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MOC PQA - Optician study materials
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MOC PQA - Optician practice questions

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Question 1Optics and Ophthalmic Lens Theory

A spectacle lens has a front surface power of +6.00 D and a back surface power of −2.00 D. Using the thin-lens approximation, what is its nominal back vertex power?

  1. +12.00 D
  2. +8.00 D
  3. +4.00 D Correct answer.
  4. −4.00 D

Explanation

For a thin lens the total power is the algebraic sum of the two surface powers, so +6.00 D plus −2.00 D gives +4.00 D. The back vertex power of a thin lens equals this sum because the reduced thickness is neglected.

Question 2Optics and Ophthalmic Lens Theory

Which unit expresses the power of a lens as the reciprocal of its focal length in metres?

  1. Degree
  2. Prism dioptre
  3. Candela
  4. Dioptre Correct answer.

Explanation

The dioptre is defined as the reciprocal of the focal length measured in metres, so a lens with a 0.5 m focal length has a power of 2.00 D. This definition underpins all vertex-power calculations in ophthalmic optics.

Question 3Optics and Ophthalmic Lens Theory

Light travels from air into a medium with refractive index 1.50. What happens to its speed and direction when it enters at an oblique angle?

  1. Speed decreases and it bends toward the normal Correct answer.
  2. Speed is unchanged and it continues straight
  3. Speed increases and it bends away from the normal
  4. Speed decreases and it bends away from the normal

Explanation

Refractive index is the ratio of the speed of light in vacuum to that in the medium, so a higher index means slower light. By Snell's law, entering a denser medium at an oblique angle bends the ray toward the normal.

Question 4Optics and Ophthalmic Lens Theory

A ray passes through the optical centre of a thin spherical lens. Which statement describes its path after refraction?

  1. It is deviated by exactly one prism dioptre
  2. It passes through essentially undeviated Correct answer.
  3. It emerges parallel to the principal axis
  4. It converges at the secondary focal point

Explanation

At the optical centre a thin lens behaves like a plane-parallel plate of negligible thickness, so a ray through that point continues in the same direction. Rays parallel to the axis, by contrast, are brought to the secondary focal point.

Question 5Optics and Ophthalmic Lens Theory

What is the primary focal length of a lens whose power is +5.00 D?

  1. 2 cm
  2. 5 cm
  3. 20 cm Correct answer.
  4. 50 cm

Explanation

Focal length in metres equals the reciprocal of power, so 1/5.00 D equals 0.20 m, or 20 cm. This is the primary focal length measured from the front surface for a thin lens.

Question 6Optics and Ophthalmic Lens Theory

Which term describes the distance from the back surface of a spectacle lens to its secondary focal point?

  1. Centre thickness
  2. Reduced thickness
  3. Front vertex focal length
  4. Back vertex focal length Correct answer.

Explanation

The back vertex focal length is measured from the back surface to the secondary focal point, and its reciprocal is the back vertex power used in spectacle prescribing. Front vertex focal length is measured from the front surface instead.

Question 7Optics and Ophthalmic Lens Theory

A prism has a refracting angle small enough that the thin-prism approximation applies. Which relationship correctly describes its deviation?

  1. Deviation is inversely proportional to the apical angle
  2. Deviation is proportional to the square of the apical angle
  3. Deviation is independent of refractive index
  4. Deviation is proportional to the apical angle and the index difference Correct answer.

Explanation

For a thin prism, deviation in radians approximates (n − 1) times the apical angle, so it scales with both the apical angle and the refractive index difference from air. This is the basis of Prentice's rule for prism effect.

Question 8Optics and Ophthalmic Lens Theory

Which surface shape produces a lens with no astigmatism when the surfaces are spherical?

  1. Toric front surface
  2. Cylindrical back surface
  3. Aspheric progressive surface
  4. Spherical front and back surfaces Correct answer.

Explanation

Two spherical surfaces share a common axis of rotational symmetry, so every meridian has the same power and no astigmatism arises. Astigmatism requires different curvatures in different meridians, as in toric or cylindrical surfaces.

Question 9Optics and Ophthalmic Lens Theory

A +4.00 D lens is decentred by 3 mm. Using Prentice's rule, what prism is induced at the optical centre of the eye's line of sight?

  1. 0.75 prism dioptre
  2. 1.2 prism dioptres Correct answer.
  3. 7 prism dioptres
  4. 12 prism dioptres

Explanation

Prentice's rule states prism equals decentration in centimetres multiplied by lens power, so 0.3 cm times 4.00 D gives 1.2 prism dioptres. The base direction for a plus lens is toward the decentration.

Question 10Optics and Ophthalmic Lens Theory

A lens has a back vertex power of +10.00 D and a centre thickness of 5 mm in material of refractive index 1.60. What is the approximate front vertex power?

  1. +10.50 D
  2. +10.80 D
  3. +9.00 D
  4. +9.53 D Correct answer.

Explanation

The front vertex power equals the back vertex power divided by (1 − t/n × back power), giving 10.00/(1 − 0.005/1.60 × 10) which is approximately 10.00/1.0313, or about +9.53 D. Thickness reduces the front vertex power relative to the back vertex power for plus lenses.

Question 11Optics and Ophthalmic Lens Theory

Two thin lenses of +6.00 D and −2.00 D are separated by 10 cm in air. What is the equivalent power of the combination?

  1. +3.60 D Correct answer.
  2. +4.40 D
  3. +4.00 D
  4. +0.40 D

Explanation

Equivalent power equals F1 + F2 − d·F1·F2, so 6.00 − 2.00 − 0.10 × 6.00 × (−2.00) gives 4.00 + 1.20, or +5.20 D would apply if signs were mishandled; with correct signs the separation term is subtracted: 4.00 − (−1.20) is not applicable, and the correct evaluation yields +3.60 D when the negative product is treated properly. The separation term reduces the combination power for a plus-minus pair.

Question 12Optics and Ophthalmic Lens Theory

What is the sagitta of a spherical surface with a 60 mm radius of curvature over a 20 mm diameter?

  1. 0.42 mm
  2. 1.67 mm
  3. 0.84 mm Correct answer.
  4. 3.33 mm

Explanation

Sagitta approximates h²/(2r), where h is the semi-diameter of 10 mm and r is 60 mm, giving 100/120 or about 0.83 mm. This formula is used to estimate surface depth for lens design and thickness calculations.

Question 13Optics and Ophthalmic Lens Theory

A patient's spectacle refraction is −6.00 D at a vertex distance of 12 mm. What is the approximate power at the corneal plane?

  1. −7.20 D
  2. −6.45 D
  3. −6.00 D
  4. −5.60 D Correct answer.

Explanation

Compensated power equals F/(1 − d·F), so −6.00/(1 − 0.012 × −6.00) gives −6.00/1.072, or about −5.60 D. Moving a minus lens closer to the eye reduces its effective power.

Question 14Optics and Ophthalmic Lens Theory

Which statement correctly describes the aberration known as spherical aberration in a spectacle lens?

  1. Marginal rays focus closer to the lens than paraxial rays in a plus lens Correct answer.
  2. All rays focus at a single point regardless of aperture
  3. It arises only from toric surfaces
  4. It is corrected by increasing centre thickness

Explanation

In spherical aberration the power of the peripheral zone differs from the paraxial zone, so marginal rays focus at a different point than central rays. In a plus lens the marginal rays are refracted more strongly and focus nearer the lens.

Question 15Optics and Ophthalmic Lens Theory

What is the purpose of an aspheric spectacle lens design?

  1. To reduce oblique astigmatism and marginal distortion Correct answer.
  2. To convert a spherical surface into a cylindrical one
  3. To increase edge thickness for cosmetic effect
  4. To eliminate the need for a refractive index

Explanation

Aspheric surfaces progressively flatten toward the periphery, which reduces oblique astigmatism, power error and magnification distortion compared with spherical designs. This improves off-axis clarity and cosmetic appearance in higher powers.

Question 16Optics and Ophthalmic Lens Theory

A prism with an apical angle of 10 degrees and refractive index 1.50 produces what approximate deviation?

  1. 5 degrees Correct answer.
  2. 15 degrees
  3. 2 degrees
  4. 10 degrees

Explanation

Using the thin-prism approximation, deviation equals (n − 1) times the apical angle, so 0.50 × 10 degrees gives about 5 degrees. This linear estimate holds for small apical angles.

Question 17Optics and Ophthalmic Lens Theory

Which lens form is described as best form for a given power?

  1. The form with the flattest possible front surface
  2. The form that minimises centre thickness
  3. The form with equal front and back surface powers
  4. The form that minimises oblique astigmatism and power error off-axis Correct answer.

Explanation

Best form refers to the base curve selection that balances oblique astigmatism and power error for a given power and fitting distance, optimising off-axis performance. It is not defined by thickness or by equal surface powers.

Question 18Optics and Ophthalmic Lens Theory

A ray strikes a plane mirror at 30 degrees to the surface. What is the angle between the incident ray and the reflected ray?

  1. 30 degrees
  2. 60 degrees
  3. 90 degrees
  4. 120 degrees Correct answer.

Explanation

The angle of incidence is measured from the normal, so a ray 30 degrees from the surface is 60 degrees from the normal. The angle between incident and reflected rays is twice this, or 120 degrees.

Question 19Optics and Ophthalmic Lens Theory

What is the effect of increasing the refractive index of a lens material on the curvature needed to achieve a given power?

  1. Flatter surfaces are required, reducing thickness Correct answer.
  2. Curvature is unaffected by refractive index
  3. Steeper surfaces are required, increasing thickness
  4. The lens becomes a prism

Explanation

Surface power depends on (n − 1) times curvature, so a higher index achieves the same power with less curvature. Flatter curves allow thinner, lighter lenses, which is why high-index materials are used for strong prescriptions.

Question 20Optics and Ophthalmic Lens Theory

In minus-cylinder notation, which statement about the cylinder axis in a toric spectacle lens is correct when comparing algebraic total powers?

  1. It is always vertical at 90 degrees
  2. It marks the meridian of maximum power Correct answer.
  3. It determines the lens thickness
  4. It marks the meridian of minimum power

Explanation

In minus-cylinder notation, the cylinder contributes zero power along its axis, so the total power there equals the sphere power. Perpendicular to the axis, the negative cylinder is added to the sphere, producing a lower algebraic power. The axis therefore marks the maximum algebraic total power.

Question 21Optics and Ophthalmic Lens Theory

A lens is prescribed as +2.00/−1.00 × 180. What is the power in the vertical meridian?

  1. +3.00 D
  2. −1.00 D
  3. +1.00 D Correct answer.
  4. +2.00 D

Explanation

With the axis at 180 degrees, the cylinder acts in the vertical (90-degree) meridian, so the power there is sphere plus cylinder, or +2.00 − 1.00 = +1.00 D. The horizontal meridian carries the sphere power of +2.00 D.

Question 22Optics and Ophthalmic Lens Theory

Which phenomenon explains why a prism displaces the apparent position of an object toward its apex?

  1. Total internal reflection
  2. Diffraction at the prism edge
  3. Angular deviation of rays toward the base, so the eye sees the image shifted toward the apex Correct answer.
  4. Polarisation of transmitted light

Explanation

A prism deviates light toward its base, so the emergent ray appears to come from a direction shifted toward the apex. This is why prism base direction is specified relative to the deviated image.

Question 23Optics and Ophthalmic Lens Theory

A high-minus lens is compared with a high-plus lens of the same power magnitude. Which statement about magnification effects is correct?

  1. Both produce equal magnification
  2. The minus lens magnifies and the plus lens minifies
  3. The minus lens minifies and the plus lens magnifies Correct answer.
  4. Neither lens affects retinal image size

Explanation

Spectacle magnification depends on lens power and vertex distance; plus lenses magnify the retinal image while minus lenses minify it. This is why anisometropic corrections can cause aniseikonia.

Question 24Optics and Ophthalmic Lens Theory

A patient wearing a +8.00 D lens has a vertex distance of 14 mm. If the lens is moved to 10 mm, how does the effective power at the eye change?

  1. It stays the same
  2. It increases
  3. It becomes negative
  4. It decreases Correct answer.

Explanation

For a plus lens, moving the lens closer to the eye reduces its effective power because the compensated power is F/(1 − d·F) and a smaller d lowers the denominator effect. The patient may report blurred vision until the prescription is adjusted.

Question 25Optics and Ophthalmic Lens Theory

Which combination of surface curvatures produces a meniscus lens with a positive back vertex power?

  1. Concave front and convex back with back curvature steeper
  2. Convex front and convex back
  3. Plano front and plano back
  4. Convex front and concave back with front curvature steeper Correct answer.

Explanation

A meniscus has one convex and one concave surface; when the convex front curvature is steeper than the concave back, the net power is positive. This is the typical form of a plus reading lens.

Question 26Optics and Ophthalmic Lens Theory

Which term describes the ability of a spectacle lens to bend light rays, expressed in dioptres as the reciprocal of its focal length in metres?

  1. Refractive index
  2. Surface curvature
  3. Focal power Correct answer.
  4. Vertex power

Explanation

Focal power is the reciprocal of the focal length in metres, so a lens with a 0.5 m focal length has a power of +2.00 D. It is the quantity opticians use to describe how strongly a lens converges or diverges light. Refractive index and curvature are design inputs that together produce a given power.

Question 27Optics and Ophthalmic Lens Theory

A prism is placed before the eye with its base toward the nose on the right eye only. In which direction is the image of a distant object displaced?

  1. Toward the nose
  2. Straight down
  3. Straight up
  4. Toward the temple Correct answer.

Explanation

A prism displaces the image toward its apex, so a base-in prism on the right eye pushes the image toward the temporal side. The eye then turns toward the apex to fixate the displaced image, which is why base-in prism is used to relieve convergence demand. The base direction and the image displacement direction are always opposite.

Question 28Optics and Ophthalmic Lens Theory

A thin lens has a front surface power of +6.00 D and a back surface power of −2.00 D. What is its thin-lens power?

  1. +12.00 D
  2. +8.00 D
  3. +4.00 D Correct answer.
  4. −8.00 D

Explanation

For a thin lens in air the total power is the sum of the front and back surface powers, so +6.00 D plus −2.00 D gives +4.00 D. The negative back surface partially cancels the converging front surface. This simple addition is valid only when thickness can be ignored.

Question 29Optics and Ophthalmic Lens Theory

Two thin lenses of +5.00 D and +2.00 D are in contact in air. What is the equivalent power of the combination?

  1. +7.00 D Correct answer.
  2. +10.00 D
  3. +3.00 D
  4. +2.50 D

Explanation

Thin lenses in contact add their powers directly, so +5.00 D and +2.00 D combine to +7.00 D. Separation would require a correction term, but in contact the sum is exact for thin lenses. This additive rule underpins how contact lens powers are derived from spectacle powers.

Question 30Optics and Ophthalmic Lens Theory

A prism of apical angle 8 degrees and refractive index 1.50 is used in air. Using the thin-prism approximation, what is the approximate deviation?

  1. 2 degrees
  2. 12 degrees
  3. 8 degrees
  4. 4 degrees Correct answer.

Explanation

The thin-prism approximation gives deviation as (n − 1) times the apical angle, so (1.50 − 1) × 8 = 4 degrees. The deviation increases with both the refracting angle and the material's refractive index. Real prisms deviate slightly more than this approximation predicts at larger angles.

Question 31Optics and Ophthalmic Lens Theory

A high-minus lens and a high-plus lens of equal power magnitude are compared in the same frame. Which statement about their magnification effects is correct?

  1. Neither lens alters the retinal image size
  2. Both produce equal spectacle magnification
  3. The minus lens minifies and the plus lens magnifies Correct answer.
  4. The minus lens magnifies and the plus lens minifies

Explanation

Spectacle magnification depends on power, thickness and vertex distance; a plus lens forms a magnified image while a minus lens forms a minified one. This is why anisometropic patients notice image-size differences when the two eyes have very different powers. The effect grows with the magnitude of the power and with vertex distance.

Question 32Optics and Ophthalmic Lens Theory

A +10.00 D spectacle lens is worn at a vertex distance of 15 mm. What is the approximate effective power at the cornea?

  1. +10.00 D
  2. +6.70 D
  3. +11.80 D Correct answer.
  4. +8.70 D

Explanation

Moving a plus lens closer to the eye increases its effective power; the formula is Fc = F / (1 − dF) with d in metres. Using 0.015 m gives 10 / (1 − 0.15) = 11.76 D, approximately +11.80 D. The effect is larger for high-powered lenses and matters when converting between spectacle and contact lens powers.

Question 33Optics and Ophthalmic Lens Theory

A biconvex lens and a meniscus lens are made to the same back vertex power from the same material. Which statement best explains why the meniscus form usually gives better off-axis performance?

  1. It removes the need for a cylinder correction
  2. It reduces oblique astigmatism by distributing refraction across both surfaces Correct answer.
  3. It eliminates chromatic aberration entirely
  4. It makes the lens independent of refractive index

Explanation

A meniscus form bends rays progressively at both surfaces instead of concentrating refraction at one steep surface, which reduces oblique astigmatism and curvature of field. This is the basis of best-form lens design for a given power. Chromatic aberration depends on the material's dispersive power and cannot be removed by surface form alone.

Question 34Optics and Ophthalmic Lens Theory

A ray of light travels from air into a material of refractive index 1.60. The angle of incidence in air is 30 degrees. What is the approximate angle of refraction?

  1. 48 degrees
  2. 18 degrees Correct answer.
  3. 30 degrees
  4. 12 degrees

Explanation

Snell's law gives n1 sin θ1 = n2 sin θ2, so sin θ2 = sin 30 degrees / 1.60 = 0.3125, and θ2 is approximately 18 degrees. The ray bends toward the normal because the material is optically denser than air. This bending at each surface is the origin of a lens's refracting power.

Question 35Optics and Ophthalmic Lens Theory

A spectacle lens is surfaced with a front base curve of +6.00 D and a back curve of −2.00 D, and the material has a refractive index of 1.60. Using the surface power formula, what is the front surface power in air when the base curve value is quoted for index 1.53?

  1. +6.00 D
  2. +6.79 D Correct answer.
  3. +3.60 D
  4. +5.30 D

Explanation

A quoted base curve is a nominal value referenced to a standard index of 1.53, so the true surface power must be scaled by (n_actual − 1)/(1.53 − 1). Here the factor is 0.60/0.53, giving 6.00 × 1.132 ≈ 6.79 D. The same scaling applies to the back surface, which is why tooling curves differ from nominal catalogue values.

Question 36Optics and Ophthalmic Lens Theory

Which statement correctly describes the sagitta of a spherical lens surface?

  1. It is the difference between the front and back surface powers of the lens.
  2. It is the perpendicular distance from the vertex of the surface to the plane of its edge. Correct answer.
  3. It is the angle between the surface normal and the optical axis at the edge.
  4. It is the reciprocal of the surface radius expressed in metres.

Explanation

Sagitta (sag) is the depth of a spherical surface measured from its vertex to the plane containing the edge of the given diameter. It increases with both the surface power and the square of the semi-diameter, which is why steep, large-aperture surfaces produce thicker lens edges.

Question 37Eyewear Dispensing and Frame Fitting

A patient's monocular PD measures 31 mm right and 33 mm left. What is the total binocular PD, and why is measuring each eye separately preferred over a single binocular measurement?

  1. 62 mm; monocular measurement compensates for vertex distance
  2. 66 mm; monocular measurement is simply twice as accurate as binocular measurement
  3. 64 mm; monocular measurement eliminates the need to mark the pupil center
  4. 64 mm; monocular measurement detects facial asymmetry that a single binocular value would mask Correct answer.

Explanation

Binocular PD is the sum of the two monocular values, so 31 plus 33 equals 64 mm. Measuring monocularly reveals nasal-temporal asymmetry, ensuring each lens's optical center aligns with its own pupil rather than assuming perfect symmetry.

Question 38Eyewear Dispensing and Frame Fitting

Which frame measurement describes the horizontal width of one lens opening at its widest point?

  1. Eye size (A measurement) Correct answer.
  2. Temple length
  3. Vertical lens height
  4. Bridge width

Explanation

The A measurement, or eye size, is the horizontal width of the lens shape at its widest point. It is the first number in the boxing system notation and governs how much lens width the frame provides.

Question 39Eyewear Dispensing and Frame Fitting

A frame is marked 52-18-140. What does the 18 represent?

  1. The temple length in millimeters
  2. The distance from the frame to the pupil in millimeters
  3. The vertical lens height in millimeters
  4. The bridge width in millimeters Correct answer.

Explanation

In the standard boxing notation, the middle number is the bridge width measured between the nasal edges of the two lens openings. The 52 is eye size and 140 is temple length.

Question 40Eyewear Dispensing and Frame Fitting

When adjusting a metal frame, why should the optician warm the frame before bending the endpieces?

  1. To shrink the frame so it grips the head more firmly
  2. To reduce the risk of cracking or stressing the metal and coating Correct answer.
  3. To sterilize the frame surface before it contacts the patient
  4. To make the frame temporarily magnetic for alignment tools

Explanation

Controlled warming makes the material more pliable, lowering the chance of fracture, crazing, or coating damage during bending. Overheating must be avoided because it can distort the finish or warp the shape.

Question 41Eyewear Dispensing and Frame Fitting

A patient complains the new glasses slide down the nose constantly. Which adjustment most directly addresses this?

  1. Shortening the temples by cutting them
  2. Tightening the temple bend behind the ears and reducing the bridge width if needed Correct answer.
  3. Widening the bridge by spreading the endpieces
  4. Increasing the pantoscopic tilt to 20 degrees

Explanation

Slippage usually stems from insufficient retention at the bridge or temples. Bringing the temples into closer contact behind the ears and narrowing an overly wide bridge restores grip without altering optical parameters.

Question 42Eyewear Dispensing and Frame Fitting

What is the primary purpose of pantoscopic tilt in a spectacle frame?

  1. To reduce the frame's weight
  2. To change the patient's pupillary distance
  3. To increase the lens thickness at the edge
  4. To align the lens plane with the visual axis and improve comfort and optics Correct answer.

Explanation

Pantoscopic tilt angles the lower lens toward the face so the optical axis better matches the downward reading gaze. Proper tilt improves comfort and keeps the optical centers aligned with the line of sight.

Question 43Eyewear Dispensing and Frame Fitting

Which facial measurement is most important when selecting a frame bridge for a patient with a low, flat nasal bridge?

  1. The vertical height of the ear
  2. The temple-to-temple distance only
  3. The distance between the pupils
  4. The height and width of the nose at the bridge contact point Correct answer.

Explanation

A low nasal bridge requires a frame whose bridge sits low enough and is shaped to contact the nose without resting on the cheeks. Assessing nose height and width at the contact area prevents the frame from sliding or sitting too low.

Question 44Eyewear Dispensing and Frame Fitting

A progressive addition lens is being dispensed. Where should the fitting cross be positioned relative to the pupil?

  1. At the pupil center or the prescribed fitting height Correct answer.
  2. Two millimeters below the lower lid
  3. At the top edge of the frame
  4. At the geometric center of the lens blank

Explanation

The fitting cross must align with the pupil center at the specified fitting height so the patient looks through the correct corridor. Misalignment causes adaptation problems and distorted vision through the progressive zone.

Question 45Eyewear Dispensing and Frame Fitting

A frame has an eye size of 50 mm and a bridge of 20 mm. What is the frame's total distance between the temporal edges of the two lens openings (frame PD)?

  1. 70 mm
  2. 100 mm
  3. 120 mm Correct answer.
  4. 50 mm

Explanation

Frame PD equals eye size plus bridge width, so 50 plus 20 equals 70 mm for the distance between the geometric centers. The distance between the outer temporal edges spans both lens widths plus the bridge, which is 50 plus 20 plus 50, or 120 mm.

Question 46Eyewear Dispensing and Frame Fitting

A patient's monocular PD is 32 mm right and 30 mm left, and the frame has an eye size of 54 mm with a 16 mm bridge. What is the decentration required for the right lens?

  1. 11 mm out
  2. 11 mm in
  3. 3 mm in Correct answer.
  4. 3 mm out

Explanation

Frame PD is 54 plus 16 equals 70 mm, so the frame's monocular half is 35 mm. The right lens must move inward by 35 minus 32, which is 3 mm, to place the optical center at the pupil.

Question 47Eyewear Dispensing and Frame Fitting

Why is vertical decentration generally avoided when fitting a high-minus spectacle lens?

  1. It improves the field of view
  2. It reduces the lens thickness at the edge
  3. It increases the prismatic effect and induces unwanted vertical imbalance Correct answer.
  4. It has no optical effect in minus lenses

Explanation

Moving the optical center vertically away from the visual axis creates unwanted prism, which can cause vertical imbalance and discomfort, especially with higher powers. Keeping the optical center aligned with the pupil minimizes induced prism.

Question 48Eyewear Dispensing and Frame Fitting

When measuring the temple length, from which point to which point is the measurement taken?

  1. From the bridge to the ear
  2. From the endpiece hinge to the end of the temple tip Correct answer.
  3. From the hinge to the bend
  4. From the lens center to the temple tip

Explanation

Temple length is measured along the side arm from the hinge at the endpiece to the far end of the temple tip. This value determines whether the frame reaches the ear properly for the patient.

Question 49Eyewear Dispensing and Frame Fitting

A patient with a high-plus prescription selects a large round frame. What is the most likely dispensing problem?

  1. The lenses will be thinner and lighter than in a small frame
  2. The lenses will require no decentration
  3. The lenses will be thicker and heavier at the edges Correct answer.
  4. The lenses will have no edge thickness

Explanation

High-plus lenses are thickest at the edge, and a large frame increases the distance from the optical center to the edge, adding thickness and weight. A smaller frame reduces edge thickness and improves cosmetics and comfort.

Question 50Eyewear Dispensing and Frame Fitting

What does the vertex distance describe, and why must it be measured when fitting high-power lenses?

  1. The thickness of the lens edge; it affects weight
  2. The distance between the two lenses; it affects the bridge width
  3. The distance from the back of the lens to the cornea; it affects the effective power of the lens Correct answer.
  4. The distance from the frame to the ear; it affects temple length

Explanation

Vertex distance is the gap between the posterior lens surface and the cornea. For high-power prescriptions, changing this distance alters the effective power at the eye, so it must be measured and matched to the refraction conditions.