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S4 Physics: Optics Practice Paper 4
S4 Physics: Optics Practice Paper 4
Paper four opens with a 55 degree reflection, a refractive index and a 1.8 m mirror distance, moves to mirror angles, 40 degrees into glass, a 1.6 critical angle and a 15 cm lens, then reverses for index and magnification, ending with total internal reflection.
Quick answer 55 degrees, n = 2.0, 3.6 m, 60 degrees, 25.4 degrees, 38.7 degrees, 22.5 cm and 0.5, 1.49 and 2.01 x 10^8 m/s, 30 cm and 2, total internal reflection.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Physics: Optics – Tutor Lesson Plan |
| Level | S4 |
02 Questions (10 items)
- FoundationA ray hits a plane mirror at 55 degrees to the normal. Find the angle of reflection.
- FoundationLight travels at 1.5 x 10^8 m/s in a medium. Find its refractive index.
- FoundationA person stands 1.8 m in front of a plane mirror. Find the distance to the image.
- FoundationA ray meets a plane mirror at 30 degrees to the normal. Find the angle between the reflected ray and the mirror.
- IntermediateLight goes from air into glass (n = 1.5) at 40 degrees. Find the angle of refraction.
- IntermediateA medium has refractive index 1.6. Find the critical angle into air.
- IntermediateA convex lens of focal length 15 cm has an object 45 cm away. Find the image distance and magnification.
- ChallengeA medium has a critical angle of 42 degrees. Find its refractive index and the speed of light inside.
- ChallengeA convex lens of focal length 10 cm has an object 15 cm away. Find the image distance and magnification.
- Error redoLight goes from water (n = 1.33) into air at 60 degrees. Find the refraction angle. A common mistake gives 49.6 degrees; why is that wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 55 degrees (reflection equals incidence) | Both angles are measured from the normal and are equal. |
| 2 | n = 2.0 (3 x 10^8 / 1.5 x 10^8) | The refractive index is the vacuum speed over the speed in the medium. |
| 3 | 3.6 m (image distance equals object distance) | The image sits as far behind the mirror, so the total is double. |
| 4 | 60 degrees (90 – 30) | The angle of incidence is from the normal, so subtract it from 90. |
| 5 | 25.4 degrees (sin r = sin 40 / 1.5) | Snell law: n1 sin theta1 = n2 sin theta2, with air taken as 1. |
| 6 | 38.7 degrees (sin c = 1 / 1.6) | At the critical angle the refracted ray runs at 90 degrees, so the right side is 1. |
| 7 | Image at 22.5 cm, magnification 0.5 (inverted, diminished, real) | Use 1/f = 1/u + 1/v, then v / u for the magnification. |
| 8 | n about 1.49, speed about 2.01 x 10^8 m/s | Invert sin c = 1/n for n, then divide the vacuum speed. |
| 9 | Image at 30 cm, magnification 2 (inverted, enlarged, real) | With the object between f and 2f, the image is beyond 2f and enlarged. |
| 10 | There is no refracted ray: 1.33 x sin 60 is about 1.15, above 1, so total internal reflection occurs | Going from dense to less dense, check sin r first; a value above 1 means no solution. |
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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