Why does 50 Ω termination result in less noise than 1 MΩ termination on the scope reading?
The first thing to consider is the Johnson noise of the resistor which cannot be eliminated. The higher the resistance the greater the noise. Reducing bandwidth will also reduce Johnson noise. So if your scope has bandwidth settings, and if you don't need high bandwidth for your signal, you can get cleaner results using the reduced bandwidth modes.

The second thing to consider is noise which couples in to the oscilloscope, particularly if it couples through the probe wiring arrangement by way of a magnetic field. The time-varying magnetic field will induce a current in the probe. The termination resistance inside the oscilloscope will convert that current to a voltage. If the termination resistor is 50 Ohms, that will lead to a much smaller voltage than if it is 1M Ohm.

In general, lower impedance termination is more resistant to noise. This is a very important concept when you encounter situations where noise immunity is required. Usually any noise coupling path will have some series resistance or fundamental power limiting just by its nature. So the lower your termination resistance, the lower the voltage due to noise coupling. Sometimes a 20 pF capacitor on a digital input can make the difference between a flaky and totally unreliable piece of junk and a rock solid product.

Often if I need to put the oscilloscope on a shunt resistor, I will use the 50 Ohm termination feature of the oscilloscope. This greatly reduces noise, and since the shunt resistance is much less than 50 Ohms (for the shunts I deal with) there is no worry of excessive current flowing into the oscilloscope, even if the shunt current may be high.



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