Limiting magnitude

Builds on: Sections 5.1 and O1.1
Reading time: 25 min
Exercises: 4
Olympiad problems: 1

On a moonless night far from towns, the faintest stars the naked eye can see are about magnitude 6. Through the telescope of IOAA 2026, stars down to about magnitude 12.7 appear, which are about 460 times fainter. Neither number is exact, since both depend on the darkness of the sky, the steadiness of the air and the eye of the observer, but the difference between them follows from the light the telescope gathers. In this section we work out that difference, and the small losses at each mirror that wear it down.

5.5.1Gaining magnitudes

Before you readMake a guess first

A telescope twice as wide as another collects four times as much light. How many magnitudes fainter can it see?

A star is visible when the light from it that reaches the eye is above some threshold. A telescope of diameter $D$ delivers $(D / d)^{2}$ times as much of the star's light to the eye as the bare pupil of diameter $d$ , by (5.1), so a star $(D / d)^{2}$ times fainter than the naked-eye limit reaches the same threshold. As Section 5.1 showed from the magnitude scale, a factor $x$ in brightness is $2.5 \log_{10} x$ magnitudes, and the factor $(D / d)^{2}$ is $2.5 \log_{10} (D / d)^{2} = 5 \log_{10} (D / d)$ magnitudes.

ResultLimiting magnitude of a telescope(5.8)

m_{lim} = m_{eye} + 5 \log_{10} \frac{D}{d}

$m_{lim}$	the faintest magnitude visible through the telescope
$m_{eye}$	the faintest magnitude visible to the naked eye from the same place, about 6 under a dark sky
$D, d$	the diameter of the objective and of the pupil of the eye, about 7 mm in the dark

The result ignores the light lost in the telescope and the darkening of the sky background at higher magnification. Real telescopes usually reach within about a magnitude of it.

Quick checkThe telescope of IOAA 2026

What is the limiting magnitude of a telescope 150 mm across, if the naked eye reaches magnitude 6.0 with a pupil of 7 mm?

The magnification helps a little as well. A star stays a point however much it is magnified, while the faint glow of the sky behind it is spread over a larger area and looks darker. A star near the limit is therefore easier to see against the darker background of a higher magnification, which is why observers searching for faint stars use a moderately high power.

5.5.2Losses in the telescope

Not all the light that reaches the objective arrives at the eye. A mirror coated with aluminium reflects about 90% of the light, and each lens surface loses a few per cent more. The fraction of the light that gets through all of them is the product of the fractions for each, and since it multiplies the brightness of every star, it shifts the limiting magnitude by $2.5 \log_{10}$ of that product. The small mirror in front of a reflector's main mirror also blocks a fraction $(d_{s} / D)^{2}$ of the incoming light, where $d_{s}$ is its diameter.

Example 5.7An old Newtonian

Each of the two mirrors of a Newtonian reflects 85% of the light. By how much does this lower its limiting magnitude?

Solution

The fraction reaching the eyepiece is $0.85 \times 0.85 = 0.72$ , so the stars look fainter by

2.5 \log_{10} \frac{1}{0.72} = 0.35 magnitudes .

AnswerThe limiting magnitude is about 0.35 magnitudes brighter than for perfect mirrors. Recoating the mirrors wins back most of it.

Common errorsSection 5.5

A brighter limiting magnitude means a smaller number. A telescope that loses light has a smaller $m_{lim}$ .
The light grows with the square of the diameter, so the factor $(D / d)^{2}$ in brightness is $2.5 \log_{10} (D / d)^{2} = 5 \log_{10} (D / d)$ magnitudes.
The losses at each mirror multiply. Two mirrors that each reflect 90% pass $0.9 \times 0.9 = 81 %$ of the light.

ExercisesSection 5.54 questions

5.5.1
What is the limiting magnitude of a 70 mm telescope, if the naked eye reaches magnitude 6.0 with a 7 mm pupil?
5.5.2
How wide must a telescope be to show stars of magnitude 14.0, for an observer who reaches magnitude 6.0 with a 7 mm pupil?
mm
5.5.3
Light passes two mirrors, each of which reflects 90% of it. By how many magnitudes do the stars look fainter than with perfect mirrors?
magnitudes
5.5.4
A small mirror 35 mm across blocks the middle of a 150 mm main mirror. How much of the incoming light does it block?

Olympiad problemsPast IOAA problems, restated1 problem

2019Better mirror coatings8 points

Adapted from IOAA 2019 (Keszthely), theory problem 4. We have restated parts (a) and (b) in our own words and kept the original data. The solution was last checked on 20 October 2026.

A student has a Cassegrain telescope whose two mirrors have aluminium coatings that reflect $ε_{1} = 91 %$ of the light.

(a) By how much does the limiting magnitude change if both coatings are replaced by premium ones that reflect $ε_{2} = 98 %$ ? (5 points)

(b) The student also uses a star diagonal, a flat mirror at 45° that turns the light towards the eye, with the same 91% coating. By how much does the limiting magnitude change if, together with the premium coatings on the two main mirrors, the diagonal is replaced by one that reflects $ε_{3} = 99 %$ ? (3 points)

Ignore any dependence on wavelength and any other losses.

I solved it without the solution

AnswerThe premium mirror coatings raise the limiting magnitude by 0.16, and with the better diagonal as well by 0.25. The original problem also asked whether the change is obvious to the eye, and the official answer says that it is. We have left that part out, since it is a matter of the observer's experience more than of calculation.