Why Fluorescence is detected @right angle to the excitation?

[Shi]

Hi everybody,
I would like to know what would be the best angle (or best geometry) to put the detector relative to the light source in order to observe the emission radiation from the fluorescent materials?
As I know the most common geometry used for fluorescence is right angle observation. Detector is placed at 90 deg to incident light to minimize the risk of transmitted or reflected incident light reaching the detector (from wiki). Why?!
If we place detector at 45 deg or at smaller angle to the source, what would be happened?

 

[mfb]

Various materials tend to reflect under shallow angles, or directly back to the source. I don't know details about it in terms of fluorescence measurements, but that effect is well-known in astronomy.

 

[Baluncore]

Detector is placed at 90 deg to incident light to minimize the risk of transmitted or reflected incident light reaching the detector (from wiki). Why?!

Near the sample, the incident beam and the emitted beam pass along the same optical path but in different directions. The source and the detector cannot be in the same place so the two beams must be combined or separated at some point.

For the right angled illumination case, a half silvered mirror mounted at 45° would combine the beams. But the half silvered mirror would reduce the detected beam to 25%, since half the illumination is lost on the first pass, before half the emission is lost on the second pass. Since the two beams have different wavelengths, a wavelength sensitive dichroic mirror is used to effectively brighten the response for the same illumination source.

I believe that right angled illumination with a 45° mirror is used because it leads to simple compact engineering and avoids the Brewster angle. https://en.wikipedia.org/wiki/Brewster's_angle

Various materials tend to reflect under shallow angles, or directly back to the source. I don't know details about it in terms of fluorescence measurements, but that effect is well-known in astronomy.

The astronomical phase curve is best explained by the relative brightness of our moon in different phases. A full moon has more surface illuminated by the Sun that is also visible from Earth. The new moon returns little light because we cannot see much of the illuminated surface. The shape of the curve is determined to some extent by surface texture. That cannot be related to the florescence microscope situation.

 

[mfb]

A full moon has more surface illuminated by the Sun that is also visible from Earth.

That is not the only effect, and it is not the effect I was discussing. The full moon is much brighter than you would expect based on illuminated area. The linked article explains those brightness excesses and opposition surges.
But I see that wavelength filters work fine to stop light from the initial light source, so this is not an issue here.