These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: mirrors are plane or spherical; f ≈ R/2; 1/sᵢ + 1/sₒ = 1/f with sₒ > 0, f > 0 for a concave mirror, f < 0 for a convex mirror, and sᵢ > 0 for a real image in front of the mirror, sᵢ < 0 for a virtual image behind it; |M| = |hᵢ/hₒ| = |sᵢ/sₒ|. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A student stands 2.5 m in front of a vertical plane mirror. She then steps 0.5 m towards the mirror. How far is she now from her image?
- (A) 2.0 m
- (B) 4.0 m
- (C) 4.5 m
- (D) 5.0 m
Answer and explanation
Answer: (B). She is now 2.5 m − 0.5 m = 2.0 m from the mirror. A plane-mirror image is the same distance behind the mirror (sᵢ = −sₒ), so the image is 2.0 m behind it. Her distance to the image is 2.0 m + 2.0 m = 4.0 m.
- (A) is her distance to the mirror, as if the image were on the glass.
- (C) subtracts her 0.5 m step from the original 5.0 m but forgets that the image also moves 0.5 m closer.
- (D) is the original separation; the image distance changes when she moves.
Question 2 (multiple choice · core)
An object is placed 15 cm in front of a concave mirror whose radius of curvature is 50 cm. Which describes the image?
- (A) Virtual, upright, enlarged, 37.5 cm behind the mirror
- (B) Real, inverted, enlarged, 37.5 cm in front of the mirror
- (C) Virtual, upright, enlarged, 21.4 cm behind the mirror
- (D) Virtual, upright, reduced, 9.4 cm behind the mirror
Answer and explanation
Answer: (A). f = R/2 = +25 cm. 1/sᵢ = 1/25 − 1/15 = (3 − 5)/75 = −2/75, so sᵢ = −37.5 cm. Negative means virtual and behind the mirror, so upright. |M| = 37.5/15 = 2.5: enlarged. This fits the rule that an object inside F gives a virtual, upright, enlarged image.
- (B) has the right distance but drops the minus sign, turning a virtual image into a real one.
- (C) uses R = 50 cm as the focal length.
- (D) uses f = −25 cm, as if the mirror were convex: 1/sᵢ = −1/25 − 1/15 = −8/75, so sᵢ = −9.4 cm. A reduced image is impossible for an object inside F of a concave mirror.
Question 3 (multiple choice · core)
An object is 30 cm in front of a convex mirror with a focal length of magnitude 20 cm. Where is the image?
- (A) 12 cm behind the mirror
- (B) 12 cm in front of the mirror
- (C) 60 cm in front of the mirror
- (D) 60 cm behind the mirror
Answer and explanation
Answer: (A). For a convex mirror f = −20 cm. 1/sᵢ = 1/(−20) − 1/30 = −3/60 − 2/60 = −5/60, so sᵢ = −12 cm: 12 cm behind the mirror (virtual, upright, |M| = 0.40).
- (B) has the right size but the wrong side. A convex mirror cannot form a real image of a real object.
- (C) uses f = +20 cm, treating the mirror as concave: 1/sᵢ = 1/20 − 1/30 = 1/60.
- (D) swaps the terms: 1/sₒ − 1/f = 1/30 − 1/20 = −1/60.
Question 4 (multiple choice · core)
An object starts at a distance 3f in front of a concave mirror of focal length f. It is moved to a distance 2f. Compared with the first image, the new image is
- (A) farther from the mirror and larger
- (B) closer to the mirror and larger
- (C) farther from the mirror and smaller
- (D) closer to the mirror and smaller
Answer and explanation
Answer: (A). At sₒ = 3f: 1/sᵢ = 1/f − 1/(3f) = 2/(3f), so sᵢ = 1.5f and |M| = 1.5f/3f = 0.5. At sₒ = 2f (the centre of curvature): sᵢ = 2f and |M| = 1. The image moves from 1.5f to 2f (farther away) and doubles in size.
- (B) gets the size right but assumes the image moves towards the mirror, like the object. For a real image from a concave mirror, the object and image move in opposite directions: as the object comes closer to the mirror, the image moves away from it.
- (C) and (D) say the image gets smaller. As a real object approaches F from outside, the real image always grows.
Question 5 (calculation and ray diagram · core)
A 2.0 cm tall object stands on the principal axis, 12.0 cm in front of a concave mirror of focal length 8.0 cm.
(a) Draw a ray diagram to scale using at least two principal rays, and mark the image. (b) Use the mirror equation to calculate the image distance. (c) Calculate the height of the image and describe the image fully.
Worked solution and suggested Marlbridge rubric
(a) The object is between F (8.0 cm) and C (16.0 cm). Parallel ray: reflects through F. Vertex ray: reflects symmetrically below the axis. Focal ray (through F before reaching the mirror): reflects parallel to the axis, 4.0 cm below it. The reflected rays meet about 24 cm in front of the mirror, 4 cm below the axis.
(b) 1/sᵢ = 1/f − 1/sₒ = 1/8.0 − 1/12.0 = 3/24 − 2/24 = 1/24, so sᵢ = +24.0 cm (real, in front of the mirror).
(c) |M| = |sᵢ/sₒ| = 24.0/12.0 = 2.00, so hᵢ = 2.00 × 2.0 cm = 4.0 cm. The image is real, inverted and enlarged, 24.0 cm in front of the mirror (beyond C).
| Point | What earns it |
|---|---|
| 1 | At least two correct principal rays from the top of the object, with arrows, meeting at the image |
| 1 | Image drawn inverted, below the axis, at about 24 cm |
| 1 | sᵢ = +24.0 cm from the mirror equation |
| 1 | hᵢ = 4.0 cm from |M| = 2.00 |
| 1 | Full description: real, inverted, enlarged (location stated) |
Common error: computing 1/12 − 1/8 and getting sᵢ = −24 cm, then calling the image virtual. Check against the diagram: an object outside F of a concave mirror always gives a real image.
Question 6 (derivation · core)
A dentist uses a small concave mirror of focal length f to view a tooth.
(a) Starting from the mirror equation, derive an expression for the magnitude of the magnification |M| in terms of sₒ and f only. (b) For f = 2.4 cm, find the object distance that gives an upright image three times the size of the tooth. Give the image position. (c) A second student finds another object distance that gives |M| = 3 with the same mirror. Find it, and explain why it is not useful to the dentist.
Worked solution and suggested Marlbridge rubric
(a) From 1/sᵢ = 1/f − 1/sₒ = (sₒ − f)/(f sₒ), so sᵢ = f sₒ/(sₒ − f). Then |M| = |sᵢ/sₒ| = |f/(sₒ − f)|.
(b) Upright means virtual, which needs sₒ < f. Then |M| = f/(f − sₒ) = 3, so f = 3f − 3sₒ, giving sₒ = 2f/3 = 1.6 cm. The image: sᵢ = (2.4)(1.6)/(1.6 − 2.4) = −4.8 cm, i.e. 4.8 cm behind the mirror (check: 4.8/1.6 = 3 ✓).
(c) The other solution has sₒ > f: f/(sₒ − f) = 3 gives sₒ = 4f/3 = 3.2 cm. Then sᵢ = +9.6 cm: a real, inverted image 9.6 cm in front of the mirror. An inverted image formed in mid-air in front of the mirror is useless for viewing a tooth close up, so the dentist holds the tooth inside F.
| Point | What earns it |
|---|---|
| 1 | Rearranges the mirror equation correctly to sᵢ = f sₒ/(sₒ − f) |
| 1 | |M| = |f/(sₒ − f)| |
| 1 | Uses “upright” to require a virtual image (sₒ < f) and gets sₒ = 1.6 cm |
| 1 | sᵢ = −4.8 cm, interpreted as 4.8 cm behind the mirror |
| 1 | Second solution sₒ = 3.2 cm, with the reason it is unsuitable (real, inverted image) |
Accept (b) done by substituting sᵢ = −3sₒ directly into the mirror equation.
Question 7 (experimental analysis · stretch)
A student places a lit object at several distances sₒ from a spherical mirror. For each one, she moves a screen until a sharp image appears on it, and records the image distance sᵢ. All distances are measured from the mirror.
| sₒ (cm) | 20.0 | 25.0 | 30.0 | 40.0 | 60.0 |
|---|---|---|---|---|---|
| sᵢ (cm) | 59.5 | 37.8 | 29.9 | 24.1 | 20.1 |
(a) Explain why a graph of 1/sᵢ (vertical) against 1/sₒ (horizontal) should be a straight line. State its expected gradient and vertical intercept. (b) Calculate 1/sₒ and 1/sᵢ for each row, plot the graph, draw a best-fit line and use it to find the focal length. (c) A classmate says: “The equation works for convex mirrors too, so we can’t tell from these data what kind of mirror it was.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Rearranging the mirror equation: 1/sᵢ = −1 × (1/sₒ) + 1/f. This has the form y = mx + b with y = 1/sᵢ and x = 1/sₒ. So the graph is a straight line with gradient −1 and vertical intercept 1/f.
(b)
| 1/sₒ (cm⁻¹) | 0.0500 | 0.0400 | 0.0333 | 0.0250 | 0.0167 |
|---|---|---|---|---|---|
| 1/sᵢ (cm⁻¹) | 0.0168 | 0.0265 | 0.0334 | 0.0415 | 0.0498 |
The points lie close to a straight line. A least-squares fit gives a gradient of −0.99 and a vertical intercept of 0.0663 cm⁻¹, so f = 1/0.0663 cm⁻¹ = 15.1 cm (a hand-drawn line should give 14.8–15.2 cm). The gradient close to −1 supports the model.
(c) The claim is wrong: the data show the mirror is concave. Every image was caught on a screen, so every image was real (sᵢ > 0), and the intercept gives a positive f. A convex mirror always forms a virtual image of a real object (sᵢ < 0), which cannot appear on a screen, and its 1/f is negative. So the data rule out a convex mirror.
| Point | What earns it |
|---|---|
| 1 | Rearranges to 1/sᵢ = −(1/sₒ) + 1/f and identifies the straight-line form |
| 1 | Gradient −1 and intercept 1/f both stated |
| 1 | Correct reciprocal values, plotted with labelled axes and units (cm⁻¹) |
| 1 | Best-fit line used (not a single data point) to find f in the range 14.8–15.2 cm |
| 1 | Rejects the claim: images on a screen are real / f positive, so the mirror is concave |
| 1 | Supports the rejection with the fact that a convex mirror forms only virtual images of real objects |
Also accept: finding f from the horizontal intercept, which is also at 1/f (x-intercept 0.0669 cm⁻¹ gives f = 14.9 cm).
How did you do?
- Q1 wrong: re-read the plane-mirror line in “The mirror equation and sign conventions” in the study guide.
- Q2, Q3 or Q5 wrong: check the sign of f first, then redo Worked examples 1 and 2.
- Q4 wrong: compare rows of the image table, or sketch both ray diagrams.
- Q6 or Q7 incomplete: work through Worked example 3, then rearrange the mirror equation into y = mx + b yourself.
Then tick off the topic checklist.