# Superposition Interference

1. Mar 1, 2009

### nubey1

1. The problem statement, all variables and given/known data
A very thin oil film(n=1.25) floats on water(n=1.33). What is the thinnest film that produces a strong reflection for visible light with a wavelength of 700nm?

2. Relevant equations
$$\lambda$$=(2nd)/(m-0.5)
d=(m-0.5)*($$\lambda$$/2n)
thinnest film has m=1

3. The attempt at a solution
d=(0.5*700)/(2*1.25)=140nm
This answer is incorrect. Is there a problem in my technique?

2. Mar 1, 2009

### Delphi51

It seems to me the film will just be a half wavelength thick so that light travelling through it to the water surface and back will be a full wavelength delayed and so will constructively interfere with waves reflected off the film surface.

Of course that will be in terms of the wavelength in the film, so you'll need a Law of Refraction formula that lets you computer the wavelength in the film.

3. Mar 2, 2009

### nubey1

I was not introduced to the law of refraction. I did look ahead in my physics book catch a glimpse of what it is about. This is my attempt at just hucking the formulas.
$$\vartheta$$=arcsin(n2/n1)=arcsin(1.25/1.33)=1.222
$$\vartheta$$*$$\lambda$$=1.222*700=855.5nm

Is this correct?

4. Mar 2, 2009

### Delphi51

If I understand this correctly, it says the wavelength is 700 nm in AIR.
In the oil film the wavelength will be reduced by a factor of 1.25 to 560 nm.
The part of the law of refraction you need is that
$$\lambda1/$$ $$\lambda2 = n2/n1$$
Use n1 = 1 for air, n2 = 1.25 for the oil.