Specification: AQA A-level Physics (7408) 3.9.1.1 Astrophysics option | Paper: 3 Section B | Data and formulae booklet: given in the exam (both forms of M)
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Angular magnification: M = angle subtended by the image at the eye ÷ angle subtended by the object at the unaided eye. It is a ratio of angles, so it has no unit.
Normal adjustment: the principal focus of the objective lens coincides with the principal focus of the eyepiece, so the final image is formed at infinity. The lenses are then a distance f₀ + fₑ apart and M = f₀ ÷ fₑ.
Objective and eyepiece: the objective is the converging lens with the long focal length that collects light and forms a real image at its principal focus; the eyepiece has the short focal length and acts as a magnifying glass to view that image.
The equation
angular magnification = angle subtended by image at eye ÷ angle subtended by object at unaided eye M = α ÷ β and in normal adjustment M = focal length of objective ÷ focal length of eyepiece M = f₀ ÷ fₑ
| Quantity | Symbol | Unit |
|---|---|---|
| angular magnification | M | no unit (a ratio) |
| angle subtended by the image at the eye | α | radians (rad) |
| angle subtended by the object at the unaided eye | β | radians (rad) |
| focal length of the objective lens | f₀ | metres (m) |
| focal length of the eyepiece lens | fₑ | metres (m) |
Rearranged: α = Mβ | β = α ÷ M | f₀ = Mfₑ | fₑ = f₀ ÷ M | length of telescope in normal adjustment = f₀ + fₑ | For a distant object, angle ≈ diameter ÷ distance in radians (small-angle approximation). Convert degrees to radians with × π ÷ 180, and f₀ and fₑ to the same unit.
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 lens of focal length 1.20 m and an eyepiece of focal length 0.040 m. Calculate its angular magnification in normal adjustment.
Show solution
F: M = f₀ ÷ fₑ
I: M = 1.20 ÷ 0.040
F: M is already the subject
A: 30 (no unit)
Q2 (Core) A planet subtends an angle of 4.0 × 10−4 rad at the unaided eye. Through a telescope, its image subtends 1.2 × 10−2 rad at the eye. Calculate the angular magnification.
Show solution
F: M = α ÷ β
I: M = 1.2 × 10−2 ÷ 4.0 × 10−4
F: M is already the subject
A: 30 (no unit)
Q3 (Core) A refracting telescope in normal adjustment has an angular magnification of 36. The objective has a focal length of 900 mm. Calculate the focal length of the eyepiece and the distance between the two lenses, in metres.
Show solution
Convert first: 900 mm = 0.900 m
F: M = f₀ ÷ fₑ
I: 36 = 0.900 ÷ fₑ
F: fₑ = 0.900 ÷ 36 = 0.025 m (25 mm)
A: fₑ = 0.025 m; separation = f₀ + fₑ = 0.900 + 0.025 = 0.925 m
Q4 (Core) Jupiter subtends an angle of 45 arcseconds at the unaided eye (1 arcsecond = 1/3600 of a degree). It is viewed through a telescope of angular magnification 120. Calculate the angle subtended by the image at the eye in radians, to 2 significant figures.
Show solution
Convert first: 45 ÷ 3600 = 0.0125°; 0.0125 × π ÷ 180 = 2.18… × 10−4 rad
F: M = α ÷ β
I: 120 = α ÷ 2.18… × 10−4
F: α = 120 × 2.18… × 10−4 = 0.0261… rad
A: 0.026 rad (2 s.f.), which is 1.5°
Watch out: leaving the calculator in degrees mode or mixing degrees and radians. Small-angle working must be in radians.
Q5 (Exam) The Moon has a diameter of 3.47 × 106 m and is 3.84 × 108 m from Earth. It is viewed through a telescope in normal adjustment with f₀ = 0.60 m and fₑ = 30 mm. Calculate the angle subtended at the eye by the image of the Moon, to 2 significant figures.
Show solution
Convert first: 30 mm = 0.030 m
Step 1 – angle subtended by the Moon at the unaided eye
β ≈ diameter ÷ distance = 3.47 × 106 ÷ 3.84 × 108 = 9.03… × 10−3 rad (about 0.52°)
Step 2 – angular magnification
M = f₀ ÷ fₑ = 0.60 ÷ 0.030 = 20
Step 3 – angle subtended by the image
F: M = α ÷ β
I: 20 = α ÷ 9.03… × 10−3
F: α = 20 × 9.03… × 10−3 = 0.180… rad
A: 0.18 rad (2 s.f.)
Q6 (Exam) At closest approach Jupiter (diameter 1.43 × 108 m) is 6.30 × 1011 m from Earth. An observer wants the image of Jupiter to subtend at least 1.0° at the eye, using a telescope whose objective has a focal length of 1.0 m. Calculate the minimum angular magnification needed and the maximum focal length of eyepiece that could be used, to 2 significant figures.
Show solution
Convert first: 1.0° = 1.0 × π ÷ 180 = 0.0174… rad
Step 1 – angle subtended by Jupiter at the unaided eye
β ≈ 1.43 × 108 ÷ 6.30 × 1011 = 2.26… × 10−4 rad
Step 2 – minimum magnification
F: M = α ÷ β
I: M = 0.0174… ÷ 2.26… × 10−4
A: M = 76.8…, so a minimum of 77 (2 s.f.)
Step 3 – maximum eyepiece focal length
fₑ = f₀ ÷ M = 1.0 ÷ 76.8… = 0.0130… m
A: 0.013 m (13 mm, 2 s.f.). A longer eyepiece focal length would give a smaller magnification.
Q7 (Exam) A student builds a telescope in normal adjustment from two converging lenses. The lenses are 1.05 m apart and the angular magnification is 20. Calculate the focal length of each lens.
Show solution
Step 1 – write two equations
f₀ + fₑ = 1.05 and M = f₀ ÷ fₑ = 20, so f₀ = 20fₑ
Step 2 – solve
I: 20fₑ + fₑ = 1.05, so 21fₑ = 1.05
F: fₑ = 1.05 ÷ 21 = 0.050 m
A: eyepiece fₑ = 0.050 m (50 mm); objective f₀ = 20 × 0.050 = 1.00 m
Q8 (Exam) A telescope has lenses of focal length 0.80 m and 40 mm. (a) Calculate its angular magnification in normal adjustment. (b) A student looks through the wrong end, so the 0.80 m lens is used as the eyepiece. Calculate the angular magnification now and describe the image.
Show solution
Convert first: 40 mm = 0.040 m
(a) F: M = f₀ ÷ fₑ
I: M = 0.80 ÷ 0.040
A: 20
(b) I: M = 0.040 ÷ 0.80
A: 0.050, so the image subtends a smaller angle than the object: it looks 20 times smaller (and is still inverted)
Watch out: the objective must be the lens with the longer focal length. Swapping them gives M = 1/20, not 20.
Q9 (Stretch) The lunar crater Copernicus has a diameter of 93 km. The Moon is 3.84 × 108 m from Earth. The unaided eye can just distinguish two points that subtend 3.0 × 10−4 rad. (a) Show that Copernicus cannot be seen as a crater with the unaided eye. (b) It is viewed through a telescope in normal adjustment with f₀ = 1.20 m and fₑ = 25 mm. Calculate the angle subtended by its image, to 2 significant figures. (c) Calculate the diameter of the smallest crater the observer could see through this telescope, to 2 significant figures. Assume the eye, not diffraction, limits what can be seen.
Show solution
Convert first: 93 km = 9.3 × 104 m; 25 mm = 0.025 m
Step 1 – (a) angle at the unaided eye
β ≈ 9.3 × 104 ÷ 3.84 × 108 = 2.42… × 10−4 rad. This is less than 3.0 × 10−4 rad, so the crater cannot be distinguished with the unaided eye.
Step 2 – (b) magnification and image angle
M = f₀ ÷ fₑ = 1.20 ÷ 0.025 = 48
α = Mβ = 48 × 2.42… × 10−4 = 0.0116… rad
A: 0.012 rad (2 s.f.)
Step 3 – (c) smallest crater
the image must subtend at least 3.0 × 10−4 rad, so β = 3.0 × 10−4 ÷ 48 = 6.25 × 10−6 rad
diameter = β × distance = 6.25 × 10−6 × 3.84 × 108 = 2400 m
A: 2.4 km (2 s.f.)
Q10 (Stretch) Ganymede orbits Jupiter at a radius of 1.07 × 109 m. Jupiter (diameter 1.43 × 108 m) is 6.30 × 1011 m from Earth. Jupiter is viewed through a telescope in normal adjustment with f₀ = 0.96 m and fₑ = 30 mm. When Ganymede is at its greatest angular distance from Jupiter, calculate, to 2 significant figures: (a) the angle between Ganymede and the centre of Jupiter at the unaided eye; (b) this angle as seen through the telescope; (c) the angle subtended by the image of Jupiter’s disc. (d) How many disc diameters from the centre of Jupiter does Ganymede appear?
Show solution
Convert first: 30 mm = 0.030 m, so M = 0.96 ÷ 0.030 = 32
Step 1 – (a) unaided angle
β ≈ 1.07 × 109 ÷ 6.30 × 1011 = 1.69… × 10−3 rad
A: 1.7 × 10−3 rad (2 s.f.)
Step 2 – (b) through the telescope
α = Mβ = 32 × 1.69… × 10−3 = 0.0543… rad
A: 0.054 rad (2 s.f.), about 3.1°
Step 3 – (c) Jupiter’s disc
unaided: 1.43 × 108 ÷ 6.30 × 1011 = 2.26… × 10−4 rad; through telescope: 32 × 2.26… × 10−4 = 7.26… × 10−3 rad
A: 7.3 × 10−3 rad (2 s.f.)
Step 4 – (d) ratio
0.0543… ÷ 7.26… × 10−3 = 7.48…, the same as 1.07 × 109 ÷ 1.43 × 108
A: about 7.5 disc diameters (2 s.f.)
Watch out: the telescope multiplies every angle by M, so ratios of angles (and the shape of the scene) do not change.
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