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2026 Statistics
Key Facts: OLE Exam
8
written examination subjects covering the full scope of optometric practice
PRC OLE Blueprint / RA 8050
75%
minimum general weighted average required to pass the OLE
RA 8050 / PRC OLE Passing Standard
70/30
written-to-practical weighting ratio for the general weighted average
PRC OLE Examination Structure
50%
minimum rating allowed in any written subject; falling below this in any subject = conditional failure
PRC OLE Passing Standard
O.D.
Doctor of Optometry degree required for OLE eligibility under RA 8050
RA 8050 Section 14
1995
year Republic Act No. 8050 (Revised Optometry Law) was enacted
RA 8050
The OLE is the Philippine PRC examination for Registered Optometrist (R.Opt.) licensure under RA 8050. It has a written component (8 subjects, 70% weight) and a clinical skills component (5 areas, 30% weight), with a 75% general weighted average passing score and no written subject below 50%. The O.D. degree from a recognized institution is the minimum academic requirement.
Sample OLE Practice Questions
Try these sample questions to test your OLE exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A ray of light traveling from water (n = 1.33) into air (n = 1.0) strikes the interface at an angle of incidence of 40°. What is the approximate angle of refraction?
A.58.5°
B.25.4°
C.40°
D.90°
Explanation: Snell's law states n₁ sin θ₁ = n₂ sin θ₂. Using 1.33 × sin 40° = 1.0 × sin θ₂ gives sin θ₂ = 1.33 × 0.6428 ≈ 0.855, so θ₂ ≈ 58.7°. When light passes from a denser (higher-index) medium into a rarer one, the refracted ray bends away from the normal, so the angle of refraction is larger than the angle of incidence.
2The critical angle for glass with a refractive index of 1.52 relative to air is approximately:
A.41.1°
B.33.2°
C.56.3°
D.48.8°
Explanation: The critical angle θc = arcsin(n₂/n₁) = arcsin(1/1.52) = arcsin(0.6579) ≈ 41.1°. At angles of incidence greater than this value, total internal reflection occurs and no light is transmitted to the rarer medium. This principle underlies optical fiber transmission and prismatic instruments in optics.
3A thin convex lens of power +5.00 D forms an image of an object placed 50 cm in front of it. Using the vergence (thin lens) equation, where is the image formed?
A.33.3 cm behind the lens
B.20 cm behind the lens
C.10 cm in front of the lens
D.Infinity
Explanation: Work in vergences. The object vergence is U = −1/0.50 m = −2.00 D. The image vergence is V = P + U = +5.00 D + (−2.00 D) = +3.00 D. The image distance is 1/V = 1/3.00 = 0.333 m = 33.3 cm behind the lens, forming a real, inverted image. Because the object (50 cm) lies beyond the focal length (20 cm for a +5.00 D lens), a real image forms on the far side of the lens.
4Which type of aberration causes the image of a point object located off the optical axis to appear as a comet-shaped blur?
A.Coma
B.Spherical aberration
C.Astigmatism
D.Distortion
Explanation: Coma is an off-axis monochromatic aberration that causes rays from oblique pencils to focus at different points depending on the zone of the lens through which they pass, producing a comet-like flare. It is the most objectionable of the five Seidel aberrations for ophthalmic purposes.
5The wavelength of light in a medium with refractive index 1.5 when the wavelength in vacuum is 600 nm is:
A.400 nm
B.600 nm
C.900 nm
D.300 nm
Explanation: The wavelength in a medium is λmedium = λvacuum / n = 600 / 1.5 = 400 nm. Light slows down in denser media; its frequency remains unchanged while wavelength decreases proportionally. This relationship underlies optical path length calculations important in interferometry and coherence optics.
6Which type of chromatic aberration results in different-sized images at different wavelengths even when all wavelengths are in focus at the same plane?
A.Lateral (transverse) chromatic aberration
B.Longitudinal (axial) chromatic aberration
C.Spherical aberration
D.Field curvature
Explanation: Lateral (transverse) chromatic aberration causes images of different wavelengths to be the same size in the axial direction but differ in their lateral magnification, resulting in colored fringes at the edges of images. It is distinct from longitudinal chromatic aberration, which places different wavelengths at different axial focal points.
7In a prism, the minimum deviation angle occurs when the ray inside the prism travels:
A.Parallel to the base of the prism
B.Perpendicular to the first refracting surface
C.Along the apex angle bisector from outside
D.Parallel to the apex
Explanation: At minimum deviation, the refracted ray inside the prism is parallel to the base, and the angles of incidence and emergence are equal. At this condition, the relationship n = sin[(A + δmin)/2] / sin(A/2) applies, where A is the apex angle and δmin is the minimum deviation angle.
8Accommodation in the human eye is primarily produced by:
A.Increase in the curvature of the crystalline lens due to ciliary muscle contraction
B.Elongation of the eyeball by extraocular muscle tension
C.Pupillary dilation to allow more light to enter
D.Relaxation of the ciliary muscle increasing zonular tension
Explanation: Accommodation is the process by which the eye increases its optical power to focus on near objects. Contraction of the ciliary muscle relaxes zonular tension, allowing the elastic crystalline lens to become more convex—particularly its anterior surface—thereby increasing refractive power. This mechanism is described by Helmholtz's theory of accommodation.
9A patient has an amplitude of accommodation of 4.00 D and a near point of clear vision at 25 cm. Where is their far point?
A.100 cm (1 meter) in front of the eye
B.At infinity
C.25 cm in front of the eye
D.50 cm behind the eye
Explanation: Amplitude of accommodation (AA) = 1/near point − 1/far point (using vergences in diopters, with distances in meters). Near point vergence = 1/0.25 = 4.00 D. Solving: 4.00 = 4.00 − 1/far point → 1/far point = 0 → far point = infinity. This means the relaxed eye focuses at infinity—the eye is emmetropic. An emmetrope with 4 D of accommodation has a near point of 25 cm (1/4.00 D) and a far point at infinity.
10Panum's fusional area refers to:
A.The region of binocular single vision surrounding the horopter within which disparate retinal images can still be fused
B.The blind spot corresponding to the optic disc in each eye
C.The central foveal zone responsible for highest visual acuity
D.The area of maximum stereoscopic depth perception beyond the horopter
Explanation: Panum's fusional area is the zone around the horopter (the set of points in space imaged on corresponding retinal points) within which small amounts of binocular disparity can still be fused into a single percept with stereoscopic depth. Objects inside Panum's area are seen singly; outside it, diplopia results.
About the OLE Practice Questions
Verified exam format metadata for PRC Optometrist Licensure Examination (OLE), Philippines is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.