A picture on the screen becomes a problem in proportion
Suppose a school projects a small printed letter through a converging lens. The letter on the source card is 3 cm high; the focused image on a screen is 6 cm high. The first useful statement is that the image is twice the height of the object. That sounds simple, but it opens several distinct mathematical questions. What corresponding lengths were compared? Where is the optical centre? What do the measured object and image distances say? Will the image fit on the available screen? Does making the picture bigger make it easier to read?
Mathematics turns a visual effect into a relationship that can be tested, reversed and bounded. The Physics tells us whether a particular sharp real image can form. Mathematics asks what the resulting geometry implies, when lengths are measured from a consistent point. It cannot guarantee sharp focus merely by making two numbers add up. The order of reasoning matters: check the physical situation; draw the corresponding triangles; calculate the geometric consequence; then decide whether the screen arrangement is feasible.
This guide continues the practical lens observations from the previous Physics volume without copying its experimental discussion. Here we will solve for missing heights and distances, analyse space and screen constraints, compare physical scale with a viewer’s impression and learn to reject impossible or unjustified answers. The goal is not to memorise a new optics formula. It is to carry the familiar ideas of similar triangles, proportion, scale, units and inequalities into an unfamiliar setting.
The official 2027 G3 Mathematics and Physics boundaries
For the first Singapore–Cambridge Secondary Education Certificate examination in 2027, the SEAB G3 subject directory lists Mathematics K310 (former O-Level 4052). The K310 Mathematics syllabus assesses numerical, algebraic, geometrical, statistical and problem-solving skills, including similar figures, ratio, scale, trigonometry and representations of real situations. Paper 1 is 2 hours 15 minutes and carries 90 marks (50%). Paper 2 is also 2 hours 15 minutes and carries 90 marks (50%); its last question focuses on applying Mathematics to a real-world scenario. Verify the specific instructions on the examination paper rather than relying on general habits about rounding or units.
Standalone Physics K323, described in the official K323 syllabus, includes thin converging lenses, real and virtual images and focal length. In this mathematical application we use a ray passing through the optical centre in the ideal thin-lens model to justify similar triangles. We do not claim that a derived projector sizing problem is an official combined examination task or that the Physics image-distance equation is a compulsory K310 formula. We will avoid assuming a particular focal length unless it is supported by the stated lens arrangement.
These worked calculations are original teaching examples. Their numerical distances are treated as specified positions of a focused real-image arrangement, not as interchangeable choices that any lens would necessarily support. If the model gives a geometric height while no focused image has been verified, the honest wording is ‘predicted under the assumed arrangement’, not ‘measured’.
Respect the owners of the three related skills
Start with Volume 0476: Physics screen tests when you need to decide whether the screen intercepts a real focused image at all. For the ray-construction mechanism, Volume 0137: converging-lens ray diagrams is the canonical Physics owner. Volume 0473: geometry in reflected-ray paths teaches measuring angles and maintaining diagram constraints, though it treats reflection rather than a lens.
The earlier Volume 0399: microscope scale diagrams owns reversed magnification in biological scale images. Its sizes relate a specimen to a represented image. This lesson instead investigates the geometry and feasibility of a projected real image and its screen. The underlying mathematics is transferable, but the task and physical evidence are not identical.
Do not use a photographic-looking drawing as a substitute for marked distances. A sketch may exaggerate the thickness of a lens or the size of a letter to make a point clear. The mathematics only begins when the relevant reference lengths are stated and their units are consistent.
Derive the useful ratio from corresponding triangles
Place the object on one side of a thin converging lens and a focused real image on the other side. In a standard idealised construction, the ray from the top of the object through the optical centre continues along a straight line. The line from the object tip to the optical centre and the corresponding line to the image tip form two triangles with a common angle at the centre, and their upright heights are perpendicular to the principal axis. Those triangles are similar.
Let u be the positive object distance from the optical centre and v the positive distance from the optical centre to the focused real-image plane. Let H be the positive object height and h the positive image height. Similarity gives h/H = v/u. We use positive sizes here and separately state that the real image is inverted. If a signed coordinate convention is introduced in a higher course, use it consistently; do not mix its signs with a school sizing problem that simply asks for dimensions.
For example, if H is 3 cm, u is 30 cm and v is 60 cm, then h is 3 times 60 divided by 30, or 6 cm. The height ratio is 2. The units cancel in the distance ratio because both distances are in centimetres. This is a geometrical consequence of the assumed focused arrangement. The numbers do not independently prove that a particular physical lens will focus at those positions.
The four checks that should precede a calculation
First, identify corresponding heights. The object’s height is not the diameter of the lens, and the image’s height is not the distance across the whole screen. A labelled drawing helps you connect the top and bottom of the same feature. Second, establish the origin of both distances. If u is measured to the face of one holder and v to a different face, the ratio no longer represents the intended similar triangles accurately.
Third, convert units before substitution. When object distance is stated in metres and image distance in centimetres, treating both numerical values as if they had the same unit produces a tenfold or hundredfold error. Fourth, ask what is actually required. The question might ask for a height, a distance, a maximum acceptable object size, or whether a choice is feasible. A correct intermediate ratio is not the final decision.
One final check concerns physical interpretation. A height of 7.5 cm can fit on a screen of 10 cm height if the image is properly positioned and the usable area genuinely is 10 cm. If the frame hides 2 cm at each edge, the available viewing region is only 6 cm. The word ‘screen’ is not automatically the same as ‘usable screen’.
Twenty-eight original geometry and decision clinics
Each clinic is an independent teaching scenario. Where an image is described as focused, treat u and v as a compatible measured or given pair for that scenario. Do not combine numerical pairs from different examples as if they belonged to the same lens at fixed focal length. For each problem, try the arithmetic yourself before reading the explanation. Record units and the final physical decision as well as the number.
01 — Find an image height
A 2.4 cm letter has a focused image when u = 36 cm and v = 54 cm.
The ratio v/u is 54/36 = 1.5. Thus h = 2.4 × 1.5 = 3.6 cm. The physical image is 3.6 cm tall and is inverted in the standard real-image arrangement. Notice that using u/v would have given 1.6 cm and reversed the meaning of magnification.
The check is to ask which triangle is larger: because the image-side distance is greater, the corresponding image height is greater too.
02 — Find an object height
The measured focused image is 7.2 cm tall. The distances are u = 45 cm and v = 60 cm.
Write 7.2/H = 60/45. Hence H = 7.2 × 45/60 = 5.4 cm. This is an inverse use of the same relationship, not a new law. Check by multiplying 5.4 by 4/3 to recover 7.2.
An answer of 9.6 cm would be the result of scaling the already magnified image in the wrong direction.
03 — Find an image distance
A 3.0 cm symbol makes a focused image 4.5 cm high at object distance 40 cm.
Use h/H = v/u, or 4.5/3 = v/40. The magnification is 1.5, so v = 60 cm. Check the result against the diagram: the image is larger than the object and farther from the lens than the object.
This is a geometrical inference conditional on the focused arrangement, not proof that the apparatus will project at 60 cm without testing.
04 — Find an object distance
The image is half the object’s height and forms at v = 28 cm.
The ratio v/u = 0.5, so u = 28/0.5 = 56 cm. A quick check gives v/u = 28/56 = 0.5. The object must be farther from the optical centre than the image plane in this case.
Half-sized does not mean the object is only half as far away. Read the ratio in the correct order.
05 — Keep measurements in one unit
The object is 0.60 m from the lens and the image plane is 30 cm away; the object is 4 cm high.
Convert 0.60 m to 60 cm. Then h = 4 × 30/60 = 2 cm. Alternatively express both distances in metres and obtain the same result. The unit of image height remains centimetres because H was in centimetres.
Dividing 30 by 0.60 without conversion gives a meaningless scale for this question.
06 — Separate sign and size
An object 5 cm high has u = 50 cm and v = 25 cm, producing a focused real image.
The size ratio is 25/50 = 0.5, so the image is 2.5 cm high. The real-image orientation is inverted. The length can be stated as 2.5 cm without putting a negative sign on a physical size; inversion should be reported in words or with a clearly defined signed coordinate.
Do not leave the reader guessing whether a negative height means a smaller object or an inverted image.
07 — Fit a rectangular screen
A 4 cm high source is magnified by 1.75 on a screen with 6 cm usable vertical height.
Its image height would be 4 × 1.75 = 7 cm. The picture does not fit vertically because 7 > 6. The excess is 1 cm in total before considering centring. Reporting 7 cm without comparing the 6 cm limit does not answer the design question.
A working projection and a usable complete projection are different success conditions.
08 — Allow a frame margin
A nominal 14 cm high display has 2 cm borders at the top and bottom. The image would be 11 cm high.
The usable vertical height is 14 − 2 − 2 = 10 cm. Because 11 cm exceeds 10 cm, the complete image cannot fit in the stated usable region. The answer should not compare 11 to 14 and conclude that it is acceptable.
Read ‘frame’ and ‘usable area’ as different quantities whenever a practical design brief distinguishes them.
09 — Maximum source letter height
The usable screen height is 9 cm and the specified focused arrangement has magnification 1.5.
For a complete image, H × 1.5 must be no more than 9. Thus H ≤ 6 cm. A 6 cm source exactly meets the mathematical limit; if the brief asks for spare margin, select a smaller value. A source of 7 cm would predict a 10.5 cm image and fail.
An inequality is more useful than a single attractive number when the choice is constrained.
10 — Minimum magnification for visibility
A display requires letters at least 5 cm high. The source card letters are 2 cm high.
Let m denote the positive height ratio. The condition is 2m ≥ 5, giving m ≥ 2.5. If a specified focused arrangement provides m = 2, it is insufficient. If it provides m = 3, the criterion is met, subject to focus, contrast and screen fit.
‘At least’ means a lower bound; do not reverse it into a maximum.
11 — Combining a minimum and maximum
The projected letter must be between 4 cm and 7 cm tall. The source letter is 2 cm high.
The required ratio is 4 ≤ 2m ≤ 7, hence 2 ≤ m ≤ 3.5. This range tells the team what geometrical arrangements could meet the size requirement if they also focus. A ratio of 4 would be too large even though a bigger image may sound desirable.
The best size is often a feasible interval, not the largest possible figure.
12 — Compare two compatible trial arrangements
Prototype A has H = 3 cm, u = 40 cm, v = 60 cm; B has the same H, u = 50 cm, v = 75 cm. Both are stated as focused.
A predicts h = 3 × 60/40 = 4.5 cm. B predicts 3 × 75/50 = 4.5 cm. The different distances still give the same ratio 1.5. Size alone cannot select one; practical space, focus quality and equipment limitations require additional information.
Different dimensions can encode the same proportional relation.
13 — Choose a compact arrangement
The available screen-to-lens distance is no greater than 50 cm. Candidate focused trials require 42, 48 and 56 cm.
The first two satisfy v ≤ 50 cm. The third does not. If the screen also needs clearance behind the holder, the brief may require a stricter bound; do not silently add or ignore a stated gap. Geometry filters feasible candidates before a recommendation is written.
A candidate with the largest image can still be infeasible because the room is too short.
14 — A maximum total bench length
A setup needs u + v ≤ 120 cm. Proposed focused pairs are (50,60), (45,80), and (60,55) cm.
Compute 110, 125 and 115 cm respectively. The first and third satisfy the stated ideal length limit; the second exceeds it. If the apparatus includes stands or protruding parts not represented by u and v, the physical bench may need additional clearance. Do not claim that the mathematical check covers equipment not included in the data.
Identify precisely which components a dimensional total represents.
15 — Calculate a spare clearance
A focused arrangement occupies 108 cm of a 130 cm usable bench, measured by the same endpoints.
The spare length is 130 − 108 = 22 cm. That is an arithmetic margin, not a guarantee that the stands will fit if the 108 cm omitted their widths. The result can support an initial layout check but should be carried into a real placement test.
A measured margin has meaning only relative to the defined boundary.
16 — Reverse a distance ratio
A source letter of 2 cm is to appear 8 cm high, and the focused image distance is given as 80 cm.
The desired magnification is 8/2 = 4. Similarity implies v/u = 4, so u = 80/4 = 20 cm. Without knowing the lens and focal condition, the mathematics does not tell us that a real focused image will form at that pair. The task provides a target geometry, not an experimental guarantee.
Flag the difference between a design ratio and a physically validated lens position.
17 — Do not confuse an area ratio with a height ratio
An image is twice the height and twice the width of a rectangular source label.
Corresponding lengths scale by 2, so its area scales by 2² = 4. A 3 cm by 2 cm source has area 6 cm²; the 6 cm by 4 cm image has area 24 cm². The area is not merely doubled. This matters when calculating how much of the screen is occupied.
Keep linear, square and cubic scales distinct instead of multiplying by the first ratio you see.
18 — Reverse from occupied area
A rectangular projected image occupies 36 cm² and is similar to a source rectangle of area 9 cm².
The area ratio is 36/9 = 4. The corresponding linear ratio is √4 = 2, assuming true similarity. If the source height was 3 cm, the image height is 6 cm. An answer of 12 cm would confuse area enlargement with length enlargement.
Use a square root to recover a linear scale factor from a two-dimensional area ratio.
19 — Orientation is not a change in area
A focused projected image is inverted, 8 cm by 3 cm, and fits on a matching rectangular screen.
Its area is 24 cm² regardless of its inverted orientation. The inversion does not make the area negative. A sign may be meaningful in a coordinate diagram, but the physical coverage of the screen is a positive measure. Compare dimensions separately if the screen is narrower in one direction.
Don’t transfer a coordinate sign into a real-world area calculation.
20 — Round only at the end
An object 2.75 cm tall projects with u = 33 cm and v = 47 cm in a stated focused arrangement.
The height is 2.75 × 47/33 = 3.91666… cm, or 3.92 cm to three significant figures if the problem calls for that accuracy. Rounding the ratio 47/33 to 1.4 before multiplication would give 3.85 cm, an avoidable intermediate-rounding error.
Keep calculator precision through the working and apply the requested final format once.
21 — A percentage increase in height
The projected letter height changes from 5 cm to 6.5 cm between two measured trials.
The increase is 1.5 cm. Relative to the original 5 cm, percentage increase is 1.5/5 × 100 = 30%. The percentage base is the earlier image height, not the new 6.5 cm. This describes the change in size, not necessarily a change in focus quality.
When stating a percentage, identify what the ‘100%’ refers to.
22 — The inverse percentage trap
A projected height decreases from 8 cm to 6 cm when the arrangement is changed.
The reduction is 2 cm, or 2/8 × 100 = 25% of the original. To recover 8 cm from 6 cm requires increasing by 2/6 × 100 = 33⅓% of the new height. A decrease of 25% and a reversal increase of 25% do not cancel.
Reverse problems often change the denominator. Show it explicitly.
23 — Distinguish length from field of view
A classroom slide is 10 cm wide, but only a 7 cm wide part is seen on a screen.
The visible fraction of the source width is 7/10 = 0.7 only if the 7 cm figure is measured in the same source-scale units; a 7 cm screen width cannot be compared directly to a 10 cm source width if the picture is magnified. Instead, scale the full source width using the image magnification, then compare to the usable screen width.
A ratio is only meaningful when its two lengths refer to corresponding scales.
24 — Test width as well as height
A 4 cm × 3 cm source rectangle is magnified by 1.5. The usable screen is 7 cm high and 5 cm wide.
The projected dimensions are 6 cm high and 4.5 cm wide if the source orientation follows those dimensions. Both fit the limits 7 and 5. If the screen is rotated or the source’s width and height were swapped, redo the comparisons using the actual orientation.
A one-dimensional check cannot establish that the entire rectangle fits.
25 — Check the missing band
A projected 8 cm image must be centred on a 12 cm usable screen.
Assuming symmetric centring, the total leftover vertical space is 12 − 8 = 4 cm, giving 2 cm top and 2 cm bottom. If the image centre is displaced by 3 cm from the screen centre, one side may be clipped even though 8 ≤ 12. Position matters as well as size.
Numerical fit is necessary but not always sufficient for a practical layout.
26 — Two candidate image sizes
Trial A produces a sharp 6 cm image; trial B produces a sharp 9 cm image. The usable screen height is 8 cm.
A fits and B does not, under the whole-image requirement. If the criterion were instead ‘show the central word’, an incomplete crop might have a different value, but that must be stated in the brief. Without it, B’s larger picture is not a better match.
Every recommendation should identify the actual success criterion.
27 — An uncertain ruler origin
The lens holder is 2 cm wide and a student measures both distances from its outer edges rather than the optical centre.
If each edge is displaced by about 1 cm from the optical centre, the ratios inferred from the reported readings can differ from those intended by the ideal model. The correction depends on which side each reading comes from; simply adding 1 cm to both without a diagram may be wrong. Draw the holder and mark every endpoint.
Systematic offset cannot be repaired by rounding results to more digits.
28 — A decision with two objectives
A school needs a complete image between 5 and 8 cm high, using no more than 70 cm of screen-side distance. A produces 6 cm at 62 cm; B produces 7.5 cm at 75 cm.
A meets both conditions. B meets the size interval but fails the distance condition. Even if B is brighter, that property cannot cancel a mandatory physical limit unless the school explicitly changes the brief. The correct choice from these two on the stated criteria is A, subject to real focus and readability checks.
Use a table of stated constraints before arguing about a single desirable feature.
A longer worked project: choose a projected label
An original classroom brief says that a display has a usable screen 12 cm high by 18 cm wide. The source text card is 4 cm high and 7 cm wide. A design team has three tested focused arrangements. Trial A produces a positive magnification ratio of 1.5. Trial B gives 2.0. Trial C gives 2.5. The brief requires the whole source label to remain visible. Which candidates survive a geometry-only first pass?
For A, the image is 6 cm high and 10.5 cm wide, so it fits. For B, it is 8 cm high and 14 cm wide, so it fits. For C, it is 10 cm high and 17.5 cm wide, so it also fits. A tempting shortcut would be to choose C immediately because it gives the largest image. But a 10 cm image leaves only 2 cm of total vertical margin, and a 17.5 cm image leaves only 0.5 cm of total horizontal margin. If the screen must be centred and the frame dimensions are exact, the horizontal margin is just 0.25 cm on each side.
Now add another condition from the school’s real brief: there must be at least 1 cm clear horizontal margin at each side to avoid clipping when visitors brush the card. This means the image’s width must not exceed 18 − 2 = 16 cm. A and B still pass; C fails. We have not changed the image physics or erased a measurement. We have changed the explicit feasibility rule, so the set of acceptable options has changed.
Finally, suppose observers can read the small print on B consistently and that on A only from a close position, while B’s projector fits the available bench. On those hypothetical observations, B has a defensible advantage. Yet the written recommendation should not claim B is universally clearer than A; it should say that B performed better on the tested criterion under the stated conditions. Mathematics eliminates impossible candidates. Scientific testing and clear writing distinguish the remaining options.
Inverse design and a feasibility inequality
Suppose the source lettering is 2.8 cm high. The screen allows at least 5 cm and at most 7 cm for the image because smaller images are unreadable and larger ones clip. The required magnification range is 5/2.8 ≤ m ≤ 7/2.8, or approximately 1.7857 ≤ m ≤ 2.5. Do not round the lower bound to 1.8 and then treat the rounded number as exact when judging a borderline arrangement.
If u is specified as 35 cm and the ideal similar-triangle geometry is used, the corresponding image-side range is 62.5 cm ≤ v ≤ 87.5 cm. These are the distances that meet the geometric size requirement under the conditional model. They do not ensure a sharp real image exists at every point in this interval for one fixed lens and one fixed object. In a real lens experiment, the focus requirement may select one distance, or none within the permitted region.
A strong solution therefore gives two separate answers: the geometric size interval and the practical verification still needed. It avoids the false claim that all values in a mathematical range are automatically physically attainable. This distinction is useful in many real-world Mathematics tasks: a feasible model must satisfy every stated constraint, not just an equation.
Accuracy, evidence and the temptation of false precision
A ruler reading to the nearest millimetre may be sensible in a classroom setup, but a visual focus judgement can be less precise than the ruler markings. A student can write 47.000 cm on a page without having evidence for the last three decimal places. Make the number of significant figures consistent with what is supplied and with the examination instruction, while avoiding premature rounding during calculation.
For the G3 Mathematics examination, read the specific question’s accuracy instruction. A common syllabus convention is to give non-exact answers to three significant figures and angles in degrees to one decimal place unless directed otherwise; exact answers can remain exact when appropriate. A practical physics observation may instead require a number of digits informed by the measuring instrument. Those two contexts should not be flattened into a single rule about ‘always three significant figures’.
If your ratio and measured heights disagree, do not immediately alter a recorded observation to force exact equality. Check whether the stated arrangement was truly focused, whether the screen and source measurements used corresponding features, whether the image was cropped, and whether distances shared one reference. The mathematical ratio is a model; the physical record is evidence. Both deserve honest treatment.
Exam practice without dependence on a remembered picture
Start with a blank axis, mark the object position and the image position, and connect the top of the object to the optical centre with the appropriate ray. Draw perpendicular heights at the two positions. Write down which angles match and which sides correspond. Only then form the proportion. This reconstructs the method when a diagram is unfamiliar instead of relying on a memorised template.
Now practise three changes in a row. Ask for a missing image height, then for the maximum source height allowed by the screen, and finally for a written decision involving a second condition such as bench length. The calculation is not difficult in isolation; the challenge is keeping every number attached to the correct question. Under examination conditions, an answer that stops after finding the ratio when the question asks for feasibility is incomplete.
An efficient final check is physical and grammatical at once: ‘The predicted image is 7.2 cm high, so it fits within the 8 cm usable height, provided the image is correctly positioned and a focused real image is obtained.’ The answer uses the number, the inequality and the necessary condition. There is no advantage in adding an unsupported promise that the screen will be bright or readable to everyone.
The route onward
For the actual assessment details use SEAB’s current 2027 G3 directory, Mathematics K310 and, for optical model context, Physics K323. If the issue is whether a real image exists or whether a screen test was conducted fairly, return to Volume 0476. If the issue is converting object-image dimensions into a ratio, work through this guide and compare the different scale context of Volume 0399.
For earlier ratio foundations, the PSLE Learning Guide offers a way back into fractions, proportions and explanation discipline. The Mathematics learning hub connects these skills to other secondary problems. For readers who need to revise the original lens construction, Volume 0137 remains its owner. The next guide focuses on explaining the resulting exhibit to visitors: clear English, honest qualifiers and captions that communicate what the test actually showed.
The aim is to see the constraint before doing the arithmetic, then answer the decision question after doing it. When the physical and numerical assumptions are visible, even a complex-looking projector problem becomes a series of small, checkable steps.
