What is the equation for constructive interference in thin film interference?

Click For Summary
SUMMARY

The discussion focuses on the equation for constructive interference in thin film interference, specifically between two glass pieces with a water film of 1 µm thickness. The refractive indices are n1=1.70 for glass and n2=1.33 for water. Key wavelengths for transmitted light are identified as 667 nm, 533 nm, 444 nm, and 381 nm, while reflected light wavelengths are 593 nm, 484 nm, 410 nm, and 355 nm. The participants emphasize that interference occurs in both transmitted and reflected waves, challenging the notion that transmitted light does not exhibit interference.

PREREQUISITES
  • Understanding of thin film interference principles
  • Familiarity with refractive indices and their implications
  • Knowledge of wavelength calculations in optics
  • Basic concepts of constructive and destructive interference
NEXT STEPS
  • Study the derivation of the thin film interference equation for transmitted light
  • Learn about the role of refractive indices in interference patterns
  • Explore the concept of phase shifts upon reflection in thin films
  • Investigate practical applications of thin film interference in optical devices
USEFUL FOR

Students and educators in physics, optical engineers, and anyone interested in the principles of light behavior in thin films and interference phenomena.

fedecolo
Messages
61
Reaction score
1

Homework Statement


Between two pieces of glass (##n_1=1.70##), there is a thin film of water (##n_2=1.33## and width ##d=1 \mu m##). If there is normal-incidence of white light on the water surface, find:
(a) which wavelenghts can be seen in the light transmitted (answer: 667 nm,533 nm,444 nm, 381 nm)
(b) which wavelenghts can be seen in the light reflected (answer: 593 nm,484 nm, 410 nm, (355 nm) )

The Attempt at a Solution


(a) I don't know how to deal with light transmitted
(b) I set the ##2d= \left(m-\frac{1}{2} \right) \cdot \frac{\lambda}{n_2}## to find the lambdas for ##m=1,2...## but I obtain only ##\lambda_{m=1}=355 nm## (below the light that can be seen).

Any help?
 
Physics news on Phys.org
fedecolo said:
(a) I don't know how to deal with light transmitted
You can assume that none of the light is absorbed by the material. So, the total amount of incoming light energy must be conserved. As less light is reflected, more light must be transmitted. And vice versa.

(b) I set the ##2d= \left(m-\frac{1}{2} \right) \cdot \frac{\lambda}{n_2}## to find the lambdas for ##m=1,2...## but I obtain only ##\lambda_{m=1}=355 nm## (below the light that can be seen).
Your equation looks OK. But, in order to get a wavelength of 355 nm, I have to let m = 8. You will need to choose values of m that yield visible wavelengths.
 
TSny said:
You can assume that none of the light is absorbed by the material. So, the total amount of incoming light energy must be conserved. As less light is reflected, more light must be transmitted. And vice versa.

Thanks. But how can I set the equation for the constructive interference in the first case? There is no interference since the light is transmitted(?)
 
fedecolo said:
Thanks. But how can I set the equation for the constructive interference in the first case? There is no interference since the light is transmitted(?)
There is always interference (constructive, destructive, or something in between) in the transmitted waves and in the reflected waves. Standard textbooks show how you get two reflected rays that interfere. See if you can show how to get two transmitted waves that interfere. Hint: One of the transmitted rays has no reflections. The other transmitted wave has more than one reflection.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
Replies
7
Views
6K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K