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When we learned more about nebulas and in particular about M42 we decided to solve the following task: find how the fact that the stars are embedded in the nebula makes their brightness fade.

As we know star brightness decreases due to absorption of its radiation by the interstellar medium. That’s why the magnitude of a star increases by following laws:

V - M V = -2,5lg(E/E 0 )
V= M V – 5 + 5lg(r) + A

” A ” is the interstellar matter absorption of the magnitude

A= V - M V + 5 – 5lg(r)

So if we know the distance “ r ” and magnitudes “ V ” and “ M V ” we can find “ A ”.

There is one open cluster in the center of M42 called “Trapezium” (Theta Orionis). Its diameter is 3 ly and its coordinates are:

R.A.= 5h 35,29m (2000,0)
Dec. = -5deg 23,27’ (2000,0)
Stars:
A= 6,7 - 7,7 mag
B= 7,9 mag
C= 5,1 mag
D= 6,7 mag

To determine the visiual magnitude of these four star (which are located in the open cluster "Trapezium") we used the Pogsun's law:

V 1 -V 2 = -2,5lg(E 1 /E 2 )

and

E 1 /E 2 =2,5 V 2 -V 1

If we now the visual magnitudes of two stars we can find the total visual magnitude of these two stars.

E 2 = E 1 / 2,5 V 2 -V 1

E total = E 1 + E 2

E total = E 1 + E 1 / 2,5 V 2 -V 1 = E 1 (1 + 2,5 V 2 -V 1 )/(2,5 V 2 -V 1 )

V 1 -V total = -2,5lg(E 1 /E total )

V total = V 1 + 2,5lg(E 1 /E total ) = V 1 + 2,5lg(E 1 / (E 1 (1 + 2,5 V 2 -V 1 )/(2,5 V 2 -V 1 )))

V total = V 1 + 2,5(V 2 - V 1 )x lg (5/7)

and V 1 >V 2.

We calculated V total = 3,8 mag . This is number is not quite accurate because there are more than these four stars (but we didn't find a definite visual magnitude for this cluster).
" M V ” is not stated so we decided to use the following law:

M V1 - M V2 = 2,5lg(L 1 /L 2 )

We know M V of the Sun so we can write the law in this form:

M V – 4,8 = -2,5lg(L/L SUN )

The total luminosity of the starts in the “Trapezium” is about 3 x 10 5 times that of the Sun.

M V = 4,8 – 2,5lg(3 x 10 5 )

M V = -8,9 mag

The distance to Theta Orion varies from 1400 to 1600 ly according to different sources. So we use the mean distance of 1500 ly which is equal to 460 pc.

We can already calculate the absorption “ A ”.

A = V - M V + 5 – 5lg(r)
A = 3,8 – (-8,89) +5 –5lg(460)
A = 4,4 mag

The hypothesis of Trumpler (an American astronomer) is that the absorption of the interstellar medium is 0,67 mag per 1 kpc. So we can expect an absorption only of about 0,3 mag for the distance of 460 pc. The rest 4,1 magnitudes are due just to M42! If the "Trapezium" hadn't been embedded in the Great nebula it would have been brighter than Rigel and Betelgeuse.