Interference and Defraction

In summary, the problem asks for the bright areas, so I used the constructive interference equation. My set it up as one part of the wave being completely reflected while another part of the wave refracted into the oil and then is reflected by the water and refracted into the air. I can't figure out what to do next.
  • #1
UAPhys03
5
0

Homework Statement


7: a: A thin film of oil (n2=1.2 and t(thickness) = 600nm) lies on top of some water (n3=1.33) after spilling from an oil tanker. What wavelengths of visible light would be bright in reflection? use n1 = 1 (air)


Homework Equations


CI (constructive interference): 2t = (m+1/2)Lambda
DI (destructive interference): 2t = mLamda
Lamda = Lamda/n

Layers of film
air= n1
oil = n2
water = n3

The Attempt at a Solution



The problem asks for the bright areas so I used the constructive interference equation. my set it up as one part of the wave being completely reflected while another part of the wave refracted into the oil and then is reflected by the water and refracted into the air. I can't figure out what to do next.

The answer is: (720nm, 480nm)
 
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  • #2
I think dsin(theta) = m(lambda) would be more useful since there are two answers.

lambda = (dsin(theta))/m

d= (1/6000)x10^(-9)

Im not sure what the n's are in your question, but the two answers and most probably due to +/- theta.

maybe 'm' in the equation has something to do with the 'n's' from the question.

Hope this helps.
 
  • #3
Hi UAPhys03,

UAPhys03 said:

Homework Statement


7: a: A thin film of oil (n2=1.2 and t(thickness) = 600nm) lies on top of some water (n3=1.33) after spilling from an oil tanker. What wavelengths of visible light would be bright in reflection? use n1 = 1 (air)


Homework Equations


CI (constructive interference): 2t = (m+1/2)Lambda
DI (destructive interference): 2t = mLamda

I don't believe these equations are correct for this case, which would explain why you are not getting the right answer. Do you see why they are not right?

Lamda = Lamda/n

Layers of film
air= n1
oil = n2
water = n3

The Attempt at a Solution



The problem asks for the bright areas so I used the constructive interference equation. my set it up as one part of the wave being completely reflected while another part of the wave refracted into the oil and then is reflected by the water and refracted into the air. I can't figure out what to do next.

The answer is: (720nm, 480nm)
 
  • #4
Ok i figured it out. I had my equations mixed up like you suggested.

So for bright reflection that would be constructive interference and since both rays undergo a pi phase change, they are in phase.

CI: 2nt = mLamda , where n =1.2 for oil

I used m = 3

which gave 480nm

and m = 2

and that gave 720nm

Thanks!
 

1. What is the difference between interference and diffraction?

Interference and diffraction are both phenomena that occur when waves interact with each other or with obstacles. Interference refers to the interaction of two or more coherent waves, resulting in either constructive or destructive interference, while diffraction refers to the bending or spreading of waves as they pass through an opening or around an obstacle.

2. How do interference and diffraction affect the behavior of light?

Interference and diffraction are both important principles in the study of optics and the behavior of light. They both demonstrate the wave nature of light and can affect the intensity and distribution of light in an optical system. Interference can produce patterns of light and dark fringes, while diffraction can cause light to bend and create patterns of light and dark spots.

3. What are some real-world applications of interference and diffraction?

Interference and diffraction have many practical applications in various fields. In the study of light, they are used in optical devices such as diffraction gratings, interferometers, and holograms. In the field of engineering, they are used in the design of radio and television antennas. They also play a role in medical imaging techniques such as X-rays and MRI.

4. How is the phenomenon of interference and diffraction explained by the wave theory of light?

The wave theory of light states that light behaves as a wave, and interference and diffraction can be explained by the superposition of waves. When two or more waves interact, their amplitudes can either add up or cancel out, resulting in constructive or destructive interference, respectively. Diffraction can be explained by the bending of waves as they pass through small openings or around obstacles, similar to how water waves bend around a rock in a stream.

5. Can interference and diffraction occur with other types of waves besides light?

Yes, interference and diffraction can occur with all types of waves, including sound waves, water waves, and electromagnetic waves. The principles of interference and diffraction can be applied to any type of wave that exhibits wave-like behavior, such as wavelength and frequency. In fact, the study of interference and diffraction has contributed to our understanding of wave phenomena in various fields of science and engineering.

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