Find Thickness of Oil Film on Water, \lambda = 550 nm

AI Thread Summary
To find the thickness of an oil film on water that appears yellow-green at 550 nm, the refractive indices of oil (1.50) and water (1.33) are used in the equation for constructive interference. The net phase difference is calculated, leading to the formula for thickness: d = (2m + 1)λ0 / 4nf. The challenge lies in determining the appropriate integer value for m, as increasing m results in greater thickness, which may exceed the interference threshold. It is noted that films thinner than 1 mm will show interference, while thicker films will not, indicating a limit to the values of m that can be used.
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Homework Statement


A thin film of oil (n = 1.50) floats on water (n = 1.33). If the film appears yellow-green (\lambda = 550 nm) when viewed at normal incidence, how thick is it?

Given:
n_{f} = 1.50
n_{water} = 1.33
\lambda_{0} = 550 nm
\theta_{i} = 0

Required:
Thickness (d)

Homework Equations


\Delta\phi_{net} = 2m\pi (Since we are looking for constructive interference)


The Attempt at a Solution



\Delta\phi_{1} = \pi since the first beam is reflected from a material with a higher index of refraction


\Delta\phi_{2} = 2dk_{f} due to path difference
k_{f} = \frac{2\pi}{\lambda_{f}}
\lambda_{f} = \frac{\lambda_{0}}{n_{f}}
Substituting everything in, we get:
\Delta\phi_{2} = \frac{4dn_{f}\pi}{\lambda_{0}}


\Delta\phi_{net} = \Delta\phi_{2} - \Delta\phi_{1}
\Delta\phi_{net} = (\frac{4dn_{f}}{\lambda_{0}}-1)\pi


So, from the equation given for constructive interference, we're left with:
(\frac{4dn_{f}}{\lambda_{0}}-1)\pi = 2m\pi
Solving for d, I get:
d = \frac{(2m+1)\lambda_{0}}{4n_{f}}

Assuming I did everything else right (a big assumption), I'm still left with one problem. What do I use for my integer (m = 0, 1, 2, 3, ...)? The question does not specifically state that I'm looking for the thinnest value for the film. All the similar examples in my book solve for the wavelength of light that you see due to a film of a given thickness. For those problems, you choose the value of m such that the resulting wavelength is in the visible range. In this case, however, I am not sure what constraints I should apply such that I get one thickness out of the problem.
 
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As you increase m value, the thickness of the film will increase. Consequently space between the interfering beams will increase and interference may not take place.
 
My book tells me that a film as thick as 1 mm does not show interference in daylight. My problem is that m > 5000 before the thickness reaches 1 mm.
 
m=0 corresponds to the normal incidence.

I think you have misunderstood what your textbook is saying about the film thickness.
rl.bhat tells us that a thicker film will not show interference.
A film thinner than 1mm will show interference.
 
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