Determining values of voltage divider resistors which interface an OpAmp and an input impedance?
Let's say we have a generator with a 10 volt output (let's call it "E1") and zero output resistance (source impedance) and we want to connect that to an amplifier with an input which can take 1 volt max (let's call it "E2"), and which has an infinite input resistance. (load impedance)

In that case we're pretty much free to do whatever we want to, so let's
say we're comfortable taking 1 milliampere from the generator and using a voltage divider to drop the generator's 10 volt output down to 1 volt for the amplifier.  

Since, from Ohm's law, \$ R = \frac{E}{I}\$, we can write:

$$ R2= \frac{E2}{I} = \frac{1V}{0.001A} = 1000\text { ohms.} $$

Now, since we started with 10 volts and we've gotten our one volt by passing 1 milliampere through a 1000 ohm resistor (let's call it "R2"), we must now get rid of the other 9 volts. 

We can do that by connecting a resistor (R1) in series with the 1000 ohm resistor, and since current in a series circuit is everywhere the same, we can use Ohm's law again to find the value of R1, like this: 

$$ R1 =\frac {E1-E2}{I} = \frac{9V}{0.001A} = 9000\text { ohms}$$  

Schematically, then, we have:

 

Now let's assume that our load impedance (let's call it "\$Z_L\$") is no longer infinite but is, instead, 1000 ohms, like this:



In that case, It'll be in parallel with R2 which will make the total resistance from E2 to GND 

$$ R_T = \frac{R2 \times Z_L}{R2 + Z_L} = 500 \text { ohms}$$ 

and we can use the voltage divider formula to find the voltage at E2:

$$ E2 = \frac{E1\times R_T}{R1 + R_T}= \frac{10V\times 500\Omega}{9000\Omega + 500\Omega} = 0.526\text{ volts} $$

A gross error, but easy to fix by just removing R2 from the divider and letting \$Z_L \$ do the work.

At the other end of the divider, the generator's output impedance will affect the value of V1 and, in the end, E2.