MYP · 12 · extended response
A prism separates colours but preserves frequencies
Electromagnetic spectrum · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Glass dispersion
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Refractive index
Horizontal axis (x): Vacuum wavelength / nm
Read the exact data as a table
| Vacuum wavelength / nm | Refractive index |
|---|---|
| 400 | 1.54 |
| 550 | 1.52 |
| 700 | 1.5 |
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. White light enters glass from air at 30° to the normal. The material's measured refractive indices at several vacuum wavelengths are supplied. Air has index 1.00 and c = 3.0×10⁸ m s⁻¹. Treat the colours independently and use Snell's law. Supplied data (Vacuum wavelength / nm; Refractive index): (400, 1.54), (550, 1.52), (700, 1.5).
- (a)
Compare the refracted angles, glass speeds and glass wavelengths for 400 nm and 700 nm light.
3 marks - (b)
Explain why recombining the spectrum can produce white light and distinguish this from mixing coloured paints.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Compare the refracted angles, glass speeds and glass wavelengths for 400 nm and 700 nm light.
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
Snell's law gives sin r = sin30°/n. The blue/violet value is asin(0.5/1.54)≈18.9° and the red value is asin(0.5/1.50)≈19.5°.
- 2
Speeds are c/n, about 1.95×10⁸ and 2.0×10⁸ m s⁻¹. Higher index gives greater bending toward the normal on entry.
- 3
Frequency is unchanged at entry, so λglass = λvacuum/n, giving about 260 and 467 nm. Dispersion separates directions because n depends on wavelength.
(b)
Explain why recombining the spectrum can produce white light and distinguish this from mixing coloured paints.
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
A second appropriately oriented prism or lens arrangement can recombine the separated spectral components into the original white-light mixture.
- 2
The first prism separates pre-existing frequencies; it does not manufacture colours or convert red photons into blue ones.
- 3
Coloured pigments absorb and scatter different bands, so pigment mixing is subtractive and does not follow the same colour rules as adding light beams.
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.
- Snell's law gives sin r = sin30°/n. The blue/violet value is asin(0.5/1.54)≈18.9° and the red value is asin(0.5/1.50)≈19.5°.
- Speeds are c/n, about 1.95×10⁸ and 2.0×10⁸ m s⁻¹. Higher index gives greater bending toward the normal on entry.
- Frequency is unchanged at entry, so λglass = λvacuum/n, giving about 260 and 467 nm. Dispersion separates directions because n depends on wavelength.
- A second appropriately oriented prism or lens arrangement can recombine the separated spectral components into the original white-light mixture.
- The first prism separates pre-existing frequencies; it does not manufacture colours or convert red photons into blue ones.
- Coloured pigments absorb and scatter different bands, so pigment mixing is subtractive and does not follow the same colour rules as adding light beams.
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