AP · AP 2 · extended response
A camera that must refocus after the object moves
AP Physics 2: Algebra-Based · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Thin-lens calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): 1/v / m⁻¹
Horizontal axis (x): 1/u / m⁻¹
Read the exact data as a table
| 1/u / m⁻¹ | 1/v / m⁻¹ |
|---|---|
| 2 | 8 |
| 4 | 6 |
| 6 | 4 |
| 8 | 2 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Choose your AP course · School year 2026–27 · May 2027 exam
Choose a course to check this question’s reviewed parts and required depth. Its original AP label and topic codes do not establish suitability for the current syllabus.
Choose reviewed AP practice →A converging thin lens forms a sharp real image of a 12 mm-high object placed 150 mm in front of it. The screen is 300 mm behind the lens. Use 1/f = 1/u + 1/v with positive distances for this real-object, real-image setup. The supplied calibration points use reciprocal distances in m⁻¹. Supplied data (1/u / m⁻¹; 1/v / m⁻¹): (2, 8), (4, 6), (6, 4), (8, 2).
- (a)
Find focal length, image height and image orientation. Describe two principal rays that locate the image.
3 marks - (b)
The object moves to 300 mm from the lens. Find the new screen distance and image height; explain what happens if it moves inside the focal length.
3 marks
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(a)
Find focal length, image height and image orientation. Describe two principal rays that locate the image.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
1/f = 1/150+1/300 = 1/100 in mm⁻¹, so f = 100 mm, agreeing with the reciprocal-axis intercept 10 m⁻¹.
- 2
Magnification magnitude is v/u = 300/150 = 2, giving image height 24 mm. A real image from a single converging lens is inverted.
- 3
Draw a ray parallel to the principal axis refracting through the far focus, and a ray through the optical centre continuing straight. Their crossing determines the image position.
(b)
The object moves to 300 mm from the lens. Find the new screen distance and image height; explain what happens if it moves inside the focal length.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
1/v = 1/100−1/300 = 1/150, so the screen must move to 150 mm behind the lens.
- 2
Magnification magnitude is 150/300 = 0.50, giving 6 mm image height. This is a quantitative focusing requirement for a camera or projector design.
- 3
With the object inside f, outgoing rays diverge and only their backward extensions meet. The virtual image can be viewed but not projected directly onto a screen.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 1/f = 1/150+1/300 = 1/100 in mm⁻¹, so f = 100 mm, agreeing with the reciprocal-axis intercept 10 m⁻¹.
- Magnification magnitude is v/u = 300/150 = 2, giving image height 24 mm. A real image from a single converging lens is inverted.
- Draw a ray parallel to the principal axis refracting through the far focus, and a ray through the optical centre continuing straight. Their crossing determines the image position.
- 1/v = 1/100−1/300 = 1/150, so the screen must move to 150 mm behind the lens.
- Magnification magnitude is 150/300 = 0.50, giving 6 mm image height. This is a quantitative focusing requirement for a camera or projector design.
- With the object inside f, outgoing rays diverge and only their backward extensions meet. The virtual image can be viewed but not projected directly onto a screen.
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