Re: Thin Films

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Referring to a question asked by a student David Abineri asked
why the interference colors are not observable in "thick films".
A very satisfactory (quantitative) explanation was provided by
Doug Craigen (see below).

A qualitative (not very satisfactory) answer could be; "because
in a thick layer maxima and minima coincide and everything is
white everywhere". Let me add that this can be demonstrated
by comparing interference fringes observed with white light
and with yellow light from a sodium lamp, for example. Use
an air wedge formed between two microscopic slides, or
between a lens and a slice, as in Newton's rings. In white
light fringes appear only where the air layer is very thin; in
yellow light they also appear where the air layer is thicker.
Ludwik Kowalski
************************************************
Doug wrote:
> This effect (also true for soap bubbles, coatings on glass, the thin
space between microscope slide, etc) is noticeable for films whose
optical thickness is of the same order as visible wavelengths.  It
occurs from the interference of light reflected off the front and back
faces of the film.  If the optical path length of the ray which passes
through the film  is an integral number of wavelengths - you have
constructive interference for those colors.  If the optical path length
is 1/2, 3/2, 5/2 etc of a given wavelength - it results in destructive
interference.
There is one qualifier - an additional phase change of pi (equivalent to
another half wavelength of path) for reflection when the incident
medium has a higher index than the reflecting medium.

To demonstrate the effect I like to use two clean microscope slides.
You can easily press them together close enough that the air gap takes
on beautiful coloration on reflection. The colors are easily seen to be
related to the thickness by the rings they form around the place where
you are pressing.  You notice that the wider the gap the more rapid the
change from ring to ring and the less distinct the colors.

Suppose the optical thickness effect is 1000 nm, where is there
constructive interference?
#*wavelength=1000

# = 1  - 1000 nm (infrared)
# = 2  - 500 nm  (visible)
# = 3  - 333 nm (UV)

For this thickness, there is one maximum interference color in the
visible region.  The color will be quite distinct.

Now suppose it is 4000 nm:
# = 1 - 4000 nm (infrared)
# = 2 - 2000 nm (infrared)
# = 3 - 1333 nm (infrared)
# = 4 - 1000 nm (infrared)
# = 5 - 800 nm (red - visible)
# = 6 - 667 nm (visible)
# = 7 - 572 nm (visible)
# = 8 - 500 nm (visible)
# = 9 - 444 nm  (blue visible)
# = 10 - 400 nm (edge of UV)
... in UV

So there are 6 interference maximum colors within the visible and of
course 5 minimum in between. So the overall color will be indistinct,
and the thicker the film is the less distinct it will become.  So by the
time you consider the microscope slide, it is so thick that it will
probably have hundreds of max and min in the red region alone. The
> same argument holds for comparing an oil slick with a bottle of oil.