NAO VOLUME 3 PRACTICE EXAM QUESTIONS AND CORRECT ANSWERS (VERIFIED ANSWERS) PLUS
RATIONALES 2026 Q&A |LATEST EXAM UPDATE 2026/2027
SECTION ONE: QUESTIONS 1–100
1. The correct transposed RX for -1.00 -0.50 x 037 is?
A. -1.50 -0.50 x 127
B. -1.50 +0.50 x 127
C. -0.50 -1.00 x 037
D. -0.50 +1.00 x 037
🟢 B. -1.50 +0.50 x 127
🔴 RATIONALE: Transposition involves combining the sphere and cylinder algebraically and changing the
cylinder sign. The new sphere is -1.00 + (-0.50) = -1.50. The new cylinder is +0.50. The new axis is 90 + 37 =
127 .
2. The correct transposition for +2.00 -2.00 x 060 is?
A. Plano -2.00 x 150
B. Plano +2.00 x 150
C. +4.00 -2.00 x 150
D. +2.00 +2.00 x 150
🟢 B. Plano +2.00 x 150
🔴 RATIONALE: The new sphere is +2.00 + (-2.00) = Plano. The new cylinder is +2.00. The new axis is 90 + 60 =
150 .
3. An uncut lens must be at least _______ mm in diameter to allow the edging of a 54 mm round lens
decentered 3 mm in.
,A. 57 mm
B. 60 mm
C. 63 mm
D. 54 mm
🟢 B. 60 mm
🔴 RATIONALE: The minimum blank size (MBS) formula is MBS = ED + 2 * decentration. With a 54 mm eye size
(effective diameter) and 3 mm decentration, MBS = 54 + 2(3) = 60 mm .
4. A +3.00 sphere lens has a front base curve of +6.25. What is the back surface curve?
A. +3.25
B. -3.25
C. -6.25
D. +9.25
🟢 B. -3.25
🔴 RATIONALE: The total power is the sum of the front and back curves. To achieve +3.00, the back curve must
be +3.00 - (+6.25) = -3.25 .
5. A -2.00 -1.00 x 045 lens has a front base curve of +6.25. What are the back surface curves?
A. -4.25 and -1.00
B. -6.25 and -8.25
C. -8.25 and -1.00
D. -6.25 and -9.25
🟢 C. -8.25 and -1.00
🔴 RATIONALE: The back surface must create the necessary powers. For the sphere meridian, -2.00 - (+6.25) =
-8.25. For the cylinder meridian, -3.00 - (+6.25) = -9.25, but the cylinder power is -1.00, so the two curves are
-8.25 and -9.25. The cylinder is -1.00 .
,6. Moving the optical center of a plus lens inward from the correct PD results in _____.
A. Base out prism
B. Base in prism
C. Base up prism
D. Base down prism
🟢 B. Base in prism
🔴 RATIONALE: For a plus lens, prism base is in the same direction as the optical center movement. Moving the
optical center inward results in base in prism .
7. Where will the optical center appear in a lensometer when neutralizing a left lens for base in prism?
A. To the right of the target center
B. To the left of the target center
C. Above the target center
D. Below the target center
🟢 B. To the left of the target center
🔴 RATIONALE: Prism power in a lensometer is measured by the displacement of the target from the center.
Base in prism on a left lens will cause the optical center to appear displaced to the left .
8. Glazing is a term used to indicate _____.
A. A tinting process for lenses
B. The process of mounting lenses into frames
C. A defect that occurs when an anti-reflective lens is tinted
D. The application of an anti-reflective coating
🟢 C. A defect that occurs when an anti-reflective lens is tinted
🔴 RATIONALE: Glazing is a specific defect characterized by a grainy or cloudy appearance that can occur when
an anti-reflective lens is tinted .
, 9. Frequently, bicentric grinding (slab-off) is used as a means of obtaining prism in the reading field to
correct induced prism imbalance. The usual correction during the fabrication process is achieved by grinding
prism in the following direction:
A. Down
B. Up
C. In
D. Out
🟢 B. Up
🔴 RATIONALE: Slab-off grinding typically adds base up prism to the lens with less plus power in the vertical
meridian to balance the prismatic effect at the reading level .
10. A patient stops in to see you for adjustment of his glasses. After careful inspection you find the only
thing wrong that there is too much pantoscopic tilt. To make the necessary adjustment you will:
A. Angle the temples upward from the end pieces
B. Angle the temples downward from the end pieces
C. Widen the bridge
D. Adjust the nose pads down
🟢 A. Angle the temples upward from the end pieces
🔴 RATIONALE: Pantoscopic tilt is adjusted by changing the angle of the temples relative to the frame front.
Raising the temples at the end pieces decreases the pantoscopic angle .
11. If the RX is +2.00 -4.00 x 135, it would be necessary to decenter the lens _____ to create 1.00 prism base in.
A. 2 mm
B. 4 mm
C. 8 mm
D. Not possible
RATIONALES 2026 Q&A |LATEST EXAM UPDATE 2026/2027
SECTION ONE: QUESTIONS 1–100
1. The correct transposed RX for -1.00 -0.50 x 037 is?
A. -1.50 -0.50 x 127
B. -1.50 +0.50 x 127
C. -0.50 -1.00 x 037
D. -0.50 +1.00 x 037
🟢 B. -1.50 +0.50 x 127
🔴 RATIONALE: Transposition involves combining the sphere and cylinder algebraically and changing the
cylinder sign. The new sphere is -1.00 + (-0.50) = -1.50. The new cylinder is +0.50. The new axis is 90 + 37 =
127 .
2. The correct transposition for +2.00 -2.00 x 060 is?
A. Plano -2.00 x 150
B. Plano +2.00 x 150
C. +4.00 -2.00 x 150
D. +2.00 +2.00 x 150
🟢 B. Plano +2.00 x 150
🔴 RATIONALE: The new sphere is +2.00 + (-2.00) = Plano. The new cylinder is +2.00. The new axis is 90 + 60 =
150 .
3. An uncut lens must be at least _______ mm in diameter to allow the edging of a 54 mm round lens
decentered 3 mm in.
,A. 57 mm
B. 60 mm
C. 63 mm
D. 54 mm
🟢 B. 60 mm
🔴 RATIONALE: The minimum blank size (MBS) formula is MBS = ED + 2 * decentration. With a 54 mm eye size
(effective diameter) and 3 mm decentration, MBS = 54 + 2(3) = 60 mm .
4. A +3.00 sphere lens has a front base curve of +6.25. What is the back surface curve?
A. +3.25
B. -3.25
C. -6.25
D. +9.25
🟢 B. -3.25
🔴 RATIONALE: The total power is the sum of the front and back curves. To achieve +3.00, the back curve must
be +3.00 - (+6.25) = -3.25 .
5. A -2.00 -1.00 x 045 lens has a front base curve of +6.25. What are the back surface curves?
A. -4.25 and -1.00
B. -6.25 and -8.25
C. -8.25 and -1.00
D. -6.25 and -9.25
🟢 C. -8.25 and -1.00
🔴 RATIONALE: The back surface must create the necessary powers. For the sphere meridian, -2.00 - (+6.25) =
-8.25. For the cylinder meridian, -3.00 - (+6.25) = -9.25, but the cylinder power is -1.00, so the two curves are
-8.25 and -9.25. The cylinder is -1.00 .
,6. Moving the optical center of a plus lens inward from the correct PD results in _____.
A. Base out prism
B. Base in prism
C. Base up prism
D. Base down prism
🟢 B. Base in prism
🔴 RATIONALE: For a plus lens, prism base is in the same direction as the optical center movement. Moving the
optical center inward results in base in prism .
7. Where will the optical center appear in a lensometer when neutralizing a left lens for base in prism?
A. To the right of the target center
B. To the left of the target center
C. Above the target center
D. Below the target center
🟢 B. To the left of the target center
🔴 RATIONALE: Prism power in a lensometer is measured by the displacement of the target from the center.
Base in prism on a left lens will cause the optical center to appear displaced to the left .
8. Glazing is a term used to indicate _____.
A. A tinting process for lenses
B. The process of mounting lenses into frames
C. A defect that occurs when an anti-reflective lens is tinted
D. The application of an anti-reflective coating
🟢 C. A defect that occurs when an anti-reflective lens is tinted
🔴 RATIONALE: Glazing is a specific defect characterized by a grainy or cloudy appearance that can occur when
an anti-reflective lens is tinted .
, 9. Frequently, bicentric grinding (slab-off) is used as a means of obtaining prism in the reading field to
correct induced prism imbalance. The usual correction during the fabrication process is achieved by grinding
prism in the following direction:
A. Down
B. Up
C. In
D. Out
🟢 B. Up
🔴 RATIONALE: Slab-off grinding typically adds base up prism to the lens with less plus power in the vertical
meridian to balance the prismatic effect at the reading level .
10. A patient stops in to see you for adjustment of his glasses. After careful inspection you find the only
thing wrong that there is too much pantoscopic tilt. To make the necessary adjustment you will:
A. Angle the temples upward from the end pieces
B. Angle the temples downward from the end pieces
C. Widen the bridge
D. Adjust the nose pads down
🟢 A. Angle the temples upward from the end pieces
🔴 RATIONALE: Pantoscopic tilt is adjusted by changing the angle of the temples relative to the frame front.
Raising the temples at the end pieces decreases the pantoscopic angle .
11. If the RX is +2.00 -4.00 x 135, it would be necessary to decenter the lens _____ to create 1.00 prism base in.
A. 2 mm
B. 4 mm
C. 8 mm
D. Not possible
