I am generally having a issue with this question as I believe it very ambiguous there been several way I have been explained how to do this, non which make really sense.
Here is the question:
A Type I delta-Cepheid star with a one month (30day) period is put at the distance of the star Vega.
(a) Determine the apparent magnitude this star would have.
(b) Determine how many times more luminous the delta-Cepheid star is compared to Vega.
(c) Assume you observe the above delta-Cepheid star at an observatory on a mountain high enough that the typical air pressure is only 60% of the conditions at sea level. If the star is at an airmass of one, your detector registers 10000 counts from the star. Determine the number of counts your detector would register if the star was positioned at a zenith distance of 60 degrees - assume the star has not changed it's apparent magnitude between the two observations.
I have displayed full question for context purposes it is part c) I am having the issue with.
Here is what I know. I know that as the air mass increases the photon count will decrease which is fine, so by using the equation
$$X=\frac{1 air mass}{cos\theta}$$
so if I was on sea level then as one air mass=100000 counts the at $60^{\circ}$ I would have an air mass of 2 which the at 2 air mass=50000 counts.
Now if I follow that logic, the at an air mass of 0.6 i should have the count rate of $$10 000/0.6=16667$$ which would be on the mountain at zenith, now as the star moves to $60^{\circ}$ my count rate would be at airmass $1.2= 8333$ counts.
But I have an issue with this because in my lecture notes i recently just found this.
so if this is the case the my calculations are wrong because at zero air mass the count rate would be:
$$100 \%\ -82 \%\ =10000$$ $$18\%\ =10000$$ $$1 \%\ =555.56$$ $100 \%\ =55556$counts
so my method above dose not clearly work because 10000/0 is undefined.
I just cant see how I can calculate the relationship.
If anyone could help it would be much apprecaited.