Thin Film Interference

[studentteacher]

I'm a little confused by what these equations are saying:

 

∆d = 2t (phase shift for neither or both waves)

∆d = 2t + 0.5 * wavelength (phase shift for only one wave)

∆d = m*wavelength (constructive)

∆d = (m + 0.5)*wavelength (destructive)

 

I think t is the thickness of the film, but then what is ∆d?

[studiot]

I'm a little confused by what these equations are saying:

 

∆d = 2t (phase shift for neither or both waves)

∆d = 2t + 0.5 * wavelength (phase shift for only one wave)

∆d = m*wavelength (constructive)

∆d = (m + 0.5)*wavelength (destructive)

 

I think t is the thickness of the film, but then what is ∆d?

 

Delta d is the optical path difference between successive rays, ie the difference between travelling twice through the film and reflecting off the top surface.

 

I will add a quick sketch.

 

Edited March 8, 2017 by studiot

[studentteacher]

Hi thanks for the reply. Is the unit of delta-d distance? If the ray is perpendicular to the surface, is dealt-d just the same as 2t?

[studiot]

Hi thanks for the reply. Is the unit of delta-d distance? If the ray is perpendicular to the surface, is dealt-d just the same as 2t?

 

It is common to measure in wavelengths rather than (micro or nano)metres.

That is why you have the factor of one half.

 

Yes delta-d (normally just delta) would be twice the thickness if viewed square on so B coincided with A.

However think about it.

What would you see in this case?

 

The whole point is that there is another partial reflection at B (which I did not draw) for oblique incidence.

And this will in turn generate another partial reflection and so on.

So you will see a succession of (weakening) fringes.

 

By the way, why was this in the homework section?

It seems like a genuine physics question to me, seeking information, not requiring you to perform a calculation or answer a question.

You would probably get more response there.

Edited March 9, 2017 by studiot

[studentteacher]

I am really just not getting this. here is a homework problem.

 

A film of oil (n=1.25) floats on water (n=1.33). What is the thinnest film that produces a strong green reflection (wavelength = 500 nm).

And,2 about anti-reflection coatings on glasses.

 

A 90-nm thick coating is applied to a lens. What must be the coating's index of refraction to be most effective at 480 nm? Assume the coating's index of refraction is less than that of the lens.

 

If the index of refraction of the coating is 1.38, what's the minimum thickness the coating should be to be most effective at 480 nm?

 

None of the formulas given above include a term for index of refraction.

Is the idea that you figure ut which of the first two equations for delta-d to use, based on the index of refraction of the materials, and then substitute into one of the second two based on whether you're working with constructive or destructive interference?

 

I think in the first hw problem, both waves of shifted, since the air-oil-water sequence is always from low n to higher n. So, solve the equation 2t=500/1.25.

 

For the second, again both waves are shifted, but we're looking at destructive interference. So, 2*90=0.5*480/n ... and n = 1.33