Homework # 1 Solutions
1. Show that an extinction of Aλ = 1 magnitude leads to a flux Fλ decrease to
40% of its original value.
1 magnitude is equivalent to a reduction of intensity of 10−2/5 = 0.398.
2. Assuming dust grains are 0.1µ m in radius, that the gas density in the ISM is
nH = 1 cm−3 , and the number density of dust grains is nd = 10−12 cm−3 and
uniformly distributed, find the column density NH of hydrogen to be expected
along a line of sight in which a source is observed to have an extinction AV = 1
magnitude. What is the mean free path of light along this line of sight? What
is the distance to the source?
The mean free path ℓ = (nσ)−1 where n is the number density of scatterers/absorbers and σ is the effective cross section. For dust interacting with
light which has a wavelength smaller than the dust size, the effective cross
section is πrd2 . Then
1
ℓ=
≃ 1.001 kpc.
nd πrd2
To get an extinction of A magnitudes requires a path length d given by
2.5 d
F (d)
=
.
A = −2.5 log10
F (0)
e ℓ
Thus d = Aℓe/2.5 = 1.09 kpc. The column density is
NH = nH d = 3.47 × 1021 A.
The book gives a coefficient of 1.8 reflecting different assumptions about the
ISM density of gas and/or dust.
3. Assume a star cluster has 200 F5 stars at the main sequence turnoff and 20
K0III giant stars. Determine the quantities MV and B − V for each type of
star singly and also for the cluster. Suppose the cluster is at a distance of 2
kpc. What is mV for the cluster?
F5 have MV = 3.3, B − V = 0.41 and MB = 3.71. K0 have MV = 0.7, B −
V = 1.02 and MB = 1.72.
The general formula
L = 10(C−M )/2.5
where C is a constant is useful. Then
Ltot,V = 200 × 10(C−3.3)/2.5 + 20 × 10(C−0.7)/2.5 = 10(C−Mtot,V )/2.5 .
The constant C can be cancelled, and one finds Mtot,V ≃ −3.25. Similarly, one
finds Mtot,B = −2.57. Then the cluster B − V = −0.68.
The apparent visual magnitude is
d
mtot,V = Mtot,V + 5 log10
= −3.25 + 5 log10 200 = 8.26.
10 pc
4. In a galaxy at a distance of d Mpc, what would be mB⊙ ? In this galaxy, what
length does 1 arcsec correspond to? If the surface brightness of this galaxy is
IB = 27 mag arsec−2 , show that the brightness of this galaxy is equivalent to
0.99 LB⊙ pc−2 . Why doesn’t this quantity depend on the distance?
6 10 d
mB⊙ = MB⊙ + 5 log10
= 5.48 + 25 = 30.48.
10 pc
The length corresponding to the angular distance θ = 1 arcsec is
6 10 d
z=θ
AU = 106 AU = 4.85 pc.
pc
Then we see the equivalence of 27 mag per square arcsecond and
L⊙
2
(4.85 pc)
× 10(30.48−27)/2.5 ≃ 1.05
L⊙
.
pc2
This relation is independent of distance because both quantities scale the same:
with distance squared. Had we chosen the distance to be different, the relation
would have worked out to be the same.