Thin film oil drop interference

[stpmphysics]

Homework Statement

Hi all, i have a question on thin film interference.

Lets say there is an oil drop on the surface of water. Light reflected from the oil drop is observed from above. Explain why interference fringes are not seen when the thickness of the oil drop increases.

Homework Equations

The Attempt at a Solution

When the thickness is increased, the width of the central fringe will increase. This narrows the width of the rest of the subsequent fringes. Therefore, interference of the light rays occur in a smaller scale till an extend where they are not resolvable by our eyes. That kept me thinking again, is the central fringe a bright or dark fringe because if its a bright fringe, there is no way the rays are not detectable by our eyes.

 

[Delzac]

This has got to do with the path difference of the light that i reflected from the oil-water boundary.

Interference occurs when light reflected from air-oil boundary interfere with the light reflected from the oil-water boundary. Whether you will see anything depends of the phase difference (i.e. also path difference). As the thickness of oil increase, the path difference of the light from oil-water, and the light from air-oil increasingly become more destructive. Hence, at some point in time you will not see any more rings.

 

[stpmphysics]

This has got to do with the path difference of the light that i reflected from the oil-water boundary.

Interference occurs when light reflected from air-oil boundary interfere with the light reflected from the oil-water boundary. Whether you will see anything depends of the phase difference (i.e. also path difference). As the thickness of oil increase, the path difference of the light from oil-water, and the light from air-oil increasingly become more destructive. Hence, at some point in time you will not see any more rings.

Thanks Delzac. Why do the light rays become more destructive when their optical path difference increases?

 

[Delzac]

Path difference * 2pi/wavelength = phase difference.

If phase difference = (m plus/minus 1/2) pi, where m is a integer, the interference will be destructive.

 

[Delzac]

And if you keep adding oil, you will see rings again.

 

[stpmphysics]

Path difference * 2pi/wavelength = phase difference.

If phase difference = (m plus/minus 1/2) pi, where m is a integer, the interference will be destructive.

Thanks again, so here the path difference, $$\triangle L=2 \mu t+\frac{1}{2}\lambda$$

It's either it equals m\lambda for constructive or (m-0.5)\lambda for destructive. How did you know its destructive ?

 

[Delzac]

If phase difference = (m plus/minus 1/2) pi, where m is a integer, the interference will be destructive. If it is m*pi, then it is constructive.

Do take note that if your ray from a optically less dense medium reflect off a optically denser medium, the phase will shift by pi.

And in your equation, what is mu and t?

 

[stpmphysics]

If phase difference = (m plus/minus 1/2) pi, where m is a integer, the interference will be destructive. If it is m*pi, then it is constructive.

Yes i know that, but why previously you said that the light rays will undergo more destructive interference? How can i know which is more dominant in this case?

Do take note that if your ray from a optically less dense medium reflect off a optically denser medium, the phase will shift by pi.

I know this too but what's the actual reason behind this?

And in your equation, what is mu and t?

mu is the refractive index of the oik and t is the thickness of the oil.

Thanks for all your replies.

 

[Dadface]

When the film thickness increases:
The fringes get closer together which can reduce the ability to see them as separate fringes.
The intensity of the rays emerging from the film reduces which in turn reduces the contrast between any fringes thereby rendering them less visible.
The lateral separation of the non normal reflected rays increases and if this separation goes beyond the diameter of the eye pupil the rays can not be focussed simultaneously so as to observe interference with the unaided eye.

 

[Delzac]

1) Well, if you said that initially you see rings, then it must be constructive interference, or at least not complete destructive interference. Thus any more increase in the thickness of oil will only increase phase difference, and the initial condition of m*pi will start change to m +/- 1/2 pi, if you keep adding it will change again to m*pi.

2) The reason why it will shift by pi phase has to do with the boundaries conditions, if not the maths will not work out. And this i only know in passing. Maybe if some of the PF mentors were to pop by (*hint hint), they will answer in more depth.

 

[Delzac]

When the film thickness increases:
The fringes get closer together which can reduce the ability to see them as separate fringes.
The intensity of the rays emerging from the film reduces which in turn reduces the contrast between any fringes thereby rendering them less visible.
The lateral separation of the non normal reflected rays increases and if this separation goes beyond the diameter of the eye pupil the rays can not be focussed simultaneously so as to observe interference with the unaided eye.

Hmmm, then is my explanation wrong?

And why does the fringes get closer too?

 

[stpmphysics]

Hmmm, then is my explanation wrong?

And why does the fringes get closer too?

yeah why the fringes get closer?

 

[Delzac]

I think i partially, after reading up, know what is happening.

In the first place, we can see rings of bright and dark fringes (monochromatic light source) because rays coming out from the sides have a bigger path difference, so you will see rings.

Now you imagine if you add more oil, the present light rays will experience a increase in path difference across the board. Thus those dark fringes will start to become bright fringes and vice versa. This gives an apparent shift in the fringes becoming closer.

And this is what i think is happening.

 

[stpmphysics]

I think i partially, after reading up, know what is happening.

In the first place, we can see rings of bright and dark fringes (monochromatic light source) because rays coming out from the sides have a bigger path difference, so you will see rings.

Now you imagine if you add more oil, the present light rays will experience a increase in path difference across the board. Thus those dark fringes will start to become bright fringes and vice versa. This gives an apparent shift in the fringes becoming closer.

And this is what i think is happening.

When the thickness of the oil increases, the path difference increases. Isn't that now the interference pattern(bright and dark fringe) are spread over a larger area and the fringe distance subsequently increases?

 

[Delzac]

When the film thickness increases:
The lateral separation of the non normal reflected rays increases and if this separation goes beyond the diameter of the eye pupil the rays can not be focussed simultaneously so as to observe interference with the unaided eye.

Like what dadface said, the pattern will spread out.

And i think i will not speculate on things i am not sure of anymore. Will wait for PF mentors to come and explain.

 

[Dadface]

Like what dadface said, the pattern will spread out.

And i think i will not speculate on things i am not sure of anymore. Will wait for PF mentors to come and explain.

Hello stpmphysics and Delzac.What I meant by my third comment is that for any particular common angle of reflection the rays reflected from the top and bottom surfaces of the film become wider apart (more spread out) for thicker films.
As for the pattern I think that closes in because the virtual images formed by reflection become wider apart for thicker films and the reflected rays responsible for interference appear to come from these virtual images.
I tried to do a search on this but found no useful information so hopefully others,more in the know, will come in and contribute.

 

[ehild]

The thicknes of the oil drop is not even. It is thicker in the middle and thinner at the edges.
We observe the interference pattern from above. This means normal incidence and reflectance.
Part of the incoming light reflects from the surface of the oil drop. Other part enters into the oil and reflects from the interface with water, then travels back through the oil again and part of it this ray steps out into the air and interferes with the first reflected ray. As the refractive index of oil is about 1.42 and that of the water is 1.33, there is pi phase change at the air-oil interface and no change of phase at the oil-water interface.
There is constructive interference if the phase difference between the interfering waves is k*2pi (k is an integer).
When traveling across the thickness of the oil drop, the phase of the wave changes by 4 π N d/λ. The phase difference between the reflected waves is 4 π N d/λ+ π, and constructive interference happens if 4 π N d/λ+ π= 2πk. At that places where d=λ/(4N)(2k-1) we get a bright circle of a certain colour if the oil drop is circular.

The directly reflected beam can only interfere with the one reflected from the water-oil interface if their path difference is shorter than the coherence length of the light. If the oil is thick the phase of these reflected waves becomes random, and there will be no interference.

ehild

 

[stpmphysics]

The thicknes of the oil drop is not even. It is thicker in the middle and thinner at the edges.
We observe the interference pattern from above. This means normal incidence and reflectance.
Part of the incoming light reflects from the surface of the oil drop. Other part enters into the oil and reflects from the interface with water, then travels back through the oil again and part of it this ray steps out into the air and interferes with the first reflected ray. As the refractive index of oil is about 1.42 and that of the water is 1.33, there is pi phase change at the air-oil interface and no change of phase at the oil-water interface.
There is constructive interference if the phase difference between the interfering waves is k*2pi (k is an integer).
When traveling across the thickness of the oil drop, the phase of the wave changes by 4 π N d/λ. The phase difference between the reflected waves is 4 π N d/λ+ π, and constructive interference happens if 4 π N d/λ+ π= 2πk. At that places where d=λ/(4N)(2k-1) we get a bright circle of a certain colour if the oil drop is circular.

The directly reflected beam can only interfere with the one reflected from the water-oil interface if their path difference is shorter than the coherence length of the light. If the oil is thick the phase of these reflected waves becomes random, and there will be no interference.

ehild

Thanks a lot ehild, what's the coherent length of the light?

 

[ehild]

The photons a light source emits are wavelets, waves with finite length. They say that a photon can interfere only with itself, as the other photons emitted by the light source have different phase constants. It is a bit strange that one part of the photon reflects from the front surface of a layer, an other part from the back interface, but it can be imagined that way. This coherence length is connected to the bandwidth of the light. You get the most clear interference and most fringes with a coherent source - a laser.

The other reason while the interference fringes disappear is when the film thickness is homogeneous. In such case you see one interference colour, but this colour depends on the angle you see the film.

ehild