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## Homework Statement

A thin (460 nm) film of kerosene (n=1,20) is spread out on water (n = 1,30).

Light hits the (horizontal) surface of the film coming (almost) perpendicularly from above.

A diver underwater, below the film.

a) Find out the visible wavelength reflected into air that has the maximum intensity.

b) Find out the visible wavelength refracted into water that has the maximum intensity.

## Homework Equations

- when light travels a path of length [itex]\ell[/itex] through a medium whose refractive index is n, it undergoes a phase shift. The emerging light has the same (opposite) phase as the incident light if the path length [itex]\ell[/itex] is an even (odd) multiple of [itex]\lambda/(2n)[/itex].
- when light reflects at the interface between two different transparent media it can undergo a [itex]\lambda/(2n)[/itex] phase shift. This happens when the refractive index of the first medium is less than the second.
- when light is refracted at the interface between two different transparent media, there is no phase shift.

## The Attempt at a Solution

a) I've been able to solve this using the above facts.

The interference occurs between two reflected rays of different order. The first is reflected by the air-kerosene interface, the second is reflected by the kerosene-water interface, and travels twice through the film. Neither undergoes a phase shift at reflection because [itex]n_{\rm air}<n_{\rm kerosene}<n_{\rm water}[/itex].

As a result, the condition for constructive interference is

[tex]2 t = m \frac{\lambda_m}{n_{\rm kerosene}}[/tex]

Upon inspection, only [itex]\lambda_2 = 552\; {\rm nm}[/itex] falls in the visible window.

b) The provided solution is motivated is equivalent to assuming that refraction is maximum when reflection is minimum. Hence

[tex]2 t = (2m +1) \frac{\lambda_m}{2 n_{\rm kerosene}}[/tex]

and basically only [itex]\lambda_2 = 442\; {\rm nm}[/itex] falls in the visible window.

**Doubts**

Despite I'm obtaining the proposed results, I am a little confused by question b).

The inference "maximum refracted intensity when minimum reflected intensity" seems to come from Fresnel equations, but it can't be that, right?

Here the reflected light has minimum intensity because of interference between two (or more) reflected rays, not because each ray is individually faint.

I think that the solution should be obtained with the same kind of reasoning used for the reflected right.

It seems to me that there is one "principal" and one "secondary" refracted ray. The second is due to a reflection at the kerosene-water interface and another reflection at the kerosene-air interface.

Only the first brings about a [itex]\lambda/(2n)[/itex] phase shift. In order to have constructive interference the path of the second ray inside the film should compensate this phase shift. Therefore

[tex]2 t = (2m +1) \frac{\lambda_m}{2n_{\rm kerosene}}[/tex]

Is this correct?