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

AI Thread Summary
The discussion focuses on calculating wavelengths for constructive interference in thin film interference involving a water film between glass. For transmitted light, the key equation is based on the condition for constructive interference, which considers the phase changes upon reflection. The user initially struggles with determining visible wavelengths for transmitted light but is reminded that interference occurs in both transmitted and reflected waves. The conversation emphasizes that even transmitted light experiences interference, with one ray reflecting off the surface and another reflecting within the film. Understanding these principles is crucial for accurately identifying visible wavelengths in both transmitted and reflected light.
fedecolo
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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?
 
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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.
 
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