Okay, let's start. Because it's recorded, I want to say also to people who listen online

that the solutions to home tasks I only accept before the beginning of the lecture. So if

you have something you should give it to me now, because I'm going to tell you how the

problem should be solved, just in case someone wants to know. The solution to the first task

after the first lecture I didn't present because it will be considered at the problem class

that Gaetano leads. But now about the second task which was about optical coherence tomography

and the task was, hopefully you remember, I just want to briefly tell how it should be

solved. So the object is, as I said, two layers with some distance given, d, between the layers.

And I didn't give you any other information, but actually what is necessary as well is

the refractive index. I apologize for not giving it, but I think, well, those who didn't

consider it probably are allowed to think that this is hair or vacuum. It's a bit unphysical,

but still. So what happens, the question was, what happens when we scan this mirror and

what, there were two questions, what should be the bandwidth of the source and what spectrum

will we observe here in the end? And the first question was very simple because we know that

as we scan this path difference, we will see first some burst of interference fringes when

the path is equal to the distance to the first layer and then another one when the path is

equal to the distance to the second layer. And so we will observe as intensity as a function

of delay, we will have first this thing and then another thing, maybe with another height

because the reflectivities of the two layers could be different. What is important is that

this width of the, I would call it actually coherence time for each case. This should

be the coherence time of the source, right? If there is one layer, the problem is known,

this is just g1, the measurement of g1, and if there are two layers, then there will be

two such distributions. And so this is the, or rather two coherence times, because I defined

coherence time as the time where the visibility decays twice on one way. But obviously this

two coherence time should be smaller than this distance, and this distance is very simple

to calculate because it's just the double path of light from one layer to another. So

this distance is two, and then if it were distance, it would be just 2d times n because

the optical path is what matters, but of course there is also speed of light, and c is here.

So this is the time delay, the time distance, the time separation between the two bursts

of interference, and the condition is that this 2dn over c is much larger, better, much

larger than this width, yeah, than the two tau coherence. And from this, from very simple

algebra, knowing the relation between the coherence time and the spectral width, I got

for instance the condition that delta lambda is much larger than lambda squared over 2d

times n. If this two is taken into account, if this two is not taken into account, there

would be 4dn, and I got, in numbers I got about 15 nanometers, or if this two is taken

into account, maybe about 30 nanometers for n about 1.5, which is just the case for glass

for instance. And the second question was much more difficult, what does the spectrum

look here? So if we put a spectral device, like a spectrometer, what we will see here?

But then I will just write it like this, I will write the output field as a function

of time t will be the input field, so here we send input field u of t, and because of

the general delay in this interferometer, there will be three contributions. There are

field reflected from the reference, let's say t minus delta t, delta t is the time that

is needed to travel here and back, then there will be the field reflected from the first

layer, let's call it R1 e of t minus delta t, and for instance plus tau 1, and then there

will be R2 e of t minus delta t plus tau 2, this corresponds to reflectivities of the

two layers and these are the three contributions. So in principle it's clear what to do, we

have to calculate the e of omega and take the squared average value. Yeah, this is of

course e plus or e minus, I use analytic signal and it's complex. So this is the strategy

but very simple considerations show that it's similar to triple slit interference, the Young

Scheme and three slits, but we can simplify it, for instance the delay to this, the tau