# Thin film interference

• fedecolo
In summary, a question was asked about finding the wavelengths of light transmitted and reflected when white light is incident on a thin film of water between two glass surfaces. The solution involves setting equations for constructive interference and assuming no light is absorbed by the material. It is important to note that there is always interference in both the transmitted and reflected waves, and standard textbooks provide examples of this.f

## 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?

(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.

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(?)

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.