How Thick Is the Oil Film for Red and No Light Interference?

In summary, the conversation discusses the calculation of the thickness of a thin oil film on a smooth, wet pavement. The film has a refractive index of 1.30 and reflects red light at 640 nm and no light at 548 nm when viewed perpendicular to the pavement. The equations and attempt made to find the thickness involve using the refractive index of air, oil, and water, as well as considering constructive and destructive interference. After some trial and error, it is determined that the thickness of the film is approximately 733.04 nm, with some discrepancy and discussion about the validity of the reasoning used to obtain this value.
  • #1
quanticism
29
3
Problem

A thin film of oil (n = 1.30) is located on a smooth, wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no light at 548 nm. How thick is the oil film?

Equations and Attempt

Refractive index:
air - 1
oil - 1.3
water - 1.33

Assumption: pavement doesn't reflect any light.

Light will undergo a 180 degree phase reversal at the air-oil and oil-water boundaries so the net phase difference in the reflected light is 0.

Thus, for constructive interference, path difference must be an integer number of wavelengths so:

2t = m (640 nm)/1.3, where m is an integer

Similarly, for destructive interference, path difference must be a 1/2 number of wavelength so:

2t = p (548nm)/(2*1.3), where p is an integer


So, I tried to find integers values for m and p for which I could get destructive and constructive interference simultaneously by finding the ratio of m and p:

m/p = 137/320

So minimum integer values for which that ratio is satisfied is m = 137 and p = 320

Subbing it back into the above for t, I obtain t = 33723nm...

A friend told me the answer should be in the domain of the visible spectrum ~(380nm-750nm). But I don't see where I made the mistake. Could someone please nudge me in the right direction?

Note: Alternately, for destructive interference, I could have said:
2t = [(m+1/2)](548nm)/1.3 = (2m+1)(548nm)/[2(1.3)] where m is the "same" integer/order as the one I used for constructive interference.

If I equate this to the eqn for constructive interference, I obtain m = 2.978. Using this, I'll get t = 733.04 nm. However, for this method to work, doesn't m need to be an integer? I guess 2.978 is close to 3 but wouldn't what I did above be more "valid"? (Assuming I didn't make some mistake)
 
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  • #2
Well, I just tried m=3 and the answer was right. Still not happy about the reasoning behind it though so I'd still appreciate any insight.
 
  • #3
quanticism said:
Thus, for constructive interference, path difference must be an integer number of wavelengths so:

2t = m (640 nm)/1.3, where m is an integer
Good.

Similarly, for destructive interference, path difference must be a 1/2 number of wavelength so:

2t = p (548nm)/(2*1.3), where p is an integer
Not so good. Use 2t = (p + 1/2)(548nm/1.3). (You want an odd number of half wavelengths.)

Set those equal:
(p + 1/2)(548nm/1.3) = m (640 nm)/1.3
(p + 1/2)(548) = m (640)
(p + 1/2)(548/640) = m

Now just start plugging in values for p, starting with 0, to see which gives an integer value for m.
 
  • #4
Thanks for the reply :)

2t = (p + 1/2)(548nm/1.3, where p = 0,1,2,...
= (2p+1)(548nm)/(2*1.3)

Ah, I see. I should have said "2t = p (548nm)/(2*1.3), where p is an odd integer". But your expression is easier to deal with so I'll use it.

Now just start plugging in values for p, starting with 0, to see which gives an integer value for m.

I think equating the two expressions and finding the ratio m/(2p+1) is more efficient

m/(2p+1) = 137/320. But for 2p+1=320, p can't be an integer.

So I'm tempted to say that with the values given by the question, it's impossible to find a thickness that gives perfect constructive + destructive interference simultaneously.

The answer that the test accepted used m=3.

m/(2p+1) = 137/320 [tex]\approx[/tex] 3/7.00729927... so m [tex]\approx[/tex] 3, p [tex]\approx[/tex] 3

Or using your method:

(p + 1/2)(548/640) = m

Now just start plugging in values for p, starting with 0, to see which gives an integer value for m.

When I plug p = 3 in, I get m = 2.996875. This method is probably better in this case because I honestly don't know how I could have gone from 137/320 to 3/7.00729927...

So this question probably gave "dodgy" values to me?

My friend got these values in his version:
Constructive interference: 640nm
Destructive interference: 512nm

m/(2p+1) = 512/(640*2) = 2/5. So m = 2, p=2. (Except he had m as the integer variable for both cases so he was able to "solve for m". Guess he was lucky that m = p in this case).

Or using Doc Al's expression.

(p + 1/2)(512/640) = m
m/(p+1/2) = 512/640= 4/5
m/(2p+1) = 2/5 (I converted p+1/2 to 2p+1 because this "lowest integer ratio" method requires the denominator and numerator to be integers)
 
  • #5




Based on the information provided, it seems that you have correctly used the equations for constructive and destructive interference to determine the thickness of the oil film. However, there may be some small errors in the calculations which could explain the discrepancy between your result and your friend's suggestion. It is also possible that there are other factors at play, such as the angle of incidence of the light or the presence of any impurities in the oil film. I would suggest double-checking your calculations and also considering any other variables that could affect the interference patterns. Additionally, it may be helpful to consult with a colleague or another scientist to review your approach and results. Overall, your attempt to solve this problem using the principles of interference is commendable and it shows a strong understanding of the topic. Keep up the good work!
 

Related to How Thick Is the Oil Film for Red and No Light Interference?

1. What is a thin film of oil interference?

A thin film of oil interference refers to the phenomenon in which a thin layer of oil on the surface of a liquid or solid material causes the reflection and refraction of light, resulting in interference patterns and changes in color.

2. How does a thin film of oil interference occur?

A thin film of oil interference occurs when light waves pass through a thin layer of oil and reflect off the top and bottom surfaces. These reflected waves interfere with each other, resulting in a change in the wavelength and intensity of the light that is perceived as a colorful pattern.

3. What factors affect the appearance of a thin film of oil interference?

The appearance of a thin film of oil interference is affected by the thickness of the oil layer, the refractive index of the oil and the material it is on, and the angle at which the light hits the surface. Changes in any of these factors can result in different interference patterns and colors.

4. What are the practical applications of thin film of oil interference?

Thin film of oil interference has various practical applications, such as in the production of anti-reflective coatings for lenses and screens, in the analysis of materials' properties, and in the creation of colorful packaging and holograms.

5. How can we control or manipulate thin film of oil interference?

Thin film of oil interference can be controlled and manipulated by adjusting the thickness of the oil layer, changing the refractive index of the oil or the material it is on, or by using polarized light. These factors can alter the interference patterns and colors produced, allowing for control and manipulation of the phenomenon.

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