How does negative feedback works?????

I have an non-inverting opamp circuit with Rf=100K and Ri=4.7K.Say Aol=10000 and i have applied an input of 1mV.

So,at t=0,Vo=0,Vf=0 and Vin=(1-0)mV.So,just after applying the input Vo should be equals to 10V.Then Vf=448.9 mV.

Now,Vin=(1-448.9)mV= -447.9mV.If the opamp amplify this input,then how it is  possible to make the voltage at both terminal same or nearly same????

A basic rule for op-amps (this assumes a perfect amplifier with infinite gain) is that the inverting and non inverting pins are at the same voltage with negative feedback applied. This implies that the output will always drive the feedback network to ensure that the two inputs are at the same voltage. Intuitively, this must be the case, since any difference in the voltage between the two pins is amplified infinitely.
Additionally, the two inputs have infinite input resistance, that is, they draw no current.
The gain with feedback and the input signal have to be low enough that the output is not saturated at one or other of the power supply rails (a characteristic of a real world op-amp) with an applied signal.
The gain equals (Rf/Ri+1), ignoring the less than infinite open loop gain, or ~22, so the output should be 22mV with 1mV input.
With 1mV on the non-inverting input, there should be 1mV on the inverting input.
In order to have 1mV on the inverting terminal, the output voltage has to be 22mV, since the resistors Rf and Ri act as a voltage divider.
Remember, you have put feedback around the amplifier, it is not operating open loop, without the resistor feedback, if you applied a 1mV signal, the output would go to 10V. Obviously feedback throws away gain, but you gain other benefits from negative feedback, which is the whole point of it.
Trying to explain this without the benefits of diagrams and lots of equations is not easy, however, there is a vast amount of info on the web on this topic, including you tube class room style demos, so have a look there for a better explanation,
cheers,
Richard

Sir,I understand that the opamp output should be 22 mV to make the input voltage same at the non-inverting & inverting terminal.But at the first moment there was some output & small amount of this output is fed back to the inverting terminal.This process continues in order to make the fed-back voltage and applied input same.We cant observe this process is happening.But it actually happens.So,at very first moment,output should be 448 mV.I want to know from there how opamp make the fed-back voltage same to the applied iput voltage????

There is a slew rate issue, since the amplifier doesn't have infinite bandwidth. It is the slew rate that limits how fast the op amp can swing, so it takes a finite time for the op-amp output to respond to the input. In the simplest case, it behaves as an embedded single pole RC network, an integrator. So if you apply a 1mV step, it takes some time for the output to swing towards the 22mV output. Once it gets to the 22mV level, it meets the feedback requirements and steadies to that value. During this time there is a voltage difference between the two inputs.
Note that over shoot, and ringing may occur since the op amp is not a perfect integrator, there are issues associated with finite band width that will even cause oscillations.
Cheers,
Richard

You have to remember that in placing a feedback network around the op amp, resulting in negative feedback, that you are now dealing with a set of rules, one of which is that the two inputs must remain at the same voltage. What you are dealing with is a servo system in effect, any difference between the two inputs drives the output through the feedback network to keep the two inputs at the same voltage.
With the voltage divider in your feedback loop, for both inputs to be at the same voltage, the output has to be ~ 22mV.
As I said, there are you tube videos on the web which go through this with the maths and circuit diagrams, go there for a good explanation,
cheers,
Richard