Significance of "thin" in thin film interference What is the significance of the word "thin" in thin film interference. Will the phenomenon not occur when the film is thick or not of order of magnitude of wavelength of light ( thin films are of order of magnitude of light). Thick films may too produce various path differences and thus varying phase differences. Then why is the phenomenon not specified for thick films?
 A: Later
I had an idea to see if it was possible to show that in the word thin is only appropriate if the source of light emits a range of wavelengths.
For light from a laser the optical paths can be very much larger than a few wavelengths of light. 
I set up the following arrangement using Blu-Tack to hold the apparatus in position.
A key element is the microscope slide which although relative cheap to buy is actually manufactured to be optically flat with the opposite faces parallel to one another within fractions of a wavelength of visible light.
This is very similar to the arrangement used to observe Newton's rings although the type of fringes produced are not quite the same.    
 
The x5 hand magnifier produces a divergent beam.
The inclined microscope slide deflects part of the laser beam down onto the bottom microscope slide.
The light which is reflected from top and bottom of that slide passes through the inclined slide and form an interference pattern on the screen.
The insert does not do justice to what is seen and I adjusted the apparatus so that I could observe the central fringe which has concentric circular fringes around it.  
To convince myself that the interference pattern was due to the the interference of light from reflections off the two parallel sides of the microscope slide I made use of a hair dryer.
Whilst heating the bottom microscope slide the fringes were seen to more into the centre and disappear due to the expansion of the glass and the reverse happened when the microscope slide was cooling down with the glass contracting.
As an aside I also found that if the batteries were not relatively new, even though the pointer was producing an intense beam of light, the light from the laser had a greater range of wavelengths and the fringes were much more difficult to observe and for me the green laser produced clearer fringes than the red laser.

Original answer
The word "thin" in thin film interference is used because the observation of such interference is "more likely"/"easier" if the film thickness is of order of a few wavelengths.
That is not to say that such interference cannot be seen when the film is hundreds, thousands, millions of wavelengths thick.
Let us suppose that there is a wedge of refractive index $\mu$ and at some position the thickness of the wedge is $t$.  
Light is reflected from the top $A$ and bottom $B$ of the wedge and at one of those reflections there is a $\pi$ phase change.  

The two beams of light superpose and the conditions for maxima and minima are 
$t_{\rm m,\lambda} = (2m +1) \dfrac {\lambda}{4\mu} $ where $m = 0,1,2,....$ for a maximum
$t_{\rm m,\lambda} = m \dfrac {\lambda}{4\mu} $ where $m = 0,1,2,....$ for a minimum
So maxima will occur when the thickness of the wedge is $\,\dfrac {\lambda}{4},\,\dfrac {3\lambda}{4},\,\dfrac {5\lambda}{4},\,.....$ and minima will occur when the thickness of the wedge is $0, \,\dfrac {\lambda}{2},\,\lambda,\, \dfrac {3\lambda}{2},\,.....$
However the fringes which are seen have a finite width as is illustrated below.

Now consider what happens if the source emits light of three wavelengths. 
 
So now wht you see is three sets of fringes which overlap one another and they only position where the same type of fringe occurs at the same position is when there is a minimum and $m=0$.  
The maxima around the $m_{\rm red} =4$ do approximately coincide.
This means that the visibility of the fringes decreases and for the whole spread of wavelengths for white you might get something like this.
 
Note the thickness of the film which in terms of everyday distances can be thought of as thin.
You cam relatively easily replicate this behaviour in the laboratory or at home using a soap film as is shown below in the still from this video of a draining soap film.  
 
Now if you use monochromatic light eg from a sodium lamp, then the optical path difference and hence the thickness of the film can be larger.
However even with sodium light the fringes can disappear because the intense yellow D-line of sodium is actually two wavelengths $589.0\,\rm nm$ and $589.6\,\rm nm$.  
So the as the order of interference increases there will be slippage of the relative positions of the two maxima and there will come a time when the maximum of one fringe system coincides with the minimum of the other fringe system.
The for an even greater thickness the slippage between maxima is exactly one order of interference and the fringe pattern will reappear so you will have $m \times 589.6 = (m+1) \times 589.0$ which gives $m \approx 1000$ and this corresponds to film thickness of approximately $0.3\,\rm mm$.
This would now no longer be consider thin film interference and much more so if you use light from a laser.
The fringes themselves can be made to have less width by allowing multiple reflections in the wedge which is equivalent to increasing the number of slits from the two for a Young's slits arrangement to many tens$^+$ thousand which is a diffraction grating 

Another consideration is the actual constriction of the wedge in that all the analysis above has assumed that the reflecting surfaces are flat to a fraction of a wavelength of light.
If this is not so this can affect whether or not an interference pattern is observed.
