Passive RC Filters
It's not uncommon to come across 2-pole passive RC filter networks in circuits where the designer has used equal values for both R and C. Presumably the intention here is to create a 2-pole filter with the same $-3dB$ frequency as the equivalent 1-pole network; I have made this mistake myself more than once.
Unfortunately, things aren't so simple.
Primarily, the issue is that the second RC network loads the first one, shifting its response.
1-Pole Passive RC Low-Pass Filter
Let's start with the detail of a 1-pole passive RC low-pass filter consisting of a series-resistor followed by a parallel-capacitor. We will call our input voltage $V_0$ and our output (across the capacitor) $V_1$. From inspection, and using the convention that $X_C$ is the reactance of the capacitor, we can say that:
$$ V_1 = V_0 \frac{X_C}{R+X_C} $$
Poles
From here, we can rearrange for the transfer function where $X_C = \frac{1}{Cs}$:
$$ T(s) = \frac{V_1}{V_0} = \frac{1}{RCs + 1} $$
Again, by inspection, we can see that this is indeed a single-pole system with a single, real-axis pole at $s = -\frac{1}{RC}$.
$-3dB$ Point
Next, we can determine the $-3dB$ point of the system by finding the point where $|T(s)| = \frac{1}{\sqrt{2}}$. To do this, we move from the Laplace domain to the Fourier domain via the substitution $s = j\omega$:
$$ T(j\omega) = \frac{1}{1+j\omega RC} $$
Our $-3dB$ point occurs at:
$$ \begin{align*} \Bigg|\frac{1}{1+ j \omega RC}\Bigg| &= \frac{1}{\sqrt{2}} \newline |1 + j \omega RC| &= \sqrt{2} \newline (1 + j \omega RC) (1 - j \omega RC) &= 2 \newline 1 + \omega^2 R^2 C^2 &= 2 \newline \omega &= \pm \frac{1}{RC} \end{align*} $$
Since we are only interested in positive frequencies for now, we can take the positive result, $\omega = \frac{1}{RC}$.
It is important to point out that, while in this case you can arrive at the result for the $-3dB$ point by applying the Laplace-Fourier substitution on the pole at $s=-\frac{1}{RC}$, this is not generally true of all filters (as we shall see).
2-Pole Equal-Value Passive Low-Pass RC Filter
What, then, of our cascaded, equal-value system?
Our filter comprises a series-resistor followed by a parallel-capacitor followed by a second series-resistor and finally a second parallel-capacitor. Let's call our input voltage $V_0$ again, but our output voltage (across the second capacitor) will be $V_2$ with $V_1$ being the intermediary voltage (across the first capacitor).
Poles
The algebra gets quite tedious here so I've hidden it under this cut...
We will use the convention that $Z_1 // Z_2$ denotes the total impedance of two parallel branches with impedance $Z_1$ and $Z_2$. By inspection, we can say that $$ \begin{align*} V_2 &= V_1 \frac{X_C}{R + X_C} \newline &= V_1 \frac{1}{RCs + 1} \end{align*} $$ and that $$ V_1 = V_0 \frac{X_C // (X_C + R)}{R + X_C // (X_C + R)} $$ First, let's investigate $X_C // (X_C + R)$, substituting for $X_C = \frac{1}{Cs}$: $$ \begin{align*} X_C//(X_C + R) &= \Bigg(\frac{1}{X_C} + \frac{1}{X_C + R}\Bigg)^{-1} \newline &= \Bigg(Cs + \frac{1}{\frac{1}{Cs} + R}\Bigg)^{-1} \newline &= \Bigg(Cs + \frac{Cs}{RCs + 1}\Bigg)^{-1} \newline &= \Bigg(\frac{Cs(RCs + 1) + Cs}{RCs + 1}\Bigg)^{-1} \newline &= \Bigg(\frac{Cs(RCs + 2)}{RCs + 1}\Bigg)^{-1} \newline &= \frac{RCs + 1}{Cs(RCs + 2)} \end{align*} $$ Next, let's deal with $R + X_C // (X_C + R)$: $$ \begin{align*} R + X_C // (X_C + R) &= R + \frac{RCs + 1}{Cs(RCs + 2)} \newline &= \frac{RCs(RCs + 2) + RCs + 1}{Cs(RCs + 2)} \newline &= \frac{RCs(RCs + 3) + 1}{Cs(RCs + 2)} \end{align*} $$ Now we can deal with our equation for $V_1$ in terms of $V_0$: $$ \begin{align*} V_1 &= V_0 \frac{X_C // (X_C + R)}{R + X_C // (X_C + R)} \newline &= V_0 \frac{RCs + 1}{Cs(RCs + 2)} \div \frac{RCs(RCs + 3) + 1}{Cs(RCs + 2)} \newline &= V_0 \frac{RCs + 1}{RCs(RCs + 3) + 1} \end{align*} $$ Finally, we can substitute this for $V_1$ in our equation for $V_2$: $$ \begin{align*} V_2 &= V_1 \frac{1}{RCs + 1} \newline &= V_0 \frac{RCs + 1}{RCs(RCs + 3) + 1} \cdot \frac{1}{RCs + 1} \newline T(s) = \frac{V_2}{V_0} &= \frac{1}{RCs(RCs + 3) + 1} \newline &= \frac{1}{R^2 C^2 s + 3RCs + 1} \end{align*} $$The transfer function is:
$$ T(s) = \frac{V_2}{V_0} = \frac{1}{R^2 C^2 s + 3RCs + 1} $$
Poles are located at
$$ R^2 C^2 s + 3RCs + 1 = 0 $$
These can be found by the quadratic equation:
$$ s = \frac{-3RC \pm \sqrt{9R^2 C^2 - 4 R^2 C^2}}{2R^2 C^2} = \frac{-3 \pm \sqrt{5}}{2RC} $$
$-3dB$ Point
As before, our $-3dB$ point occurs at $|T(s)| = \frac{1}{\sqrt{2}}$ or $\frac{1}{|T(s)|} = \sqrt{2}$:
$$ \begin{align*} | R^2 C^2 s + 3RCs + 1 | &= \sqrt{2} \newline | (1 - R^2 C^2 \omega^2) + j 3 R C \omega | &= \sqrt{2} \newline (1 - R^2 C^2 \omega^2)^2 + 9 R^2 C^2 \omega^2 &= 2 \newline 1 - 2 R^2 C^2 \omega^2 + R^4 C^4 \omega^4 + 9 R^2 C^2 \omega^2 &= 2 \newline R^4 C^4 \omega^4 + 7 R^2 C^2 \omega^2 - 1 &= 0 \end{align*} $$
We can use the quadratic equation to solve for $\omega^2$:
$$ \begin{align*} \omega^2 &= \frac{-7 R^2 C^2 \pm \sqrt{49 R^4 C^4 + 4 R^4 C^4}}{2R^2 C^2} \newline &= \frac{-7 \pm \sqrt{53}}{2R^2 C^2} \end{align*} $$
Since $\sqrt{53} > 7$ we can see that the only real solution for $\omega$ is
$$ \omega = \sqrt{\frac{-7 + \sqrt{53}}{2 R^2 C^2}} = \frac{\sqrt{\sqrt{53} - 7}}{\sqrt{2} RC} $$
In numbers, this is $\omega \approx 0.374 \frac{1}{RC}$.
Note that this $-3dB$ point corresponds to neither of our poles, and if you used $\omega = \frac{1}{RC}$ to find the critical frequency of this filter, you would be out by more than a factor of 2.
Conclusion
While it may be tempting to try to double the steepness of a passive RC filter by simply cascading two identical networks, it's important to be aware of the effects this has on the response.
That said, we can see from the fact that our poles are safely in the LHP that this filter topology is nicely stable. This, combined with the cheapness of its construction makes it a viable candidate for many electronics applications where simple and cheap are the key requirements of the design.