I am not entirely sure what ``saturation effect'' means in this case.
But let us go back to the expression for the Cherenkov angle:
$$
\cos\theta = \frac{1}{n \beta}
$$
we can obtain two limits:
first, on the minimum $\beta$ to have Cherenkov effect:
$$
\beta_{\rm th} = \frac{1}{n},
$$
and - as you noticed - the smaller $n$, the lower the velocity of the particle we can detect;
but also on the maximum aperture angle of the Cherenkov cone, in case the particle is ultrarelativistic $\beta \rightarrow 1$:
$$
\cos\theta_{\rm max} = \frac{1}{n}.
$$
I thought the ``saturation'' might be related to this second point, the fastest particles will generate cones with apertures:
\begin{equation}
\begin{split}
\theta_{\rm max}^{\rm RICH} &= \arccos \left( \frac{1}{1.0014} \right) \approx 3.03^{\circ}, \\
\theta_{\rm max}^{\rm RICH2} &= \arccos \left( \frac{1}{1.0005} \right) \approx 1.81^{\circ},
\end{split}
\end{equation}
In the two different detectors. For ultra-relativistic particles you will have smaller rings in the detector with the smaller $n$. If you receive many particles it would be easier to ``saturate'' (to create more overlapping rings, maybe?) the detector with the larger $n$.
Just my speculation though!
EDIT
After the useful suggestions of @dukwon I understood the saturation was not due to the number of particles hitting the detector but by the impossibility to distinguish their momentum when their $\beta \rightarrow 1$. Therefore I created this plot. Here I show the values of $\theta_{\rm C}$, i.e. the Cherenkov angle for different values of the particle energy. I consider three particles (kaon, pion and muon) and two mediums with the indexes, $n_1=1.0014$ and $n_2=1.0005$ as in @Sito's original question.
I placed the vertical dashed line by eye at the point where I am not capable of distinguishing anymore kaons and pions angles. As you can see in the material with the worst (higher) refraction index this happens already at $80\,{\rm GeV}$. In the material with the better (lower) refraction index you can push the indistinguishability of kaons and pions at $E > 100\,{\rm GeV}$. and that's exactly why $n_2=1.0005$
postpones the saturation effect
.
Thanks @dukwon!
@Sito, here the snippet if you want to reproduce the figure
import numpy as np
import matplotlib.pyplot as plt
def beta(E, mec2):
gamma = E / mec2
return np.sqrt(1 - 1 / np.power(gamma, 2))
def theta(beta, n):
return np.arccos(1 / (beta * n))
mec2_kaon = 493.68 # MeV / c^2
mec2_pion = 139.57 # MeV / c^2
mec2_muon = 105.65 # MeV / c^2
E = np.logspace(np.log10(5e3), 6, 100) # MeV
beta_pion = beta(E, mec2_pion)
beta_kaon = beta(E, mec2_kaon)
beta_muon = beta(E, mec2_muon)
n_1 = 1.0014
n_2 = 1.0005
fig, ax = plt.subplots()
plt.semilogx(E, theta(beta_kaon, n_1), color="crimson", ls="-", label=r"$K, n_1=1.0014$")
plt.semilogx(E, theta(beta_pion, n_1), color="crimson", ls="--", label=r"$\pi, n_1=1.0014$")
plt.semilogx(E, theta(beta_muon, n_1), color="crimson", ls="-.", label=r"$\mu, n_1=1.0014$")
plt.semilogx(E, theta(beta_kaon, n_2), color="k", ls="-", label=r"$K, n_2=1.0005$")
plt.semilogx(E, theta(beta_pion, n_2), color="k", ls="--", label=r"$\pi, n_2=1.0005$")
plt.semilogx(E, theta(beta_muon, n_2), color="k", ls="-.", label=r"$\mu, n_2=1.0005$")
plt.axvline(8e4, color="crimson", ls=":")
plt.axvline(1.1e5, color="k", ls=":")
plt.xlabel("E / MeV")
plt.ylabel(r"$\theta_{\rm C}\,/\,rad$")
plt.legend()
plt.show()
fig.savefig("cherenkov_angles.png")
