- #1
learningastronomy
- 15
- 3
Summary:: An image was taken with a ##60## second exposure time of a 6th magnitude star and the signal to noise ratio was detected to be ##S/N = 20##.
a. What should the exposure time be if you wanted a ##S/N = 100##?
b. Now calculate the ##S/N## if it were a 2nd magnitude star for a ##10## second exposure.
For part a I got the following:
Let ##S = \mu t##, where ##\mu## is the count and ##t## is time, therefore we have $$S/N = 20$$ $$\frac{\mu t}{\sqrt(\mu t)} = 20$$ $$(\frac{\mu t}{\sqrt(\mu t)})^2= 20^2$$ $$\frac{\mu^2 t^2}{\mu t} = 400$$ therefore the count ##\mu## is ##400##. Therefore in order to get ##S/N = 100## we have $$S/N = 100$$ $$\frac{\mu t}{\sqrt(\mu t)} = 100$$ $$(\frac{400t}{\sqrt(400t)})^2= 100^2$$ $$\frac{400^2 t^2}{400t} = 10000$$ now solving for ##t## I got ##1500## seconds.
But for part b of the question, do I apply the same logic even though the magnitude of the star is different? Or will the logic be different?
a. What should the exposure time be if you wanted a ##S/N = 100##?
b. Now calculate the ##S/N## if it were a 2nd magnitude star for a ##10## second exposure.
For part a I got the following:
Let ##S = \mu t##, where ##\mu## is the count and ##t## is time, therefore we have $$S/N = 20$$ $$\frac{\mu t}{\sqrt(\mu t)} = 20$$ $$(\frac{\mu t}{\sqrt(\mu t)})^2= 20^2$$ $$\frac{\mu^2 t^2}{\mu t} = 400$$ therefore the count ##\mu## is ##400##. Therefore in order to get ##S/N = 100## we have $$S/N = 100$$ $$\frac{\mu t}{\sqrt(\mu t)} = 100$$ $$(\frac{400t}{\sqrt(400t)})^2= 100^2$$ $$\frac{400^2 t^2}{400t} = 10000$$ now solving for ##t## I got ##1500## seconds.
But for part b of the question, do I apply the same logic even though the magnitude of the star is different? Or will the logic be different?
