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

In summary, a thin film of oil with a refractive index of 1.50 floating on water with a refractive index of 1.33 appears yellow-green at normal incidence with a wavelength of 550 nm. Using the equation for constructive interference, the thickness of the film can be calculated to be (2m+1)*lambda_0/(4*n_f), where m is an integer that determines the thickness. However, for interference to occur, the film must be thinner than 1 mm.
  • #1
maybetoday
2
0

Homework Statement


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

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

Required:
Thickness (d)

Homework Equations


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


The Attempt at a Solution



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


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


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


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

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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 

What is the purpose of finding the thickness of an oil film on water?

The purpose of finding the thickness of an oil film on water is to understand the behavior and properties of the oil film, which can provide valuable information for various applications such as oil spill detection and environmental monitoring.

What is the significance of the wavelength (λ = 550 nm) in this experiment?

The wavelength (λ = 550 nm) is significant because it falls within the visible light spectrum and is commonly used for measuring the thickness of thin films. It is also the wavelength at which the human eye is most sensitive, making it easier to detect changes in the oil film thickness.

What materials and equipment are needed to conduct this experiment?

To conduct this experiment, you will need a light source, a spectrophotometer or spectrometer, a container of water, and a thin layer of oil (such as cooking oil) on the surface of the water. You may also need a polarizer and a light sensor for more precise measurements.

How is the thickness of the oil film on water determined?

The thickness of the oil film on water can be determined by using the principles of interference and diffraction. When light passes through the oil film, it undergoes constructive and destructive interference, creating a pattern of light and dark fringes. By measuring the distance between these fringes, the thickness of the oil film can be calculated using the known wavelength of light.

What are the potential sources of error in this experiment?

Potential sources of error in this experiment include variations in the thickness of the oil film, external factors such as air currents or vibrations, and limitations of the equipment used. It is important to control these variables as much as possible and take multiple measurements to ensure accurate results.

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