Optics and Refraction MCQs With Worked Solutions
Optics questions are seldom lost because the formula is unknown. They are lost because metres become centimetres, a sign convention is silently dropped, or the candidate answers the power meridian instead of the written cylinder axis. Use the same sequence every time: write the known quantity with its unit, select the relationship, do the arithmetic, then ask whether the optical direction makes sense.
This is postgraduate exam revision, not a clinical prescription guide. Refraction conventions and exact clinical decisions vary with the setting and reference; the questions below test durable principles rather than a patient-specific plan.
The 30-second method before you touch the options
- Convert focal length or distance to metres.
- Put a sign beside a converging or diverging quantity before calculating.
- Keep the answer in dioptres (D), metres, centimetres or prism dioptres as asked.
- For cylinders, sketch the two principal meridians. The written axis has zero cylinder power; the cylinder acts 90° away.
- Reject an answer that has the wrong physical direction even if the number looks familiar.
The calculation items use the usual thin-lens, paraxial exam model. A real spectacle correction involves vertex distance, eye position and other factors; the model is still what most single-best-answer questions are asking you to apply.
30 worked optics and refraction MCQs
1. A thin positive lens has a focal length of 50 cm. What is its power?
Answer: +2.00 D. Convert first: 50 cm = 0.50 m. Lens power is F = 1/f, so 1/0.50 = +2.00 D. A positive lens is converging, so a minus result should immediately look wrong.
2. A +4.00 D lens has what focal length?
Answer: 0.25 m (25 cm). Rearrange the same relationship: f = 1/F = 1/4. Do not write 0.25 cm: focal length in this calculation is in metres.
3. Two thin lenses, +2.00 D and +3.00 D, are in contact. Their combined power is:
Answer: +5.00 D. For thin lenses in contact, powers add algebraically. The word algebraically matters: +3 + −2 would be +1.00 D, not +5.00 D.
4. A −5.00 D lens in contact with a +2.00 D lens produces:
Answer: −3.00 D. The net effect remains diverging because the negative power has the larger magnitude. Use the sign, not a verbal memory of “adding lenses”.
5. The unit of vergence is:
Answer: the dioptre. Vergence is the reciprocal of the distance in metres. In sign-based questions, converging and diverging vergences have opposite signs; state the convention your paper or text uses before combining them.
6. Parallel rays incident on a positive lens emerge:
Answer: converging to its posterior focal point. Parallel rays have zero vergence. A positive lens adds positive convergence in the standard sign convention. This is the physical picture behind the power formula.
7. An object beyond the focal length of a convex lens gives an image that is usually:
Answer: real and inverted. The precise size depends on object position. The trap is choosing “virtual, erect”, which is the familiar image behaviour of a concave lens, not a convex lens with the object outside its focal length.
8. A concave lens forms an image that is:
Answer: virtual, erect and diminished. A diverging lens makes the rays appear to arise from a focal point on the object side. This is why the image cannot be projected on a screen.
9. One prism dioptre means that a ray is displaced by:
Answer: 1 cm at 1 metre. This definition is more useful than trying to convert prism dioptres into degrees during an SBA. The prism convention is a displacement measure, not the refractive power of a spherical lens.
10. Light passing from a denser to a rarer medium can undergo total internal reflection when:
Answer: the angle of incidence exceeds the critical angle. The direction of travel is essential. Total internal reflection does not occur simply because the incident angle is “large”; the ray must be travelling from the optically denser medium to the rarer one.
11. In the schematic eye, which surface provides the larger share of total refractive power?
Answer: the cornea. The air–tear/cornea interface has a much larger refractive-index change than the internal ocular interfaces. Exam figures commonly approximate corneal power at about 43 D and total eye power at about 60 D; treat those as model values rather than measurements for an individual eye.
12. In an emmetropic eye viewing a distant target with accommodation relaxed, the principal focus lies:
Answer: on the retina. “Distant” is the clue: incoming rays are treated as parallel. Emmetropia describes the focus position in that relaxed state, not unaided acuity in every lighting or retinal condition.
13. In axial myopia, parallel rays are focused:
Answer: in front of the retina. The common mechanism is an eye that is too long for its optical power. A minus lens diverges incoming light and moves the focus posteriorly to the retinal plane in the simple optical model.
14. In hyperopia, parallel rays with accommodation relaxed would be focused:
Answer: behind the retina. A plus lens adds convergence. Do not equate hyperopia with a child’s manifest refraction alone: accommodation can mask some hyperopia during measurement.
15. The principal optical action of cycloplegia during refraction is to remove:
Answer: accommodation. That is why it can reveal hyperopic refractive error that accommodative effort was partly compensating for. In an examination, answer the mechanism first; drug choice and technique are separate questions.
16. During accommodation, ciliary muscle contraction causes the crystalline lens to become:
Answer: more curved and more powerful. In the Helmholtz description, ciliary contraction reduces zonular tension, allowing the elastic lens to round up. The practical memory aid is near target → increased optical power.
17. Presbyopia is primarily caused by:
Answer: an age-related reduction in accommodative amplitude. It is not simply “hyperopia in old age”. The near point recedes because the lens–zonule system can no longer increase power as effectively for near viewing.
18. A patient with no crystalline lens has an optical state classically described as:
Answer: aphakia with marked hyperopia. Removing the lens removes a substantial part of the eye’s power and its accommodative mechanism. This is a principle question, not a request to choose a correction.
19. In minus-cylinder notation, the written axis denotes:
Answer: the meridian with zero cylinder power. The cylindrical component exerts its stated power in the meridian 90° away. Draw a cross if needed; this prevents the very common axis-versus-power-meridian error.
20. Transpose +2.00 −3.00 × 180 into plus-cylinder form.
Answer: −1.00 +3.00 × 90. Add sphere and cylinder: +2 + −3 = −1. Reverse the cylinder sign: −3 becomes +3. Rotate the axis by 90 degrees: 180 becomes 90. Verify by calculating the power in both principal meridians; they must match before and after transposition.
21. The spherical equivalent of −4.00 −2.00 × 180 is:
Answer: −5.00 D. Use sphere + (cylinder/2): −4 + (−2/2) = −5. Spherical equivalent is a summary value; it does not preserve the directional nature of astigmatism.
22. In regular astigmatism, the two principal meridians are:
Answer: perpendicular. Regular astigmatism has two principal meridians at right angles with different powers. An irregular cornea need not fit that simple two-meridian cylinder model cleanly.
23. The conoid of Sturm is located:
Answer: between the two line foci of an astigmatic optical system. One line focus forms first, then the interval of Sturm, then the second line focus. The circle of least confusion lies within that interval and is the conceptual basis for spherical-equivalent thinking.
24. Chromatic aberration occurs because:
Answer: refractive index varies with wavelength. Shorter wavelengths are refracted more strongly than longer wavelengths in ordinary dispersive optical media. This makes colour focus at slightly different planes rather than at one perfect point.
25. Spherical aberration in a simple positive spherical lens means that:
Answer: peripheral and paraxial rays do not share exactly the same focus. In the classic positive-lens description, marginal rays are brought to focus more strongly than paraxial rays. The examinable point is the failure of a single sharp axial focus, not a particular instrument setting.
26. A +5.00 D spectacle lens is moved closer to the cornea. Ignoring any other change, its effective power at the corneal plane becomes:
Answer: slightly greater plus. A plus lens moved closer to the eye becomes effectively stronger; a minus lens moved closer becomes effectively less minus. In most exam questions this matters only when the prescription is high or when the stem explicitly gives a vertex distance. Do not apply a vertex conversion to every ordinary refraction item.
27. Anisometropia means:
Answer: unequal refractive state in the two eyes. It is a refractive relationship, not a statement about unequal visual acuity. The exam may then ask separately about aniseikonia, spectacle magnification or amblyopia risk; those are consequences or associations, not the definition.
28. The near point of accommodation is the nearest point that can be seen clearly when:
Answer: maximum accommodation is exerted. It moves farther from the eye as accommodative amplitude falls. Do not confuse it with the far point, which is the conjugate point for relaxed accommodation.
29. A pinhole may improve visual acuity in an uncorrected refractive error mainly because it:
Answer: reduces the blur circle by admitting a narrower pencil of rays. It increases depth of focus, but it also reduces retinal illumination. Therefore “pinhole improves” is an optical clue in an exam stem, not a complete clinical conclusion.
30. The far point of an emmetropic eye is:
Answer: at infinity. With accommodation relaxed, parallel rays from a distant object focus on the retina. In myopia the far point is finite and in front of the eye; that is the optical reason a myopic eye can see some nearer targets without a minus correction.
Three calculation traps worth revising twice
The centimetre trap
Every dioptre calculation starts with metres. A focal length of 25 cm is 0.25 m, so the answer is +4 D. If you put 25 into 1/f, you have answered a different question.
The axis trap
The axis is a direction, not an amount of cylinder power. For −2.00 × 180, the 180-degree meridian receives zero cylinder; the 90-degree meridian receives −2.00 D. Keep sphere separate until the end.
The vertex-distance trap
For modest prescriptions, the thin-lens SBA model often ignores vertex distance. At high powers, moving a lens from the spectacle plane to the corneal plane changes its effective power. First identify which plane the question specifies; do not transpose a prescription mechanically without that detail.
Use this set as retrieval practice
Do one pass without a calculator. On the second pass, write one-line error labels: unit, sign, axis, image, or concept. A list of error labels is more useful than re-reading all 30 explanations after every session.
For mixed retrieval practice, use the verified general ophthalmology MCQs. If your weakness is the underlying optics chapter rather than a calculation, see Optics & Refraction Notes. ICO / FICO candidates can also place this topic in their wider ICO / FICO preparation plan. These are study resources, not a substitute for the current syllabus or your designated text.
Sources and scope
The calculations and terminology were checked against the sources below on 18 August 2026. The sources support the optical principles; none is used to claim that a study method, product or question set changes examination results.
- Musa MJ, Zeppi M. Spectacle Correction of Ametropias. StatPearls, NCBI Bookshelf — refractive-error and spectacle-correction terminology.
- StatPearls: Hyperopia, NCBI Bookshelf — relaxed-focus position and hyperopic optical concepts.
- AAO EyeWiki: Myopia — myopic refractive state and axial mechanism overview.
- AAO EyeWiki: Hyperopia — hyperopic refractive state overview.
- AAO EyeWiki: Presbyopia — age-related loss of accommodation overview.
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