Low Level Photodetector opamp LTC1150
You may need a high V/I transimpedance  amplifier (TIA)  gain using this configuration, such as this example \$V_o/I_{in}=10^9 \Omega=[\mu V/\mu A]\$ This implies a feedback \$R_f = 10^9\Omega \$. 

This creates problems  for; reduced BW, DC bias current offset gain and stray  EMI very high impedance induced noise.

If that R was used instead, the input pF and stray capacitance pF, might only be a few pF in total, but this RC product results in 1000 times slower rise time and Bandwidth, BW than the offered T network feedback.

Also, any input bias current even in the low nA range could be amplified to >1Vdc output as an error voltage.

To prevent this all these effects, the T feedback ratio reduces the negative feedback  voltage by this divider ratio from the output by 1000:1.  That,  automatically causes any forward voltage gain  to be 1000x. 

The feedback is further reduced to a current source by the large 1M to act as a current to voltage conversion to the  input for error feedback.  

The forward gain is now \$ Vout/Iin= 1M\Omega * 1000  = 10^9\Omega \$ .

The shunt cap across the 1M has an RC value =15us chosen to match the RC value of the sensor and cable capacitance 
That is just a scope probe 10M input impedance matches the cable low capacitance to the scope 1M//xx pF input impedance to flatten the tuned probe frequency response.  

Except now it does a better job than \$10^9*C\$ because stray capacitance would exceed the  0.015 pF requirement, which cannot be achieved.

Short Answer

The rule of thumb to reduce stray EMI is try to lower the input impedance on an unbalanced source. Now it is only 1M.

So you might think, why not go further with a million to one attenuation in the T, to increase the BW further using 1k for RF.  Maybe, but then we must consider noise currents and other details, which we can discuss another time.