Specification: AQA A-level Physics (7408) 3.9.1.4 Astrophysics option | Paper: 3 Section B | Data and formulae booklet: θ ≈ λ/D is given in the exam; collecting power ∝ D² is not, so learn it
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Rayleigh criterion: two point sources are just resolved when the central maximum of the diffraction pattern of one coincides with the first minimum of the other. The minimum angular resolution is θ ≈ λ ÷ D, with θ in radians.
Collecting power: a measure of the ability of a telescope to collect radiation. It is proportional to the area of the objective, so collecting power ∝ D² (diameter²).
Quantum efficiency (of a CCD): the percentage of incident photons that liberate an electron. A CCD has a quantum efficiency of about 80%, compared with about 1% for the eye.
The equation
minimum angular resolution ≈ wavelength ÷ diameter of objective θ ≈ λ ÷ D and collecting power ∝ (diameter)² collecting power ∝ D²
| Quantity | Symbol | Unit |
|---|---|---|
| minimum angular separation that can be resolved | θ | radians (rad) |
| wavelength of the radiation | λ | metres (m) |
| diameter of the objective (lens, mirror or dish) | D | metres (m) |
Rearranged: D = λ ÷ θ | λ = θD | ratio of collecting powers = (D₁ ÷ D₂)² | A smaller θ means better resolving power. Convert nm (× 10−9), cm and mm to metres, and keep θ in radians (× π ÷ 180 from degrees).
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A telescope has an objective of diameter 0.20 m and is used with light of wavelength 5.5 × 10−7 m. Calculate its minimum angular resolution, to 2 significant figures.
Show solution
F: θ ≈ λ ÷ D
I: θ ≈ 5.5 × 10−7 ÷ 0.20
F: θ is already the subject
A: 2.8 × 10−6 rad (2 s.f.)
Q2 (Core) Telescope A has a mirror of diameter 8.0 m and telescope B has a mirror of diameter 2.0 m. How many times greater is the collecting power of A than that of B?
Show solution
F: collecting power ∝ D²
I: ratio = (8.0 ÷ 2.0)² = 4.0²
F: ratio is already the subject
A: 16 times
Watch out: answering 4. Collecting power depends on area, so it goes with diameter squared.
Q3 (Core) A radio telescope dish of diameter 76 m observes hydrogen emission of wavelength 21 cm. Calculate its minimum angular resolution in radians and in degrees, to 2 significant figures.
Show solution
Convert first: 21 cm = 0.21 m
F: θ ≈ λ ÷ D
I: θ ≈ 0.21 ÷ 76
F: θ is already the subject
A: 2.76… × 10−3 rad = 2.8 × 10−3 rad (2 s.f.); × 180 ÷ π = 0.16° (2 s.f.)
Q4 (Core) An astronomer needs to resolve two features separated by 1.0 × 10−6 rad using light of wavelength 500 nm. Calculate the minimum diameter of the telescope objective.
Show solution
Convert first: 500 nm = 5.00 × 10−7 m
F: θ ≈ λ ÷ D
I: 1.0 × 10−6 = 5.00 × 10−7 ÷ D
F: D = 5.00 × 10−7 ÷ 1.0 × 10−6
A: 0.50 m
Q5 (Exam) The Hubble Space Telescope has a mirror of diameter 2.4 m. Calculate the diameter of the smallest feature on the Moon it could resolve using light of wavelength 500 nm, to 2 significant figures. (distance to the Moon = 3.84 × 108 m)
Show solution
Convert first: 500 nm = 5.00 × 10−7 m
Step 1 – minimum angular resolution
F: θ ≈ λ ÷ D
I: θ ≈ 5.00 × 10−7 ÷ 2.4 = 2.08… × 10−7 rad
Step 2 – size of feature (small-angle approximation)
size = θ × distance = 2.08… × 10−7 × 3.84 × 108
A: 80 m (2 s.f.)
Q6 (Exam) A small optical telescope of diameter 0.10 m observes light of wavelength 550 nm. Calculate the diameter of radio dish that would give the same resolving power at a wavelength of 21 cm, to 2 significant figures, and explain why radio telescopes are so large.
Show solution
Convert first: 550 nm = 5.50 × 10−7 m; 21 cm = 0.21 m
Step 1 – resolution of the optical telescope
θ ≈ 5.50 × 10−7 ÷ 0.10 = 5.5 × 10−6 rad
Step 2 – dish diameter for the same θ
F: D = λ ÷ θ
I: D = 0.21 ÷ 5.5 × 10−6
A: 38 181… m = 38 km (2 s.f.)
Radio wavelengths are about 105 times longer than visible wavelengths, so for similar resolving power the dish must be far larger. Even the largest single dishes resolve much worse than small optical telescopes. Large dishes also give a much greater collecting power for very weak radio sources.
Q7 (Exam) Telescope A has a diameter of 0.30 m and is used with the eye (quantum efficiency about 1%). Telescope B has a diameter of 0.60 m and is fitted with a CCD (quantum efficiency 80%). Both point at the same faint star for the same time. How many times more photons are detected using telescope B?
Show solution
Step 1 – collecting power ratio
F: collecting power ∝ D²
I: ratio = (0.60 ÷ 0.30)² = 4
Step 2 – detection ratio
photons detected ∝ photons collected × quantum efficiency, so ratio of quantum efficiencies = 80 ÷ 1 = 80
A: 4 × 80 = 320 times more photons detected
Q8 (Exam) Two stars have an angular separation of 2.2 × 10−6 rad. They are observed with a telescope of objective diameter 0.25 m, first through a red filter (650 nm) and then through a blue filter (450 nm). Determine whether the stars are resolved with each filter.
Show solution
Convert first: 650 nm = 6.50 × 10−7 m; 450 nm = 4.50 × 10−7 m
F: θ ≈ λ ÷ D
I: red: θ ≈ 6.50 × 10−7 ÷ 0.25 = 2.6 × 10−6 rad; blue: θ ≈ 4.50 × 10−7 ÷ 0.25 = 1.8 × 10−6 rad
F: compare each θ with the separation of 2.2 × 10−6 rad
A: red: 2.6 × 10−6 > 2.2 × 10−6, so not resolved; blue: 1.8 × 10−6 < 2.2 × 10−6, so resolved
Watch out: a larger θ means worse resolution. The stars are resolved only if their separation is at least θ.
Q9 (Stretch) A binary star system is 25 pc from Earth. The two stars are 20 AU apart, at right angles to the line of sight. (a) Calculate the minimum diameter of telescope needed to resolve them using light of wavelength 550 nm, to 2 significant figures. (b) Show that the stars cannot be resolved by the unaided eye, which has a pupil diameter of 7.0 mm. (1 pc = 3.08 × 1016 m, 1 AU = 1.50 × 1011 m)
Show solution
Convert first: 25 pc = 25 × 3.08 × 1016 = 7.70 × 1017 m; 20 AU = 20 × 1.50 × 1011 = 3.00 × 1012 m; 550 nm = 5.50 × 10−7 m
Step 1 – angular separation (small-angle approximation)
θ ≈ 3.00 × 1012 ÷ 7.70 × 1017 = 3.89… × 10−6 rad
Step 2 – (a) minimum diameter
F: θ ≈ λ ÷ D, so D = λ ÷ θ
I: D = 5.50 × 10−7 ÷ 3.89… × 10−6
A: 0.141… m = 0.14 m (2 s.f.)
Step 3 – (b) the eye
θ ≈ 5.50 × 10−7 ÷ 7.0 × 10−3 = 7.9 × 10−5 rad (2 s.f.), about 20 times larger than 3.89… × 10−6 rad, so not resolved
Q10 (Stretch) The Lovell radio telescope has a dish of diameter 76 m. The FAST radio telescope in China has a dish of diameter 500 m. (a) Calculate the ratio of the collecting power of FAST to that of Lovell, to 2 significant figures. (b) Calculate the minimum angular resolution of FAST at a wavelength of 21 cm, to 2 significant figures. (c) By what factor is this worse than the resolution of a 0.10 m optical telescope at 550 nm, to 2 significant figures?
Show solution
Convert first: 21 cm = 0.21 m; 550 nm = 5.50 × 10−7 m
Step 1 – (a) collecting power
ratio = (500 ÷ 76)² = 43.2…
A: 43 (2 s.f.)
Step 2 – (b) resolution of FAST
F: θ ≈ λ ÷ D
I: θ ≈ 0.21 ÷ 500
A: 4.2 × 10−4 rad (2 s.f.)
Step 3 – (c) comparison
optical: θ ≈ 5.50 × 10−7 ÷ 0.10 = 5.5 × 10−6 rad; factor = 4.2 × 10−4 ÷ 5.5 × 10−6 = 76.3…
A: about 76 times worse (2 s.f.)
Watch out: a bigger dish wins on both collecting power and resolution, but wavelength matters too. Comparing resolution across wavelengths needs λ ÷ D, not D alone.
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