MYP · 11 · extended response
A mirror test with the wrong zero
Light and optics · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Student's raw angle readings
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Reflected reading from surface / °
Horizontal axis (x): Incident angle from normal / °
Read the exact data as a table
| Incident angle from normal / ° | Reflected reading from surface / ° |
|---|---|
| 20 | 70 |
| 35 | 55 |
| 50 | 40 |
| 65 | 25 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A student reports the supplied pairs of incident and reflected ray angles. The incident angle is measured from the normal, but the reflected reading is measured from the mirror surface. Both rays lie in a plane perpendicular to the mirror. Angle-reading uncertainty is ±2°. Supplied data (Incident angle from normal / °; Reflected reading from surface / °): (20, 70), (35, 55), (50, 40), (65, 25).
- (a)
Correct the reflected readings and decide whether the evidence supports the law of reflection.
3 marks - (b)
Describe a ray diagram and an improved procedure. The mirror rotates by 5° with the incident ray fixed: predict the reflected-ray rotation.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Correct the reflected readings and decide whether the evidence supports the law of reflection.
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
An angle from the surface is complementary to an angle from the normal, so r = 90°−raw reading.
- 2
The four corrected angles match the incident angles. Plotting corrected r against i would give unit gradient and zero intercept within the stated uncertainty.
- 3
The raw decreasing relationship is not a violation of reflection; it results from mixing reference directions.
(b)
Describe a ray diagram and an improved procedure. The mirror rotates by 5° with the incident ray fixed: predict the reflected-ray rotation.
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
Draw the mirror, a normal at the incidence point, and incident and reflected rays on opposite sides of the normal with equal angles.
- 2
Mark two well-separated points on each ray, draw thin lines, measure both angles from the same normal and repeat for a range of incident angles; use a narrow low-power beam without eye exposure.
- 3
A mirror rotation of 5° rotates its normal by 5°. Since reflection preserves equality about the normal while the incoming ray is fixed, the outgoing ray rotates by 2×5° = 10°.
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.
- An angle from the surface is complementary to an angle from the normal, so r = 90°−raw reading.
- The four corrected angles match the incident angles. Plotting corrected r against i would give unit gradient and zero intercept within the stated uncertainty.
- The raw decreasing relationship is not a violation of reflection; it results from mixing reference directions.
- Draw the mirror, a normal at the incidence point, and incident and reflected rays on opposite sides of the normal with equal angles.
- Mark two well-separated points on each ray, draw thin lines, measure both angles from the same normal and repeat for a range of incident angles; use a narrow low-power beam without eye exposure.
- A mirror rotation of 5° rotates its normal by 5°. Since reflection preserves equality about the normal while the incoming ray is fixed, the outgoing ray rotates by 2×5° = 10°.
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