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

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Homework Help Overview

The problem involves determining the thickness of a thin oil film floating on water, which appears yellow-green at a wavelength of 550 nm. The refractive indices of the oil and water are provided, and the context suggests a focus on constructive interference in light reflection.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the concept of phase shifts in reflection and path differences to derive a formula for thickness. They express uncertainty about the appropriate integer value for m in the context of constructive interference.
  • Some participants question the implications of increasing m on the film thickness and its visibility in daylight, suggesting that higher values may lead to a lack of observable interference.
  • Others clarify that a film thickness of 1 mm is significant in the context of interference visibility, indicating a threshold for effective interference patterns.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying assumptions regarding the relationship between film thickness and observable interference. There is no explicit consensus on the value of m or the implications of film thickness on interference visibility.

Contextual Notes

Participants note that the problem does not specify constraints on the integer m, leading to uncertainty in determining the thinnest film thickness. Additionally, the discussion references a threshold thickness of 1 mm for observable interference, which may influence the choice of m.

maybetoday
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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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