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S6 Physics: Medical Physics – Endoscope Fibres, Lasers and Imaging Limits (10 Questions)
S6 Physics: Medical Physics – Endoscope Fibres, Lasers and Imaging Limits (10 Questions)
The final paper raises the difficulty: total internal reflection and the critical angle, intensity after focusing, the trade-off between frequency and resolution, the shortest X-ray wavelength, CT dose in everyday terms, T2 decay, fractionated radiotherapy, radiotracer activity, attenuation in decibels, and an error-correction item on how PET actually detects its signal.
Quick answer Indices 1.62 and 1.52 give a critical angle of 69.8 degrees; 8.0 W focused to 0.50 mm gives 4.1 times 10 to the 7 watts per square metre, forty thousand times sunlight; 5 MHz gives 0.31 mm and 10 MHz gives 0.15 mm; 80 kV gives 80 keV and a minimum wavelength of 15.5 pm; 8.0 mSv equals 3.3 years of background, and three scans about 10 years; with T2 of 50 ms the signal at 100 ms is 13.5 percent; 60 Gy in 30 fractions is 2.0 Gy each, taking 20 minutes at 0.10 Gy per minute; 500 MBq after three half-lives is 62.5 MBq; 5 MHz over 4 cm loses 10 dB, leaving one tenth; and two 511 keV photons total 1.022 MeV with a mass of 1.82 times 10 to the minus 30 kg.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Medical Physics: Imaging, Radiation Dose and Radiotherapy |
| Level | S6 |
02 Questions (10 items)
- FoundationAn endoscope fibre has a core of refractive index 1.62 and a cladding of 1.52. Find the critical angle and explain how light travels far along a bent fibre.
- FoundationA continuous laser of 8.0 W is focused to a spot 0.50 mm in diameter. Find the average intensity in watts per square metre and compare it with sunlight at about 1.0 times 10 to the 3.
- FoundationThe speed of sound in tissue is 1540 m/s. Find the wavelength in mm for 5.0 MHz and 10 MHz probes, and say which resolves better and at what cost.
- FoundationAn X-ray tube runs at an accelerating voltage of 80 kV. Find the maximum photon energy in keV and the minimum wavelength using h c over e V, in picometres.
- IntermediateAn abdominal CT gives an effective dose of 8.0 mSv while natural background is about 2.4 mSv per year. How many years of background does one scan equal, and three scans in a year?
- IntermediateA tissue in MRI has a T2 of 50 ms and its signal follows S equals S nought times e to the minus t over T2. What percentage of the signal remains at 100 ms?
- IntermediateA course delivers a total dose of 60 Gy in 30 fractions at a dose rate of 0.10 Gy per minute. Find the dose per fraction and the irradiation time for each fraction.
- ChallengeA patient is given 500 MBq of technetium-99m, half-life 6.0 hours. Find the activity remaining after 18 hours and the percentage that has decayed.
- ChallengeUltrasound attenuation in tissue is about 0.50 dB per cm per MHz. For a 5.0 MHz wave travelling 4.0 cm one way, find the attenuation in dB and the fraction of intensity left.
- Error redoA student writes that PET detects positrons directly and explains the signal with two photons travelling in the same direction. Identify both errors and calculate the total energy of the two 511 keV photons in MeV and the equivalent mass in kg.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The sine of the critical angle is 1.52 over 1.62, which is 0.938, giving about 69.8 degrees; any ray hitting the boundary beyond that angle is totally internally reflected with no energy lost into the cladding, so the light follows the bend | Total internal reflection needs light going from the denser to the less dense medium at an angle greater than the critical angle. |
| 2 | The radius is 0.25 mm, so the area is about 1.96 times 10 to the minus 7 square metres and the intensity is 8.0 divided by that, about 4.1 times 10 to the 7, roughly forty thousand times sunlight | Focusing squeezes the same power into a tiny area, so intensity is set by the area rather than by the power alone. |
| 3 | The wavelength is 1540 divided by 5.0 times 10 to the 6, which is 0.31 mm, and 1540 divided by 10 times 10 to the 6, which is 0.15 mm; the 10 MHz probe resolves better but attenuation rises with frequency so it penetrates less | The resolution limit is roughly one wavelength, so higher frequency gives finer detail but shallower penetration. |
| 4 | The maximum energy is 80 keV, and the minimum wavelength is 6.63 times 10 to the minus 34 times 3.00 times 10 to the 8 divided by 1.60 times 10 to the minus 19 times 8.0 times 10 to the 4, about 1.55 times 10 to the minus 11 m, that is 15.5 pm | The shortest wavelength occurs when an electron gives all its kinetic energy to one photon, and a higher voltage means harder, more penetrating X-rays. |
| 5 | One scan equals 8.0 divided by 2.4, about 3.3 years of background, and three scans give 24 mSv, about 10 years | Expressing effective dose as years of natural background is the easiest comparison to explain to a patient. |
| 6 | The ratio is e to the minus 100 over 50, which is e to the minus 2, about 0.135, that is 13.5 percent | A longer T2 means slower decay, which is why different tissues appear at different brightness at the same instant. |
| 7 | Each fraction is 60 divided by 30, which is 2.0 Gy, and the time is 2.0 divided by 0.10, namely 20 minutes | Fractionation splits the same total dose so that normal tissue has time to repair between sessions. |
| 8 | 18 divided by 6 is three half-lives, so the activity is 500 divided by 8, which is 62.5 MBq, and the decayed fraction is 1 minus one eighth, that is 87.5 percent | After three half-lives only one eighth remains, so imaging must be completed soon after injection. |
| 9 | The attenuation is 0.50 times 5.0 times 4.0, which is 10 dB; every 10 dB is a factor of ten, so one tenth of the intensity remains | The decibel is a logarithmic unit, so adding decibels multiplies or divides intensities and 10 dB means one tenth. |
| 10 | PET detects the 511 keV photons from annihilation, because a positron travels only about 1 mm in tissue before annihilating, not the positron itself; the two photons must leave back to back at 180 degrees to conserve momentum, not in the same direction. The total energy is 1.022 MeV, about 1.64 times 10 to the minus 13 J, giving a mass of about 1.82 times 10 to the minus 30 kg | A positron survives only briefly, so the detectable signal is the opposing pair of photons, which is why PET relies on coincidence counting. |
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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