Problems for Chapter 4

Builds on: Sections 4.1 to 4.9
Reading time: 45 min
Exercises: 0
Olympiad problems: 3

Olympiad problems on light rarely ask for one formula at a time. The past problems in the earlier sections of this chapter each used the ideas of one or two sections. The three in this section join them up. The first follows the fading of Supernova 1987A with the magnitude scale, an exponential law and the limiting magnitude of a telescope. The second puts an observer inside a star cluster and asks how bright the cluster looks from there, which needs the inverse-square law and surface brightness together. The third finds the temperature of a star from its light at two wavelengths with the Planck law.

4.10.1From textbook to olympiad

In the first problem below, the supernova faded from magnitude 3 to the limit of the naked eye in 265 days, and we are asked how long it stayed within reach of a small telescope. The method of Section S2 takes us from the chapter to the solution in these steps.

We start by drawing it. A sketch of magnitude against time, with time across and magnitude up the side, shows the supernova falling from 3 at the start to 6 after 265 days, and a later point where it reaches the telescope's limit.
Next we name the physics. The brightness falls by the same factor in each day, $B = B_{0} e^{- t / τ}$ . By (4.2), a magnitude is $- 2.5 \log_{10}$ of a brightness, so $m - m_{0} = 2.5 \log_{10} (e) t / τ = 1.086 t / τ$ . The magnitude grows in a straight line with time, and the sketch is a straight line.
Then we work backwards from the target. The last day of visibility is the day the line reaches the telescope's limiting magnitude, so we need that limit, which comes from comparing the light the telescope collects with the light the eye collects, as in Section 5.5.
We use every given. The transmission of 70 per cent reduces the light the telescope delivers, and the diameter of the pupil sets the light the eye collects on its own.
Earlier parts are hints. Part (a) finds $τ$ , but part (b) does not really need it. Since the magnitude grows in a straight line, 3 magnitudes took 265 days and 9.64 magnitudes take $265 \times 9.64 / 3$ days, whatever $τ$ is.
We finish with a check. The supernova stayed visible in the telescope more than three times as long as to the eye, which is reasonable for a telescope that gathers several hundred times as much light, a gain of 6.6 magnitudes.

Olympiad problemsPast IOAA problems, restated3 problems

2017The fading of Supernova 1987A15 points

Adapted from IOAA 2017 (Phuket), theory problem 6. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

Supernova 1987A was at its brightest, magnitude +3, on about 15 May 1987. It then faded, and by 4 February 1988 it could no longer be seen with the naked eye, whose limit is magnitude +6. Assume that its brightness $B$ fell exponentially with time $t$ , $B = B_{0} e^{- t / τ}$ , where $B_{0}$ and $τ$ are constants.

(a) Find $τ$ in days. (5 points)

(b) Find the last day on which the supernova could have been seen with a telescope 15.24 cm across that passes 70 per cent of the light falling on it. Take the diameter of the pupil of the eye as 0.6 cm. (10 points)

I solved it without the solution

AnswerThe time constant is about 96 days, and the supernova could be seen in the telescope until about 12 September 1989.

2012A biologist inside a globular cluster

Adapted from IOAA 2012 (Rio de Janeiro), theory problem L1. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

An astronomer on Earth observes a globular cluster of angular diameter $α$ , in radians, at distance $D$ . It holds $N$ stars, all of the same absolute magnitude $M_{0}$ , spread evenly through a sphere. A biologist is at the centre of the cluster.

(a) What is the difference between the combined magnitude of all the stars as the biologist measures it and as the astronomer measures it?

(b) What diameter of telescope would the astronomer need to see the cluster with the same total brightness that the biologist sees with the naked eye? Take the pupil of the eye as 6 mm.

I solved it without the solution

AnswerThe biologist sees the cluster $2.5 \log_{10} (12 / α^{2})$ magnitudes brighter than the astronomer does, a telescope of diameter $0.021 / α$ metres would make up the difference, and through a field of diameter $α$ the biologist would see $- 2.5 \log_{10} (12 \sin^{2} (α / 4) / α^{2})$ , about 0.31 magnitudes fainter than the astronomer.

2014A temperature from two wavelengths

Adapted from IOAA 2014 (Suceava), theory problem P7. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

The light of a star is measured in two narrow bands of the same width, at wavelengths $λ_{1}$ and $λ_{2}$ , and the flux for each unit of wavelength is $I_{1}$ in the first and $I_{2}$ in the second. Treat the star as a blackbody obeying the Planck law.

(a) Write down the equation that gives the star's temperature $T$ from these measurements.

(b) Find $T$ explicitly when $h c / λ k T$ is much more than 1 at both wavelengths.

(c) When $h c / λ k T$ is much less than 1 at both wavelengths and $I_{1} = 2 I_{2}$ , how are $λ_{1}$ and $λ_{2}$ related?

I solved it without the solution

AnswerThe general equation is the ratio of two Planck curves. When $h c / λ k T$ is large, $T = h c (λ_{1} - λ_{2}) / (k λ_{1} λ_{2} \ln [(I_{1} / I_{2}) (λ_{1} / λ_{2})^{5}])$ , and when it is small, $I_{1} = 2 I_{2}$ requires $λ_{2} = 2^{1 / 4} λ_{1} \approx 1.19 λ_{1}$ .