How Thick Is the Oil Film on Wet Pavement to Appear Red?

In summary, the conversation discusses the thickness of an oil film on a wet pavement, which appears predominantly red when viewed perpendicular to the pavement. The participants discuss using the formula 2nt=(m+1/2)wavelength to solve for the thickness, but encounter difficulty with the second equation. It is suggested to switch the two equations and use different unknowns for m and m' to find the ratio and solve for m and t.
  • #1
cmilho10
20
0
A thin film of oil (n = 1.10) is located on a smooth, wet pavement. When viewed perpendicular to the pavement, the film appears to be predominantly red (640 nm) and has no component of wavelength 569 nm. How thick is the oil film?


I have tried using 2nt=(m+1/2)wavelength but am having absolutely no luck in getting the right answer
 
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  • #2
640 fits your formula.
569 would fit the formula with the same m, but without the 1/2.
Then solve for m and t.
 
  • #3
OK So:

2.2t=(m+1/2)(640nm) and 2.2t=m(569 nm)

How do I solve for m?
I tried solving the second equation for t and then plugging it into the first but still no luck...
 
Last edited:
  • #4
The 569 formula could use m+1 instead of m, so that m will be positive.
But then m should be an integer. I tried using m+N, but still can't get an integer for m. There must be a wrong number somewhere.
 
  • #5
The oil film rests on water. That's where the "wet pavement" comes in. So the reflections from the front and back surfaces are both "external" reflections: at the top surface, the light goes from n=1 to n=1.1 (n increases, so it's external), and at the bottom surface, the light goes from n=1.1 to n=1.33 (again n increases so again it's external). Therefore there is no phase difference from the reflections. The phase difference comes completely from the path difference. You need to switch the two formulas around:

[tex]2nt = m(640 nm)[/tex]

[tex]2nt = (m+1/2)(569 nm)[/tex]

This gives you pretty nearly an integer for m.
 
  • #6
cmilho10 said:
OK So:

2.2t=(m+1/2)(640nm) and 2.2t=m(569 nm)

How do I solve for m?
I tried solving the second equation for t and then plugging it into the first but still no luck...

Just a general warning about this type of question: the two equations do not necessarily contain the same integer "m"! So the safe thing to do is to give them different names, m and m'.

So there are really 3 unknowns (t, m and m'). Since you have only two equations, it might sound impossible to solve. But the fact that two of the unknowns are integers restrict the solutions and helps to solve the problem (although there are still an infinite number of solutions!)

There are two possible approaches:

a) In the first equation, plug in several values of m (0, 1, 2, etc) and calculate t. You get a list of possible t. Do the same thing with other equation (try several values of m' and calculate t). Looking at your two lists, you should quickly spot a value of t that appears in both. That`s your answer. (notice that if you kept going, you would find other matches but usually the prof wants the smallest value).

b) Isolate t in one equation and plug in the other. Solve for the ratio m/m'. Usually it is an obvious rational number. For example you might get 1.33 in which case you know that m=4 and m'=3. (of course, it could also be 8 and 6 or 12 and 9, etc but again, one wants the smallest value, usually).


Hope this helps .

Pat
 

1. What is interference in thin films?

Interference in thin films is an optical phenomenon that occurs when light waves reflect off the top and bottom surfaces of a thin film. These reflected waves interfere with each other, resulting in a pattern of light and dark bands called interference fringes.

2. How does the thickness of a thin film affect interference?

The thickness of a thin film is directly related to the number of interference fringes that are produced. Thicker films will produce more fringes, while thinner films will produce fewer fringes. This is because the distance between the two reflecting surfaces determines the phase difference between the reflected waves.

3. What is the difference between constructive and destructive interference in thin films?

Constructive interference occurs when the peaks of the two reflected waves align, resulting in a brighter interference fringe. Destructive interference occurs when the peaks of one wave align with the troughs of the other wave, canceling each other out and creating a dark fringe.

4. How is the color of a thin film determined by interference?

The color of a thin film is determined by the wavelength of light that is reflected off the film. This is determined by the thickness of the film and the refractive indices of the materials it is made of. Different thicknesses and materials will result in different colors.

5. What are some real-world applications of interference in thin films?

Interference in thin films is used in a variety of applications, such as anti-reflective coatings on eyeglasses, color filters for LCD screens, and anti-counterfeiting measures on banknotes. It is also used in scientific research to study the properties of materials and in engineering to measure the thickness of thin films.

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