A Level Physics 25.1 Standard Candles MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 25.1 of the Cambridge AS and A Level Physics 9702 syllabus, Standard Candles – an A Level (A2) topic in the Astronomy and Cosmology unit. Practise 30 original multiple choice questions on luminosity and intensity, the inverse square law for light, and how standard candles such as Cepheid variables and type Ia supernovae are used to measure the distances to faraway galaxies.
Formula recap
- Intensity at distance d from a star of luminosity L: I = L / (4πd2)
- Distance to a standard candle: d = √(L / (4πI))
- 1 light-year ≈ 9.5 × 1015 m
See the full A Level Physics Formula Sheet for every equation you need for 9702.
Luminosity and Intensity
Q1. What is meant by the luminosity of a star?
- (A) The power received from the star per unit area at the observer
- (B) The total power radiated by the star in all directions
- (C) The power radiated per unit area of the star’s surface
- (D) The apparent brightness of the star as seen in the night sky
Show Answer
Answer: B – Luminosity is the star’s total power output, radiated in all directions, measured in watts. It is a property of the star itself. The power received per unit area at the observer is the intensity, not the luminosity.
Q2. The intensity of the radiation from a star, measured at an observer, is defined as
- (A) the power received per unit area, with unit W m−2
- (B) the total power received, with unit W
- (C) the energy received per unit area, with unit J m−2
- (D) the power received per unit volume, with unit W m−3
Show Answer
Answer: A – Intensity is the power arriving per unit area of the detector, so its unit is the watt per square metre (W m−2). It is sometimes called the observed brightness or flux of the star.
Q3. An astronomer observes the same star from a new location that is much further away from the star. Which statement is correct?
- (A) The luminosity of the star has decreased, because the star looks dimmer
- (B) The luminosity of the star has increased, because less light is lost on the way
- (C) The luminosity of the star is unchanged, but the observed intensity is smaller
- (D) Both the luminosity and the observed intensity are unchanged
Show Answer
Answer: C – The star radiates exactly the same total power wherever the observer stands, so its luminosity cannot change. The light is spread over a larger area at the greater distance, so the intensity (power per unit area) received is smaller and the star appears dimmer.
Q4. Star P appears brighter in the night sky than star Q. What can be concluded about the luminosities of the two stars?
- (A) Star P is definitely more luminous than star Q
- (B) Star Q is definitely more luminous than star P
- (C) The two stars have equal luminosities
- (D) Nothing definite – star P might appear brighter simply because it is closer to Earth
Show Answer
Answer: D – Apparent brightness depends on both luminosity and distance. A dim, nearby star can outshine a brilliant, distant one, so no conclusion about luminosity is possible from brightness alone; the distances are also needed.
Q5. One light-year is
- (A) the time taken for light to travel from the Sun to the Earth
- (B) the distance travelled by light in one year, about 9.5 × 1015 m
- (C) the distance travelled by light in one second, about 3.0 × 108 m
- (D) a distance of 9.5 × 1015 km
Show Answer
Answer: B – Despite its name, the light-year is a unit of distance, not time. Working: distance = speed × time = (3.0 × 108 m s−1) × (365.25 × 24 × 3600 s) ≈ 9.5 × 1015 m.
Q6. A telescope detector of area 0.50 m2 receives starlight of intensity 4.0 × 10−8 W m−2. What is the power received by the detector?
- (A) 8.0 × 10−8 W
- (B) 4.0 × 10−8 W
- (C) 2.0 × 10−8 W
- (D) 1.25 × 107 W
Show Answer
Answer: C – Power = intensity × area = (4.0 × 10−8) × 0.50 = 2.0 × 10−8 W. This is just the definition of intensity rearranged.
Q7. The luminosity of a star is determined by
- (A) the size of the star and the temperature of its surface
- (B) the distance of the star from the Earth
- (C) the sensitivity of the telescope used to observe it
- (D) the amount of dust between the star and the Earth
Show Answer
Answer: A – Luminosity is an intrinsic property of the star: a larger, hotter star radiates more total power. Distance, dust and the telescope affect only the intensity we measure, never the star’s actual output.
Q8. Stars A and B have exactly the same luminosity. Star A is much closer to the Earth than star B. Which statement is correct?
- (A) Star B appears brighter than star A
- (B) The two stars appear equally bright
- (C) Star A has the greater luminosity
- (D) Star A appears brighter, even though the luminosities are equal
Show Answer
Answer: D – Equal luminosities spread over a smaller sphere give a larger intensity at the closer distance, so star A appears brighter. Apparent brightness is not a direct measure of luminosity.
Q9. Why does the intensity of starlight decrease as the light travels further from the star (assuming no absorption)?
- (A) Each photon gradually loses energy as it travels
- (B) The same total power is spread over the surface of an expanding sphere, whose area increases with distance
- (C) Empty space slowly absorbs the light
- (D) The photons slow down as they move away from the star
Show Answer
Answer: B – The total power crossing each sphere around the star is the same (nothing is lost), but the sphere’s surface area 4πd2 grows with distance, so the power per unit area – the intensity – falls.
Q10. The luminosity of the Sun is 3.8 × 1026 W. This means that the Sun
- (A) delivers a power of 3.8 × 1026 W to the Earth alone
- (B) radiates a total energy of 3.8 × 1026 J in its whole lifetime
- (C) radiates a total energy of 3.8 × 1026 J every second, in all directions
- (D) radiates a power of 3.8 × 1026 W from each square metre of its surface
Show Answer
Answer: C – Power is energy per unit time, so a luminosity of 3.8 × 1026 W means 3.8 × 1026 joules radiated every second in all directions. The Earth intercepts only a tiny fraction of this total.
The Inverse Square Law
Q11. A star of luminosity L radiates uniformly in all directions. The intensity of its radiation at a distance d from the star is
- (A) I = L / (4πd2)
- (B) I = L / (4πd)
- (C) I = Ld2 / (4π)
- (D) I = 4πd2 / L
Show Answer
Answer: A – At distance d the power L is spread evenly over a sphere of area 4πd2, so the power per unit area is I = L / (4πd2). This is the inverse square law for light.
Q12. A star has luminosity 4.0 × 1026 W. What is the intensity of its radiation at a distance of 1.0 × 1017 m from the star?
- (A) 3.2 × 10−7 W m−2
- (B) 3.2 × 10−11 W m−2
- (C) 3.2 × 10−9 W m−2
- (D) 1.0 × 10−9 W m−2
Show Answer
Answer: C – I = L / (4πd2) = (4.0 × 1026) / (4π × (1.0 × 1017)2) = (4.0 × 1026) / (1.26 × 1035) = 3.2 × 10−9 W m−2.
Q13. A planet moves from its star to a new orbit at twice the distance. The intensity of the starlight falling on the planet becomes
- (A) one half of its original value
- (B) one quarter of its original value
- (C) twice its original value
- (D) one eighth of its original value
Show Answer
Answer: B – Intensity follows an inverse square law, I ∝ 1/d2. Doubling d makes d2 four times larger, so the intensity falls to 1/4 of its original value.
Q14. An observer moves to a distance 10 times greater from a star. The observed intensity becomes
- (A) 1/10 of the original value
- (B) 1/20 of the original value
- (C) 1/50 of the original value
- (D) 1/100 of the original value
Show Answer
Answer: D – I ∝ 1/d2, and 102 = 100, so at 10 times the distance the intensity is 1/100 of the original value. This rapid fall is why distant stars look so faint.
Q15. At a distance d from a star, the observed intensity is I. At what distance from the star will the observed intensity be I / 9?
- (A) 3d
- (B) 9d
- (C) d / 3
- (D) 81d
Show Answer
Answer: A – Since I ∝ 1/d2, reducing the intensity by a factor of 9 requires d2 to increase by a factor of 9, so the distance must increase by √9 = 3, giving 3d.
Q16. Radiation from a star is measured to have intensity 8.0 × 10−10 W m−2 at a distance of 5.0 × 1016 m from the star. What is the luminosity of the star?
- (A) 2.5 × 1023 W
- (B) 2.5 × 1025 W
- (C) 2.5 × 1027 W
- (D) 4.0 × 1024 W
Show Answer
Answer: B – Rearranging I = L / (4πd2) gives L = 4πd2I = 4π × (5.0 × 1016)2 × (8.0 × 10−10) = 4π × 2.5 × 1033 × 8.0 × 10−10 = 2.5 × 1025 W.
Q17. Stars X and Y are at the same distance from the Earth. The luminosity of X is 4 times the luminosity of Y. How do their observed intensities compare?
- (A) The intensity of Y is 4 times that of X
- (B) The intensity of X is 2 times that of Y
- (C) The intensity of X is 16 times that of Y
- (D) The intensity of X is 4 times that of Y
Show Answer
Answer: D – At the same distance, the factor 4πd2 is identical for both stars in I = L / (4πd2), so the intensity is directly proportional to luminosity: 4 times the luminosity means 4 times the intensity.
Q18. Star M has luminosity 2L and is at distance 2d from an observer. Star N has luminosity L and is at distance d. The intensity of star M, compared with the intensity of star N, is
- (A) the same
- (B) twice as large
- (C) half as large
- (D) one quarter as large
Show Answer
Answer: C – IM = 2L / (4π(2d)2) = 2L / (16πd2), while IN = L / (4πd2). The ratio IM / IN = 2/4 = 1/2, so M appears half as bright: the doubled distance (factor 4 reduction) outweighs the doubled luminosity (factor 2 increase).
Q19. A star has luminosity 3.8 × 1026 W. What is the intensity of its radiation at a distance of 2.0 × 1016 m?
- (A) 7.6 × 10−8 W m−2
- (B) 7.6 × 10−10 W m−2
- (C) 1.9 × 10−8 W m−2
- (D) 3.0 × 10−7 W m−2
Show Answer
Answer: A – I = L / (4πd2) = (3.8 × 1026) / (4π × (2.0 × 1016)2) = (3.8 × 1026) / (5.03 × 1033) = 7.6 × 10−8 W m−2.
Q20. The intensity of light from a star is measured as 6.4 × 10−12 W m−2. If the observer could move to one quarter of the present distance, the intensity would be
- (A) 1.6 × 10−12 W m−2
- (B) 1.0 × 10−10 W m−2
- (C) 2.6 × 10−11 W m−2
- (D) 4.0 × 10−13 W m−2
Show Answer
Answer: B – Reducing the distance to 1/4 multiplies the intensity by 42 = 16 (inverse square law). New intensity = 16 × 6.4 × 10−12 = 1.0 × 10−10 W m−2.
Standard Candles and Measuring Distances
Q21. In astronomy, a standard candle is
- (A) any object whose distance from the Earth is already known
- (B) any object whose observed intensity is the same from all galaxies
- (C) an object whose luminosity is known, or can be determined independently of its distance
- (D) any unusually bright star in the night sky
Show Answer
Answer: C – If the luminosity L of an object is known, then measuring its observed intensity I lets us calculate its distance from d = √(L / (4πI)). Objects with known luminosity act as “standard candles” for measuring cosmic distances.
Q22. Cepheid variable stars can be used as standard candles because their luminosity can be found from
- (A) their period of pulsation, using the period-luminosity relation
- (B) their apparent brightness alone
- (C) the distance to their host galaxy
- (D) the time taken for their light to reach the Earth
Show Answer
Answer: A – Cepheids brighten and dim with a regular period, and the longer the period, the greater the luminosity (the period-luminosity relation). Measuring the period – which does not depend on distance – gives L, and then the measured intensity gives the distance.
Q23. A Cepheid variable has luminosity 1.0 × 1031 W and is observed to have intensity 2.0 × 10−13 W m−2. What is its distance from the Earth?
- (A) 2.0 × 1018 m
- (B) 4.0 × 1042 m
- (C) 2.0 × 1024 m
- (D) 2.0 × 1021 m
Show Answer
Answer: D – d = √(L / (4πI)) = √((1.0 × 1031) / (4π × 2.0 × 10−13)) = √(1.0 × 1031 / 2.5 × 10−12) = √(4.0 × 1042) = 2.0 × 1021 m (about 210,000 light-years). Option B is the value before taking the square root.
Q24. Type Ia supernovae are used as standard candles for very distant galaxies because they
- (A) occur very frequently in every galaxy
- (B) are extremely luminous, with a consistent peak luminosity, so they can still be detected at enormous distances
- (C) are located close to the Earth
- (D) pulsate with a period that reveals their luminosity
Show Answer
Answer: B – A type Ia supernova briefly outshines a whole galaxy and reaches a known, consistent peak luminosity. Because it is so luminous, its light is still measurable after crossing billions of light-years, making it a standard candle for the most distant galaxies. Regular pulsation is the property of Cepheids, not supernovae.
Q25. Dust between a standard candle and the Earth absorbs some of the light on its journey. If this absorption is ignored, the distance calculated for the object will be
- (A) too large, because the object looks dimmer and is therefore assumed to be further away
- (B) too small, because the object looks dimmer and is therefore assumed to be nearer
- (C) unaffected, because absorption does not change the intensity
- (D) unaffected, because absorption changes the luminosity instead
Show Answer
Answer: A – Absorption reduces the measured intensity I. In d = √(L / (4πI)), a smaller I gives a larger calculated d, so the distance is overestimated. Correcting for dust absorption is an important limitation of the standard candle method.
Q26. A standard candle has luminosity 5.0 × 1029 W and observed intensity 4.0 × 10−14 W m−2. What is its distance?
- (A) 1.0 × 1019 m
- (B) 3.2 × 1020 m
- (C) 1.0 × 1021 m
- (D) 1.0 × 1042 m
Show Answer
Answer: C – d = √(L / (4πI)) = √((5.0 × 1029) / (4π × 4.0 × 10−14)) = √(5.0 × 1029 / 5.0 × 10−13) = √(1.0 × 1042) = 1.0 × 1021 m.
Q27. Standard candles used to measure the distances of the most remote galaxies must be exceptionally luminous because
- (A) distant galaxies are moving too fast for dimmer objects to be seen
- (B) dust blocks all light from moderately bright objects
- (C) telescopes cannot detect any light from distant galaxies at all
- (D) intensity falls with the inverse square of distance, so only extremely luminous objects still give a measurable intensity at such distances
Show Answer
Answer: D – Because I = L / (4πd2), doubling the distance quarters the intensity. At distances of billions of light-years, only objects with enormous luminosities – such as type Ia supernovae – produce an intensity large enough to detect and measure.
Q28. An astronomer observes a Cepheid variable in a distant galaxy and measures both its pulsation period and its observed intensity. How is the distance to the galaxy found?
- (A) The period alone gives the distance directly
- (B) The period gives the luminosity from the period-luminosity relation, and this luminosity is combined with the measured intensity in d = √(L / (4πI))
- (C) The intensity alone gives the distance, since all Cepheids appear equally bright
- (D) The distance is found by timing how long the light takes to arrive
Show Answer
Answer: B – Two steps are needed: first the period reveals the true luminosity L (period-luminosity relation), then the measured intensity I is used with the inverse square law to calculate d. Neither the period nor the intensity on its own is enough.
Q29. Because of absorption by dust, the measured intensity of a standard candle is only half of the value it would have at its true distance. If the absorption is ignored, the calculated distance is approximately
- (A) 1.4 times the true distance
- (B) 2 times the true distance
- (C) 0.5 times the true distance
- (D) equal to the true distance
Show Answer
Answer: A – Since d ∝ 1 / √I, halving the measured intensity multiplies the calculated distance by √2 ≈ 1.4. The object is placed about 40% further away than it really is.
Q30. Two Cepheid variables with identical pulsation periods are observed in galaxies G1 and G2. The Cepheid in G2 appears 100 times dimmer than the Cepheid in G1. How do the distances of the two galaxies compare?
- (A) G2 is 100 times further away than G1
- (B) G2 is the same distance away as G1
- (C) G2 is 100 times nearer than G1
- (D) G2 is 10 times further away than G1
Show Answer
Answer: D – Identical periods mean identical luminosities (period-luminosity relation), so both are the same standard candle. With I ∝ 1/d2, an intensity 100 times smaller means d2 is 100 times larger, so G2 is √100 = 10 times further away.
Keep practising: try more topic sections on the Physics MCQs hub and revise every 9702 equation with the A Level Physics Formula Sheet.
These are original practice questions written for revision. They are not Cambridge past paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.