# Photometry Questions

Here are some questions to get you started. Additional questions will be added over the next few days so check back soon. Click on the blue answer tags to get answers (and worked solutions for some questions).

1. Canopus, α Carinae, has an apparent magnitude of -0.62 whilst Mimosa, β Crucis, has an apparent magnitude of +1.25.
a) Which appears brightest?
b) Which is intrinsically more luminous?

a) Canopus is brighter than Mimosa as it has a lower value for apparent magnitude (-0.62 b) With only apparent magnitude information provided it is impossible to state which of the two stars is intrinsically more luminous. You would need additional information such as absolute magnitude, distance, or spectral and luminosity classes.
2. Sargas, θ Scorpius, has an apparent magnitude of +1.86 and an absolute magnitude of -2.75 whilst Mintaka, δ Orionis, has an apparent magnitude of +2.25 and an absolute magnitude of -4.99.
a) Which star is intrinsically more luminous?
b) How many more times luminous is it than the other star?
c) Which star is more distant?

a) Mintaka is intrinsically more luminous as it has a lower (more negative) value of absolute magnitude, -4.99 < -2.75.
b) Now to compare two stars' luminosities we can rewrite equation 4.5 to give:
LM/LS = 100(MS - MM)/5

so substituting in:

LM/LS = 100(-2.75 - (-4.99))/5
LM/LS = 1000.448
LM/LS = 7.9
∴ Mintaka is about 8 × more luminous than Sargas.

c) Rather than actually calculate the exact distance to each star based on its apparent and absolute magnitudes we can simply determine the distance modulus, m - M for each. Mintaka has a distance modulus of +2.25-(-4.99) = 7.24 whilst Sargas has a distance modulus = +1.86-(-2.75) = 4.61. As Mintaka has a higher distance modulus (7.24 > 4.61) it is more distant than Sargas. Published values give a distance of 83 ± 6 parsecs to Sargas and 281 ± 65 parsecs to Mintaka.

3. η Cen has an apparent magnitude of +2.29 and an absolute magnitude of -3.02. How far away is it?

Using M = m - 5 log(d/10) we can rewrite it so;

d = 10 (m - M + 5)/5

so substituting in:

d = 10 (2.29 - (-3.02) + 5)/5

d = 10 2.062

∴ distance to η Cen = 115 parsecs.
4. The star ε Eridanus has an apparent visual magnitude of V = 3.73 and an apparent blue magnitude of B = 4.61.
a) What is its colour index?
b) What is its approximate spectral class?

a) Colour index, CI = B-V so for ε Eri CI = 4.61 - 3.73 = +0.88.
b) A colour index of +0.88 would place ε Eri in the K spectral class. (It is actually a K2V star). Refer here to review this.
5. Calculate the missing information and complete this table:

Photometry question 5 table.
Star m M Distance Modulus d
(pc)
V B Colour Index Approx Spectral Class
Kruger 60   11.76 9.79 11.25
Aldebaran,
α Tau
0.87     20 0.87   +1.54
τ Cet        3.64   4.21 +0.72 G
Bellatrix,
γ Ori
1.64 -2.72 4.36       -0.22

Star m M Distance Modulus d
(pc)
V B Colour Index Approx Spectral Class
Kruger 60 9.79 11.76
-1.97
4.04
9.79
11.25
+1.46
M
Aldebaran,
α Tau
0.87 -0.64
+1.51
20
0.87
2.41
+1.54
K
τ Cet  4.14 6.34
-2.20
3.64
4.14
4.21
+0.72
G
Bellatrix,
γ Ori
1.64 -2.72
+4.36
74.5
1.64
1.42
-0.22
B
6. The Sun has an absolute magnitude of +4.8.
a) What would be its apparent magnitude if it were at the distance of &alpha Cen (1.3 pc)?
b) Would it still be visible to our unaided eyes?
c) What distance would it have to be to be just detectable to the unaided eye?

a) Rewriting equation 4.2 we get:

m = M + 5 log(d/10) so
m = 4.8 + 5 log(1.3/10)
m = 4.8 + (-4.43)

mSun = 0.37

b) An apparent magnitude of +0.37 would make the Sun easily visible in the night sky to unaided eyes.
c) A star just visible to the unaided eye has an apparent magnitude of about +6.5. For a star such as the Sun with an absolute magnitude of +4.8, this corresponds to a distance, d of:

d = 10 (m - M + 5)/5
d = 10 (6.5 - 4.5 + 5)/5
d = 10 1.4

d = 25 parsecs.
7. The Sun has a luminosity of 3.86 × 1026 W and an effective temperature of 5.78 × 103 K. Calculate its diameter. (use a value of σ = 5.67 × 10-8 W m-2 K-4 for Stefan's constant.

Using equation 4.6: L ≈ 4πR2σT4 then
r2 = L/(4πσT4
r = (L/(4πσT4)1/2
and ∴ diameter = 2r = 2(L/(4πσT4)1/2
so diameter = 2r = 2(3.86 × 1026/(4π × 5.67 × 10-8 × (5.78 × 103)4))1/2

and ∴ Sun's diameter = 1,390,000 km (or 1.39 × 109 m).
8. If a star is 20 parsecs away and has an apparent magnitude of +6, what is its absolute magnitude?

Rather than try and immediately use a formula to solve this let us consider the underlying concepts.
This is basically an application of the inverse-square law of light.
If the star was at at distance of 10 pc it would be twice as close as at 20 pc and therefore four times as bright. A difference of 2.5× in brightness is roughly one magnitude difference and a 6.25× difference in brightness (2.5)2 corresponds to two magnitudes difference. 4× difference in brightness is thus about 1½ magnitudes difference. If its actual apparent magnitude at 20 pc is +6 its apparent magnitude at 10 pc (ie its absolute magnitude by definition) is 6 - 1½ = 4½. Note the magnitude difference is subtracted because the star is closer and therefore brighter at 10 pc.
If we use equation 4.3 we have:

M = m - 5 log(d/10) so;
M = 6 - 5 log(20/10)

which gives:

M = 4.49

which is the same as the answer obtained by the reasoning above.

9. Will an M-class star appear brighter or dimmer when viewed through a red filter than a blue filter? Justify your answer.

An M-class star is a cool, reddish star with a positive colour index. A red filter lets red light but not blue light through. The star will thus have more light coming through a red filter than through a blue filter. It will therefore appear brighter when viewed through a red filter than a blue filter. Its apparent red magnitude, R will thus be lower than its apparent blue magnitude, B.
10. a) Why is spectroscopic parallax useful?
b) Outline how it is applied.
c) Discuss its limitations.

a) Spectroscopic parallax is a useful way to determine an approaximate distance to a star that is too far away to have its trigonmetric parallax measured directly.
b)
1. If we take a spectrum of a star we can determine:
1. Its spectral class.
2. Its luminosity class.
2. Using photometry we can measure the apparent magnitude, m, mV or V for the star.
3. If we use B and V filters we can also determine the blue apparent magnitude, B and thus determine the star's colour index, CI = B - V.
4. Knowing either the star's spectral class or colour index allows us to place the star on a vertical line or band along a Hertzsprung-Russell Diagram. If we also know its luminosity class we can further constrain its position along this line, that is we can distinguish between a red supergiant, giant or main sequence star for example.
5. Once we know its position on the HR diagram we can infer what its absolute magnitude, M should be by either reading off across to the vertical scale of the HR diagram or looking it up from a reference table. A main sequence (luminosity class V) star with a colour index of 0.0 (ie A0 V) has an absolute magnitude of +0.9 for example.
6. Now knowing m from measurement and inferring M we can use the distance modulus equation:
m - M = 5 log(d/10) (4.2)
to find the distance to the star, d, in parsecs.

c) Spectroscopic parallax provides an imprecise result due to uncertainties in absolute magnitudes of stars of specific spectral and luminosity classes of about 0.7 to 1.25 magnitudes. This results in a distance uncertainty of 1.4 to 1.8× for individual stars. Other factors such as interstellar reddening also need to be corrected for if using colour index as a basis for estimating spectral class. Absorption and scattering also affect measured values of m.

11. Describe three advantages of using a CCD detector in place of a photographic plate on a large professional reflecting telescope.

There are several advantages in using a photoelectric detector such as a CCD rather than photographic plates on a modern large reflecting telescope. Three of these are:
1. High quantum efficiency. Modern astronomical CCDs may have peak QEs of >90% and >60% over a wide spectral waveband. This means they are more efficient than photographic plates when imaging very faint sources.It also means that exposure times can be much shorter if going to the same magnitude limit as a photographic plate.
2. Linear response. Unlike photographic plates that suffer from reciprocity failure (increasing lack of sensitivity with increased exposure time), CCDs have a linear response to exposure time and incident light. Thus twice the light on a CCD causes double the charge in the pixels. CCDs are linear to about 0.1% making them well-suited to photometric applications.
3. Broad spectral response. CCDS are more sensitive in the red part of the spectrum than most photographic emulsions. They are now also sensitive in the blue and UV parts of the spectrum. They therefore can work with a broader range of filters than photographic emulsions making them more suitable for multi-filter (eg 5-colour, UBVRI) photometry.
Note there are other advantages. Review them here.
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