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S4 Physics refraction paper 4: ten common slips
S4 Physics refraction paper 4: ten common slips
Ten items rewritten from the usual slips: multiplying instead of dividing, angles not taken from the normal, apparent depth taken as real depth.
Direct answers Items 01 to 03 retry Snell law and speed, 04 to 06 cross-media work, critical angle and apparent depth, 07 to 09 fibres, wavelength and two-media critical angle, and 10 multiplying by the index.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Physics Refraction and Total Internal Reflection |
| Level | S4 |
02 Questions (10 items)
- FoundationLight goes from air (n = 1.00) into water (n = 1.33) at 45 degrees to the normal. Find the angle of refraction.
- FoundationA glass has refractive index 1.5. Find the speed of light inside it.
- FoundationLight travels at 1.8 x 10^8 m/s in a medium. Find its refractive index.
- IntermediateLight in glass (n = 1.5) meets a glass-water boundary (water n = 1.33) at 50 degrees. Find the angle in the water.
- IntermediateA glass has refractive index 1.6. Find the critical angle for light leaving it into air.
- IntermediateA pool is 2.4 m deep and the water has refractive index 1.33. How deep does the bottom look from directly above?
- ChallengeA fibre has core index 1.50 and cladding index 1.45. Find the critical angle at the core-cladding boundary.
- ChallengeA monochromatic light of wavelength 589 nm in air enters glass of refractive index 1.52. Find its wavelength in the glass.
- ChallengeLight passes from a medium of index 1.8 into one of index 1.2. Find the critical angle.
- Error redoLight enters glass (n = 1.5) from air at 20 degrees. A student writes 30.9 degrees; find the correct angle.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 32.1 degrees | 1.00 x sin 45 = 1.33 x sin r, so sin r = 0.707 / 1.33 = 0.532 and r is about 32.1 degrees, both angles taken from the normal. |
| 2 | 2.00 x 10^8 m/s | n = c / v gives v = 3.0 x 10^8 / 1.5 = 2.0 x 10^8 m/s; a larger index always means a slower speed. |
| 3 | 1.67 | n = 3.0 x 10^8 / 1.8 x 10^8 = 1.67; the answer carries no unit because it is a ratio of two speeds. |
| 4 | 59.8 degrees | 1.5 x sin 50 = 1.33 x sin r, sin r = 1.149 / 1.33 = 0.864, so r is 59.8 degrees, larger than the incident angle as expected going into the less dense medium. |
| 5 | 38.7 degrees | sin C = 1 / 1.6 = 0.625, so C is about 38.7 degrees; a critical angle only exists when light leaves a denser medium. |
| 6 | 1.8 m | Apparent depth is real depth divided by n: 2.4 / 1.33 is about 1.8 m, so the bottom looks shallower than it is. |
| 7 | 75.2 degrees | Going from denser to less dense, sin C = n2 / n1 = 1.45 / 1.50 = 0.967, so C is about 75.2 degrees and larger angles are totally reflected. |
| 8 | 388 nm | Wavelength is 589 / 1.52, about 388 nm; the frequency is unchanged, so the light keeps its colour. |
| 9 | 41.8 degrees | Neither medium is air, but sin C = n2 / n1 = 1.2 / 1.8 = 0.667 still applies, giving 41.8 degrees; a critical angle exists whenever n2 is below n1. |
| 10 | 13.2 degrees | sin r = sin 20 / 1.5 = 0.228, so r is about 13.2 degrees; the slip was multiplying by n, and entering glass the ray must bend towards the normal, so r is below 20 degrees. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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