Lacquer Thickness Measurements

[Falstaff]

I am using a spectrophotometer (at normal incidence) to obtain an interference pattern from a thin, plano, lacquered lens. The lens is lacquered on both sides, and I am unable to view anything other than the combined effect of both interference patterns.

My question is, in general, if I ignore the beating effects, and take wavelength measurements of successive peaks and/or troughs; can I use the same analysis procedure as from a single layer pattern to find a useful estimate of the lacquer thickness?

(I have tried creating imaginary data mathematically, then analysing it by different means including a Fast Fourier Transform. So far the results have been more or less reasonable, but sometimes smaller or larger than either actual film thicknesses.) Is this the best I can hope for, or is there a better method?

[Websta]

We use a spectrometer to measure this (with mixed results)
We use peak and trough measurements and a formula, which i do not have with me currently (msg me for the formula and I will try find it) to get a thickness in microns.

The big issues we find are that the 2 indeces need to be sufficiently different to get a proper response from the spectrometer, and also make sure that you sandblast the back side of the lens and adrken it sufficiently to get a good reflection.

Let me know if you need more info.

[Falstaff]

:)Thanks for your reply. I have tried shot blasting one side and blacking it out, so that I only get one reflection. It leads to a good result, but one thing puzzles me. Using the formula

peak wavelength X integer = 2 X index X thickness

as the basis for my statistical analysis, leads me most definitely to non integer values, but which differ by integers. I have met this before when a phase difference is introduced into the analysis. But in this case I cannot see why it is there.

Best Regards,

Falstaff

[chip anderson]

Tried change in curvature measurements with Newton's rings?

[Websta]

Hi Falstaff,

You lost me at thanks:(.........I am no expert when it comes to understanding how the integers and wavelengths correspond in terms of their measured transmittances/reflectances.

I am not familiar with the formula you have. The one I have seems more simple.

Sorry i can not offer much more advice, but i am sure one of the other members (read: experts;)) here will guide you more than i can.

Good Luck.

[Falstaff]

Sorry it has taken a while for me to reply. I have been on holiday.
I would be interested to know what formula you use, if it is not too much trouble, I know that there are several methods for working out lacquer thicknesses, and that they do not always reach exactly the same answer, but generally agree to the first decimal place. My spectrophotometer has its own built in analysis which takes account of dispersion and absorption, but I do not often trust it for all sorts of reasons.

Best Regards,

Falstaff:)

[Falstaff]

Sorry, the reply is late. I have been on holiday.
No I have never tried that. It sounds interesting, and a useful check on our incoming quality.

Best Regards,

Falstaff

[Mauro.Airoldi]

the correct procedure is the follows:
1sand

[Mauro.Airoldi]

sorry...
the correct procedure is the follows:
1 sand blast the lens concave side (to have only CX th)
2 measure in reflaction by a spectrophotometer
3 pick the peak (or valley)
4 apply the formula: th= (lamb max*lamb min) / material index*2* (lamb max*lamb min)
example: lamb max=600
lamb min= 550
index material = 1.49
th= 2214 nanometer =2.21 micron
My suggestion is to measure different peack and make a media

Mauro

[Falstaff]

Thanks. This seems to work quite well, although I get slightly different answers when I choose different peaks or troughs. I have been working on a computer procedure to find the most likely position of a peak when the data is noisy. This may be part of my problem.

Best Regards,

Falstaff

[Rafael]

Thank you for the formula Mauro.Airoldi.
Works fine.
But there is a litte algebraic sign mistake in the formula.

The Formula should be:

th= (lamb max*lamb min) / material index*2* (lamb max-lamb min)

[Mauro.Airoldi]

You right, it is lambda max - lambda min
sorry for mistake

[Falstaff]

Thank you for pointing that out. It has not been a problem for me because I realised that the formula can be derived from :

m X wl = 2 X n X t,

where m is an integer, wl is the wavelength of a peak, n is the index of the film and t is the lacquer thickness. (This formula is the standard physics text book one for thin films at normal incidence.)

If you take two consecutive peaks (wl1 and wl2) in the interference pattern, they will have m values exactly 1 apart. This gives two simultaneous equations allowing m to be eliminated.

This gives :

film thickness = (1 / (2 X n)) X (wl2 X wl1) / (wl2 - wl1)

I have been working on a statistical analysis of the wavelengths of several consecutive peaks or troughs, but it is difficult to take dispersion into account.

Thanks to all those who helped put me on track. Falstaff.:cheers: