Interaction of dust and starlight - extinction

[henrco]

Hi,

I haven't posted an Astronomy question before. I did see some Astronomy questions in this section so I posted it here. But if it's in the wrong place, for future reference, please let me know where it should be posted.

1) The first problem I need guidance on is the value I have obtained for the optical depth. It seems very high, but checking my calculations carefully I'm sure I haven't made a mistake.

2) The other problem I have is if I am correct in obtaining the final answer by subtracting the change in magnitude from the stars apparent magnitude. Since I need to obtain the magnitude without dust I subtracted the change in magnitude value.

Any guidance very welcome!

1. Homework Statement
A hypothetical interstellar cloud consists of a large number of identical dust grains of radius0.1 μm and which have an extinction efficiency factor 0.75 at wavelengths corresponding to the standard V filter. A star is observed through this cloud and its apparent magnitude is V = 14.2 mag.
The column density of the sightline is Ndust = 5.8 × 10^14 m2.

Suppose that all the grains ofdust were to be removed from the sightline. What would the star’s
apparent V magnitude then be?

Homework Equations

The Attempt at a Solution

 

[mfb]

Something went wrong with powers of 10 towards the end.
1.13*10^-5 is not 1.13*10^-5%, and the opposite is not 99.99%.

The 14.87 should be the change in magnitude, not the new magnitude (note that it does not depend on the old magnitude). The difference between that and the magnitude is not the difference in magnitude.

 

[henrco]

Thank you for the response and guidance.
i) I saw my mistake regarding powers of 10 and corrected it.
ii) The change in magnitude comes out to 14.87

Finally I need to calculate the change in apparent magnitude, the current apparent magnitude is 14.2
and the change in magnitude is 14.87. However I'm hitting a metal block trying to work this out.

Appreciate any guidance to help me calculate the new apparent magnitude.

 

[mfb]

The dust makes the star dimmer, so the star is brighter than we see it. Does that mean you have to add or subtract the two numbers?

 

[henrco]

Thank you for the prompt reply.

The lower the value for apparent magnitude the brighter the object appears.

So I would say you subtract the change in magnitude. So 14.2 - 14.87 = -0.67

 

[mfb]

Right.

 

[henrco]

Thank you.