• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Thin film reflection of light

  • Thread starter DottZakapa
  • Start date
65
4
Homework Statement
Light with wave length 648 nm in air is incident perpendicularly from air on a film 8.76 micrometers think and with refractive index 1.35. par of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface where the fil is again in contact with air. How many waves are contained in its round trip through the film? What is the phase difference between these two parts of the light as they leave the film?
Homework Equations
thin film
Homework Statement: Light with wave length 648 nm in air is incident perpendicularly from air on a film 8.76 micrometers think and with refractive index 1.35. par of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface where the fil is again in contact with air. How many waves are contained in its round trip through the film? What is the phase difference between these two parts of the light as they leave the film?
Homework Equations: thin film

the light incident on the film, having an index of refraction lower than the film has a phase difference of ##\frac{λ}{ 2}## the reflection from the second surface has no phase difference, therefore the formula for thin films in order to see a reflection would be

##2t=\frac{λ}{ 2}##

i don't understand how to answer the question... can anyone give some help?
 

Charles Link

Homework Helper
Insights Author
Gold Member
2018 Award
4,385
1,857
These problems are most easily done by introducing the optical path length which is ## OPL= 2nt =2(1.35)(8.76 \cdot 10^3 ) ## nm, rather than converting the wavelength in air to its wavelength in the material. Meanwhile ## \lambda =648 ## nm. The question should read "how many wavelengths..." and not how many waves. That calculation is of course quite simple, and is ## x=\frac{OPL}{\lambda} ##. ## \\ ## For the next part, the phase change from the reflection off the front surface is ## \pi ##, (which does correspond to half of a wavelength), but the correct way of saying the phase change is to call it ## \pi ## (radians).## \\ ## For the second question, they could be more descriptive, because the problem involves the interference of coherent light waves. Whatever result you get for ## x ##, you subtract off any integer, and just take any remaining fraction and multiply by ## 2 \pi ##. I did the arithmetic and this one turns out to be almost a round number. I'm going to bend the rules here, because I really can't explain the concepts properly without pretty much giving the answer at the same time. The arithmetic I did gave me ## x=36.5 ##. That is the answer to the first part of the problem, but when you answer the question about phase difference, the 36 gets dropped, (it accounts for a total phase of ## (36)(2 \pi) ##, but integers of ## 2 \pi ## have no effect on the interference). This leaves ## \phi_{back \, surface}= (.5)(2 \pi) =\pi ##, while ## \phi_{front \, surface}= \pi ##, so that there is a phase difference of ## \phi=0 ## between the two reflected waves, and thereby constructive interference. IMO, this problem is good for an example problem, but it is asking a lot of the students in an introductory class to solve it on their own.
 

Want to reply to this thread?

"Thin film reflection of light" You must log in or register to reply here.

Related Threads for: Thin film reflection of light

  • Posted
Replies
1
Views
8K
Replies
3
Views
4K
Replies
3
Views
14K
Replies
1
Views
906
Replies
6
Views
436
  • Posted
Replies
8
Views
2K
Replies
9
Views
12K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top