8:24 AM
Ok, let's try to explain it in a simple possible way: "Virtual ground" is simply a "voltage copy" of the real ground; this is the essence of the virtual ground phenomenonâ€¦
So, to create such a "copy", you simply need a voltage follower. Connect its input to the real ground and take the virtual ground from its output. The simplest voltage follower is an op-amp whose output is connected to its inverting input. Thus, the output voltage is subtracted from the input voltage at the non-inverting input... the difference is amplified many times and appears at the output... and so on so forthâ€¦
This can be explained in a human-friendly manner by personalizing the op-amp. Figuratively speaking, the op-amp "compares" the voltage at the non-inverting input with the voltage of the inverting input and changes it until the former becomes equal to the latter...
Read the latter as: Figuratively speaking, the op-amp "compares" the voltage at the inverting input with the voltage of the non-inverting input and changes it until the former becomes equal to the latter.

8:39 AM
From this perspective, every op-amp circuit with negative feedback contains a follower (another wisdom which cannot be found in "reputable" sources). Even the inverting amplifier contains a non-inverting follower inside - the voltage of the inverting inputs, as above, copies the voltage of the non-inverting input... and, if the latter is zero (real ground), the former will be zero as well (virtual ground)...
If we put a resistor (R2) in the connection (negative feedback loop) between the output and the inverting input, the op-amp will continue keeping a zero voltage of the virtual ground...
But where is the input voltage here? We include it to this following system as a "disturbance" (another wisdom). For this purpose, we connect the "disturbing" input voltage source through a resistor (R1) to the virtual ground. The op-amp reacts to this "intervention" by raising its output voltage... and we use its "reaction" as an amplified output voltage. But there is a virtual ground as before...
Ohh... I have to prepare for the next meeting with my favorites. We will discuss, as yesterday, diode circuits... a very interesting topic... Nice to see you later...

3 hours later…
12:06 PM
The equipment in my lab is quite old. There is a need for modern digital oscilloscopes and functional generators. But because I am an "old war horse", I can handle any situation...
In terms of dressing, these students now have the freedom to dress as they wish. And at the time I dressed modestly. But in general, Bulgarians and especially women pay attention to their appearance ... also to the home and the car ... I think this is typical for small nations ...

2 hours later…
1:49 PM
@Circuitfantasist Many thanks for your detailed explanation of the idea of opAmp virtual ground, using the analogy of the "voltage copy" of a real ground. I have never heard of the thing "voltage copy" so I think I need some time to digest. It would be nice if you can give a block diagram or schematic of the opAmp show the resistors R1, R2 etc. Or you can refer to one relevant chapter in you WikiBook of opAmp.
Nothing urgent at all. I found that your explanation is a bit too hard for me now, because I have forgotten many of the basic concepts, and your descriptions like the following is indeed very confusing: "... Even the inverting amplifier contains a non-inverting follower inside, ..." So I am reading the Wikipedia to refresh or learn new things. I need to study carefully the deeper ideas of "common mode", "negative feedback", "small signal transistor model".
I indeed have the feeling of first time to learn how to drive a car, or swimming, when the teacher says driving or swimming is "easy", but to an newbie, that is a very different thing.
In the mean time, I am trying to learn how to use python to import the matplotlib to plot the simple raw data of my first tunnel diode experiment. I am making a programming log in my blog, but you need to bother to understand anything for now. I will show you more when I can really plot something. The link is this:
tunneldiode.blogspot.com/2020/12/…. Bed time now. See you tomorrow or day after tomorrow.
Ah, yes, I forgot to mention that my first impression of your lab equipment such as power supplies, are desktop models, so of course more impress than my cheapy Amazon hobbyist models. However, when I read further, I found the computers in your lab are very out of date. So I agree with you that you need to upgrade some of the equipment, such as the digital oscilloscope.

7 hours later…
9:24 PM
I will try to explain the inverting amplifier in another nonconventional way (you can find millions of conventional ways on the web). The general idea of this circuit solution is to set two voltages (input VIN and output VOUT) in some proportion, e.g., VOUT/VIN = 100. Then we say that VIN is amplified 100 times (with gain of 100). VIN is an external voltage; so we have to change VOUT so that it is always 100 times higher than VIN...
For this purpose, we compare the two voltages by a network of two resistors in series whose resistances R1 and R2 are in proportion R2/R1 = 100. This is a passive summing circuit with weighted inputs (1:100) that can be thought of as a kind of "electrical scales" or a "mechanical lever" (Bob Duhamel has such a movie)...
In 2008, my students and I created a Wikibooks story about this "elegant simplicity". At the end of the story, we showed how an inverting amplifier is made by this network...
Let's continue this amazing story about the legendary circuit... To compare the two voltages, we have to subtract them. So, VOUT should be with opposite polarity. Then we change VOUT until the result of the comparison (subtraction) - the voltage of the middle point between the resistors, becomes zero... and we do it continuously. This point behaves as a ground although it is not a ground; that us why it is called "virtual ground"...
In the circuit of an inverting amplifier, the op-amp does this donkey work". It "observes" the virtual ground and changes its VOUT so that to keep zero voltage at this point.
As a result, VOUT/VIN = -R2/R1 = - 100.