Question about scintilation detector

[Drao92]

Hi guys,
I have some questions about this detector.
So i have a project about this detector, i made some experiments, made some graphs, answering to some questions, but one questions is struggling me.
How can we avoid to calculate the solid angle, i mean in what situations?
In my lab documents i have this formula: omega`/4*pi=1/2(1-1/(sqrt(1+b^2/4D^2))
where D is the distance from detector and radioactive source, b is the diametre of circular window of the detector.
However, i used a radioactive source with bigger diametre than b so i must mutiple that formula with 8D^2/d^2(1-1/(sqrt(1+d^2/4D^2)) where d is the diametre of the source
So omega/4*pi=omega`*4pi/the above formula^.
So, this is my answer: When the distance is almost 0, 8D^2/d^2(1-1/(sqrt(1+d^2/4D^2)) becomes 0, so the omega angle is 0.
Its my answer good? Are there more answers ?
Sorry for my english, but its not my first language.
Thanks for any help!

 

[turin]

The radioactive source is at the center of a make-believe sphere, whose surface area is 4πD².

The detector window has an area of (π/4)b².

The ratio of the solid angle of your detector window Ω to the solid angle for the entire sphere 4π is approximately the ratio of the area of your detector window to the surface area of the sphere (assuming b << D):

Ω/(4π) ≈ (b/(4D))²

If I make the same approximation in the formula that you have (assuming that I have understood it correctly), then I get the same result. (I expanded the reciprocal of the square root in Taylor series assuming that b²/(4D²) is small.) I'm not quite sure what you're asking, about the approximation, or why the solid angle is important in the first place.

The solid angle is important because it basically tells you what to multiply your count rate by in order to determine the activity of the source.