupillary distance is 62 mm. How much must each lens be decentred, and in which direction?
- A.2 mm in on each lens
- B.4 mm in on each lens✓ Answer
- C.4 mm out on each lens
- D.8 mm in on each lens
The frame pupillary distance is the A measurement plus the distance between lenses, so 52 plus 18 equals 70 mm. Total decentration is the frame pupillary distance minus the wearer's pupillary distance, which is 70 minus 62, or 8 mm; that total is shared between two lenses, so each lens is decentred 4 mm inward because the wearer's eyes sit closer together than the geometric centers of the frame.
Source: American Board of Opticianry Basic Certification exam content outline, centration and the boxing system; System for Ophthalmic Dispensing, decentrationReport a problem with this question
17. A frame has an eyesize (A measurement) of 54 mm, a distance between lenses of 16 mm and an effective diameter of 58 mm, and the wearer's distance pupillary distance is 62 mm. Allowing 2 mm for edging, what is the minimum blank size required?
- A.68 mm✓ Answer
- B.66 mm
- C.60 mm
- D.64 mm
Minimum blank size equals the effective diameter plus twice the decentration per lens plus the edging allowance. The frame pupillary distance is 54 plus 16, or 70 mm, so total decentration is 70 minus 62, which is 8 mm, giving 4 mm per lens. That yields 58 plus 8 plus 2, which equals 68 mm. Decentration counts twice because the blank must reach beyond the shape on the side the optical center is moved away from.
Source: System for Ophthalmic Dispensing, minimum blank size and the boxing systemReport a problem with this question
18. A -8.00 D spherical lens was prescribed at the vertex distance used during refraction, but the finished frame sits 4 mm closer to the eye than that. What power keeps the effective power at the eye unchanged?
- A.-8.00 D, because vertex distance does not affect the power a lens delivers
- B.-7.75 D, because a minus lens moved closer to the eye needs less minus✓ Answer
- C.-7.00 D, because the change equals the distance moved expressed in diopters
- D.-8.25 D, because a minus lens moved closer to the eye needs more minus
Compensated power is the original power divided by one minus the product of the distance moved in metres and the power. Here that is -8.00 divided by one minus the product of 0.004 and -8.00, which is -8.00 divided by 1.032, giving approximately -7.75 D. The underlying rule is that greater vertex distance makes a lens act more plus, so bringing a minus lens closer to the eye means less minus is required; compensation becomes clinically important only in higher powers.
Source: Ophthalmic Lenses and Dispensing, vertex distance and lens effectivityReport a problem with this question
19. A wearer has OD -1.00 DS and OS -4.00 DS, and the reading point lies 10 mm directly below both distance optical centers. What vertical prismatic imbalance exists between the two eyes at the reading point?
- A.5.00 prism diopters
- B.0.50 prism diopters
- C.2.50 prism diopters
- D.3.00 prism diopters✓ Answer
Prentice's rule is applied to each eye separately at the vertical meridian: 1.0 cm times 1.00 D gives 1.00 prism diopter for the right eye and 1.0 cm times 4.00 D gives 4.00 prism diopters for the left. Both are minus lenses viewed below the optical center, so both produce base-down prism, and prisms with the same base in front of the two eyes cancel, leaving 4.00 minus 1.00, or 3.00 prism diopters of imbalance.
Source: System for Ophthalmic Dispensing, vertical imbalance in anisometropia; Prentice's rule, standard ophthalmic opticsReport a problem with this question
20. A wearer has OD -6.00 DS and OS -2.50 DS and reads 8 mm below the distance optical centers. If conventional slab-off is used to relieve the vertical imbalance, on which lens is it ground and how much imbalance must it neutralize?
- A.On the right lens, neutralizing 6.8 prism diopters, because the two lenses' prismatic effects are added together
- B.On the left lens, neutralizing 2.8 prism diopters, because the less minus lens is always the one that is slabbed
- C.On the right lens, neutralizing 2.8 prism diopters, because it is the more minus lens and shows that much extra base-down effect✓ Answer
- D.On the left lens, neutralizing 1.4 prism diopters, because the imbalance is always shared equally between the two lenses
At 0.8 cm below the optical centers the right lens induces 0.8 times 6.00, or 4.8 prism diopters base down, and the left lens induces 0.8 times 2.50, or 2.0 prism diopters base down, leaving 2.8 prism diopters of imbalance. Conventional slab-off is bicentric grinding that adds base-up prism to the reading area, so it is placed on the more minus lens, here the right, to offset its greater base-down effect; a reverse slab adds base-down prism and is ground on the more plus lens instead.
Source: System for Ophthalmic Dispensing, slab-off and reverse slab-off for vertical imbalanceReport a problem with this question
21. A frame is adjusted with 10 degrees of pantoscopic tilt. Applying the dispensing rule of lowering the optical center 1 mm for every 2 degrees of tilt, where should the distance optical center be placed relative to the pupil center, and why?
- A.5 mm below the pupil center, so the line of sight stays perpendicular to the lens surface and no unwanted power or prism is induced✓ Answer
- B.10 mm below the pupil center, because the optical center must drop 1 mm for every degree of tilt
- C.Level with the pupil center, because pantoscopic tilt changes only the cosmetic appearance of the frame
- D.5 mm above the pupil center, because tilting the frame raises the wearer's visual axis on the lens
Ten degrees of tilt divided by 2 degrees per millimetre calls for a 5 mm drop of the optical center. Lowering it keeps the wearer's line of sight passing through the optical center and perpendicular to the lens surface; if the line of sight instead crosses a tilted lens obliquely, the lens acts as though extra sphere and cylinder had been added and unwanted prism appears.
Source: System for Ophthalmic Dispensing, pantoscopic tilt and optical center placementReport 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 →

1. A spectacle prescription reads OD +1.75 -0.75 x 180, OS +2.00 -0.50 x 010, Add +1.50 OU, with 2 prism diopters base down before the right eye. Which statement correctly describes what this prescription specifies?

2. A prescription is written in plus-cylinder form as +1.50 +2.25 x 065. Written in minus-cylinder form, the same lens is:

3. A prescription is written in minus-cylinder form as -5.00 -1.75 x 110. Expressed in plus-cylinder form, it becomes:

4. A distance prescription for the right eye reads +2.00 -3.50 x 090. After determining the power present in each principal meridian, how should this eye's astigmatism be classified?

5. A lens is found to have +1.00 D of power in the 030 meridian and -2.00 D of power in the 120 meridian. Written in minus-cylinder form, this lens is:

6. A prescription reads -1.25 -2.00 x 180. What total power is present in the vertical (090) meridian of that lens?

7. What is the spherical equivalent of the prescription -4.75 -2.50 x 015?

8. A right lens is made to the prescription plano +2.00 x 180. Using the standard relationship in which none of the cylinder is effective at the axis, half of it is effective 45 degrees away and all of it is effective 90 degrees away, what total power is present in the 045 meridian?

9. A wearer's distance prescription for the right eye is -3.75 -1.00 x 095 with an Add of +2.50. What single-vision near prescription should be ordered for that eye?

10. A wearer's left distance prescription is +1.25 -0.50 x 070 with an Add of +2.00. If a single-vision pair is ordered for computer use at half the Add power, what should the left lens be?

11. A right lens is a +4.00 D sphere, and the wearer's line of sight passes through a point 5 mm above the optical center of that lens. What prismatic effect does the wearer experience?

12. A left lens is a -5.00 D sphere and the wearer looks through a point 3 mm nasal to its optical center. What prismatic effect is induced?

13. A prescriber orders 1.50 prism diopters to be obtained by decentering a -2.50 D sphere. By how many millimetres must the optical center be displaced from the wearer's line of sight?

14. A finished pair of lenses is verified as 1.25 prism diopters base up in the right lens and 0.75 prism diopters base down in the left lens. What total vertical prismatic effect does the wearer experience?

15. A single lens must be made carrying 4.00 prism diopters base in together with 3.00 prism diopters base up. What is the magnitude and general direction of the single resultant prism?

16. A frame has an eyesize (A measurement) of 52 mm and a distance between lenses of 18 mm, and the wearer's distance pupillary distance is 62 mm. How much must each lens be decentred, and in which direction?

17. A frame has an eyesize (A measurement) of 54 mm, a distance between lenses of 16 mm and an effective diameter of 58 mm, and the wearer's distance pupillary distance is 62 mm. Allowing 2 mm for edging, what is the minimum blank size required?

18. A -8.00 D spherical lens was prescribed at the vertex distance used during refraction, but the finished frame sits 4 mm closer to the eye than that. What power keeps the effective power at the eye unchanged?

19. A wearer has OD -1.00 DS and OS -4.00 DS, and the reading point lies 10 mm directly below both distance optical centers. What vertical prismatic imbalance exists between the two eyes at the reading point?

20. A wearer has OD -6.00 DS and OS -2.50 DS and reads 8 mm below the distance optical centers. If conventional slab-off is used to relieve the vertical imbalance, on which lens is it ground and how much imbalance must it neutralize?

21. A frame is adjusted with 10 degrees of pantoscopic tilt. Applying the dispensing rule of lowering the optical center 1 mm for every 2 degrees of tilt, where should the distance optical center be placed relative to the pupil center, and why?