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: every lens is thin and surrounded by air; 1/sᵢ + 1/sₒ = 1/f; f > 0 for a converging lens and f < 0 for a diverging lens; sᵢ > 0 is a real image on the far side, sᵢ < 0 a virtual image on the object’s side; magnification |M| = |hᵢ/hₒ| = |sᵢ/sₒ| is a size ratio, and for a single lens a real image is inverted and a virtual image is upright. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A real object is placed in front of a diverging lens. Which description of the image is correct for every object position?
- (A) Real, inverted and reduced
- (B) Virtual, upright and reduced
- (C) Virtual, upright and enlarged
- (D) Real or virtual, depending on whether the object is inside the focal length
Answer and explanation
Answer: (B). A diverging lens spreads rays out, so they never meet: the image is always virtual. Traced back, the rays meet between the lens and the near focal point, closer than the object, so the image is upright and smaller. With f < 0 and sₒ > 0, 1/sᵢ = 1/f − 1/sₒ is always negative, and |sᵢ| < sₒ gives |M| < 1.
- (A) describes a converging lens with the object beyond 2f.
- (C) describes a converging lens used as a magnifier (object inside f).
- (D) is true of a converging lens, not a diverging one. A diverging lens has no object position that gives a real image.
Question 2 (multiple choice · core)
An object is 20 cm from a converging lens of focal length 15 cm. What is the image distance sᵢ?
- (A) +60 cm
- (B) +8.6 cm
- (C) −60 cm
- (D) +5.0 cm
Answer and explanation
Answer: (A). 1/sᵢ = 1/15 − 1/20 = (4 − 3)/60 = 1/60 cm⁻¹, so sᵢ = +60 cm: a real image on the far side, inverted and 3 times the size of the object (|M| = 3.0).
- (B) adds the reciprocals (1/15 + 1/20 = 7/60), giving 8.6 cm.
- (C) has the right size but the wrong sign. With sₒ > f for a converging lens, the image is real, so sᵢ is positive.
- (D) subtracts the distances (20 − 15). The thin-lens equation works with reciprocals.
Question 3 (multiple choice · core)
A converging lens forms a sharp, real image of a lamp on a screen. A student covers the top half of the lens with a card. What happens to the image on the screen?
- (A) The top half of the image disappears.
- (B) The bottom half of the image disappears.
- (C) The whole image is still there but dimmer.
- (D) The image moves to a new position, closer to the lens.
Answer and explanation
Answer: (C). Light from every point of the lamp reaches every part of the lens. The uncovered half still collects rays from every point and brings them to the same image points. Only half as much light arrives, so the image is dimmer.
- (A) and (B) assume each part of the lens makes one part of the image. The lens is not a window onto one part of the object.
- (D) The focal length and sₒ are unchanged, so by the thin-lens equation sᵢ is unchanged.
Question 4 (multiple choice · core)
An object is 40 cm from a converging lens of focal length 10 cm, and a real image forms. The object is then moved to 15 cm from the lens. How does the image change?
- (A) It moves farther from the lens, and its height increases by a factor of 6.
- (B) It moves farther from the lens, and its height increases by a factor of 2.7.
- (C) It moves closer to the lens, and its height increases by a factor of 6.
- (D) It moves farther from the lens, and its height increases by a factor of 2.25.
Answer and explanation
Answer: (A). At 40 cm: sᵢ = 1/(1/10 − 1/40) = 13.3 cm and |M| = 13.3/40 = 0.333. At 15 cm: sᵢ = 1/(1/10 − 1/15) = 30.0 cm and |M| = 30/15 = 2.00. The image moves from 13.3 cm to 30.0 cm (farther) and its height grows by 2.00/0.333 = 6.
- (B) uses the ratio of object distances, 40/15 = 2.7. Image height depends on sᵢ/sₒ, and sᵢ changes too.
- (C) gets the size factor right but the wrong direction. As the object approaches F, the real image moves away from the lens (Figure 3 in the guide).
- (D) uses only the ratio of image distances, 30/13.3 = 2.25, and forgets that sₒ also changed.
Question 5 (calculation · core)
An object 1.5 cm tall stands 30 cm in front of a diverging lens with a focal length of magnitude 20 cm.
(a) Calculate the image distance. (b) Calculate the magnification and the image height. (c) Describe the image: real or virtual, upright or inverted, and where it is. (d) Without calculating, state what happens to the image size as the object is moved closer to the lens.
Worked solution
- (a) f = −20 cm. 1/sᵢ = 1/f − 1/sₒ = −1/20 − 1/30 = −5/60 = −1/12 cm⁻¹, so sᵢ = −12 cm.
- (b) |M| = |sᵢ/sₒ| = 12/30 = 0.40. hᵢ = 0.40 × 1.5 cm = 0.60 cm.
- (c) Virtual (sᵢ < 0), so upright for a single lens; reduced (|M| < 1); 12 cm from the lens on the same side as the object.
- (d) It gets larger, but stays smaller than the object. (At 10 cm, for example, |M| = 0.67.) Far away, the image shrinks and approaches F.
Suggested mark points (4): 1 for using f = −20 cm and getting sᵢ = −12 cm; 1 for |M| = 0.40 and hᵢ = 0.60 cm; 1 for a full description that matches the sign of sᵢ; 1 for “larger but still reduced” in (d).
Common error: using f = +20 cm, which gives sᵢ = +60 cm and an enlarged real image. A diverging lens cannot do that.
Question 6 (translation between representations · core)
A converging lens has focal length f. An object starts at 4f from the lens and is moved steadily to 1.2f.
(a) Sketch a graph of image distance sᵢ (vertical) against object distance sₒ (horizontal) for this range. Mark the values of sᵢ at sₒ = 4f, 2f and 1.2f. (b) Use your graph to explain how the size of the image changes during the motion. (c) A student says: “If I keep moving the object until it is at 0.8f, the image will just keep moving farther away on the same side.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) From 1/sᵢ = 1/f − 1/sₒ: at sₒ = 4f, sᵢ = 1.33f; at 2f, sᵢ = 2f; at 1.2f, sᵢ = 6f. The graph is a curve that rises ever more steeply as sₒ decreases toward f, and flattens toward sᵢ = f for large sₒ. It passes through (2f, 2f).
(b) |M| = |sᵢ/sₒ|. As sₒ falls, sᵢ rises, so the ratio grows in both ways: |M| goes from 0.33 at 4f, to 1.0 at 2f, to 5.0 at 1.2f. The real, inverted image gets steadily larger (and farther from the lens).
(c) The claim is wrong. As the object reaches f, sᵢ grows without limit: the rays leave the lens parallel and no image forms. Inside f (at 0.8f), 1/sᵢ = 1/f − 1/(0.8f) < 0, so sᵢ = −4f: the image is virtual, on the object’s side, upright and enlarged (|M| = 5). The image does not continue on the same side; it “jumps” from far on one side to the other side.
| Point | What earns it |
|---|---|
| 1 | Curve shape: decreasing sᵢ as sₒ increases, steep near sₒ = f, levelling toward sᵢ = f |
| 1 | Correct marked values: about 1.3f, 2f and 6f |
| 1 | Links image size to sᵢ/sₒ and states that the image grows as the object approaches f |
| 1 | States that at sₒ = f no image forms (parallel rays, sᵢ → ∞) |
| 1 | Shows sᵢ < 0 at 0.8f (for example sᵢ = −4f) and concludes the image is virtual, on the object’s side, so the claim is false |
Accept a ray-diagram argument for (c) in place of the calculation if it shows the refracted rays diverging.
Question 7 (constructed response · stretch)
A lamp illuminates a slide 2.0 cm tall. A screen is fixed 90 cm from the slide. A converging lens of focal length 20 cm can be placed anywhere between them.
(a) Show that there are two lens positions that give a sharp image on the screen, and find them. (b) For each position, find the image height and describe the image. (c) A second student moves the screen so it is only 70 cm from the slide. Show that no lens position now gives a sharp image, and state the minimum slide-to-screen distance for this lens.
Worked solution and suggested Marlbridge rubric
(a) For a real image on the screen, sₒ + sᵢ = 90 cm, so sᵢ = 90 − sₒ. Then 1/sₒ + 1/(90 − sₒ) = 1/20. Multiply through by 20 sₒ (90 − sₒ): 20(90 − sₒ) + 20 sₒ = sₒ(90 − sₒ), so 1800 = 90 sₒ − sₒ², or sₒ² − 90 sₒ + 1800 = 0. The discriminant is 90² − 4(1800) = 900, so sₒ = (90 ± 30)/2: sₒ = 30 cm or sₒ = 60 cm from the slide.
(b)
- sₒ = 30 cm: sᵢ = 60 cm, |M| = 60/30 = 2.0, hᵢ = 4.0 cm. Real (sᵢ > 0), so inverted; enlarged (4.0 cm tall).
- sₒ = 60 cm: sᵢ = 30 cm, |M| = 30/60 = 0.50, hᵢ = 1.0 cm. Real, so inverted; reduced (1.0 cm tall).
The two positions are mirror images of each other: swapping sₒ and sᵢ still satisfies the thin-lens equation.
(c) With D = 70 cm: sₒ² − 70 sₒ + 1400 = 0. The discriminant is 70² − 4(1400) = −700 < 0, so no real solution: no lens position works. In general the equation is sₒ² − D sₒ + D f = 0, which needs D² − 4Df ≥ 0, so D ≥ 4f = 80 cm. At exactly 80 cm there is one position, sₒ = sᵢ = 40 cm = 2f, with a same-size image.
| Point | What earns it |
|---|---|
| 1 | Uses sₒ + sᵢ = 90 cm with the thin-lens equation |
| 1 | Reaches the quadratic sₒ² − 90 sₒ + 1800 = 0 (or equivalent) |
| 1 | Both solutions, 30 cm and 60 cm |
| 1 | Image heights 4.0 cm and 1.0 cm, both inverted and real |
| 1 | Shows a negative discriminant (or equivalent reasoning) for D = 70 cm |
| 1 | Minimum distance 4f = 80 cm, with a justification |
Accept trial and error in (a) if both positions are found and checked in the thin-lens equation. Accept for (c) an argument from the sᵢ–sₒ graph that sₒ + sᵢ is smallest (4f) at sₒ = 2f.
How did you do?
- Q1 or Q5 wrong: re-read the diverging-lens part of “The three principal rays” and Figure 2 in the study guide. Check that f is negative.
- Q2 wrong: redo Worked example 1, taking care to subtract the reciprocals.
- Q3 wrong: read the first misconception again.
- Q4 or Q6 wrong: study the image table and Figure 3.
- Q7 incomplete: start from sₒ + sᵢ = D and build the quadratic one step at a time.
Then tick off the topic checklist.