Interference of Light waves HELP

In summary, the conversation is about finding the thickness of a thin layer of oil on a glass plate using the observed interference patterns of light of varying wavelengths. The formula for constructive and destructive interference is discussed, and the need to find the correct value for m in the second equation through trial and error is mentioned.
  • #1
yoshima
4
0
I've been trying to get the anwser to this question for awhile. Any help is appreciated. :confused:

1. A glass plate (n=1.61) is covered with a thin, uniform layer of oil (n=1.24). A light beam of variable wavelength from air is incident normally on the oil surface. Observation of the reflected beam shows destructive interference at 549 nm and constructive interference at 727 nm. From this information, calculate the thickness of the oil film.

thanks
 
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  • #2
find the phase difference

Show what you've tried and where you are stuck.

Here's a hint to get you started: Light will reflect off the surface of the oil and the surface of the glass. These two reflections will interfere. What phase difference must they have to interfere constructively? destructively? How does the phase difference depend on the thickness of the oil?
 
  • #3
the solution that i though would be right

they way I've bee doing it is by using the formula for both destructive and constructive interference.

Constructive inter. --> 2nt=m*(lambda)c
destructive inter. --> 2nt=(m+1/2)*(lambda)d

Because the thickness of the film is the same --> 2nt = 2nt

:. (m+1/2)/(m) = (lambda)c/(lambda)d = 727/549

and from here solve for mc and sub that into the constructive formula 2nt=m*(lambda)c and you get the thickness.. but teh anwser is not right.

Is there another way of solving this problem or am i solving the wrong way.
 
  • #4
yoshima said:
Constructive inter. --> 2nt=m*(lambda)c
destructive inter. --> 2nt=(m+1/2)*(lambda)d
In your second equation, what you have as "m + 1/2" could also be m + 3/2, m + 5/2, etc. You need to find the value that makes m an integer. Trial and error!
 

Related to Interference of Light waves HELP

1. What is interference of light waves?

Interference of light waves is the phenomena where two or more light waves interact with each other and create a new wave. This can result in either constructive interference, where the waves reinforce each other and create a brighter light, or destructive interference, where the waves cancel each other out and create a darker region.

2. How does interference of light waves occur?

Interference of light waves occurs due to the superposition principle, which states that when two waves overlap, the displacement of the resulting wave is equal to the sum of the individual displacements of the two waves. This results in either constructive or destructive interference, depending on the phase difference between the waves.

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

Constructive interference occurs when two waves are in phase, meaning that their crests and troughs align, resulting in a brighter light. Destructive interference occurs when two waves are out of phase, meaning that their crests and troughs do not align, resulting in a darker region.

4. What are some real-life applications of interference of light waves?

Interference of light waves is commonly used in technology such as anti-reflective coatings on glasses and camera lenses, as well as in the production of holograms. It is also the basis for interferometry, a technique used in science and engineering to make precise measurements.

5. How does the wavelength of light affect interference?

The wavelength of light plays a crucial role in interference, as it determines the phase difference between the waves. If the two interfering waves have the same wavelength, the interference pattern will be consistent. However, if the wavelengths are different, the interference pattern will vary and can result in complex patterns.

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