The maximum intensity for light transmitted through a thin film

  • #1
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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?
 

Answers and Replies

  • #2
RPinPA
Science Advisor
Homework Helper
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The inference "maximum refracted intensity when minimum reflected intensity" seems to come from Fresnel equations, but it can't be that, right?

It's just conservation of energy. If the energy doesn't come back, it must go forward. At the first interface, that means it's in the refracted beam. So the maximum energy refracted into the kerosene would be when reflection is at a minimum. But that's not the whole story. As you say, reflection and refraction occur at the kerosene-water interface as well. You need to consider the energy that makes it through both interfaces.

But since you're considering the effect of both interfaces to figure out when the energy coming back from BOTH reflections is a minimum, then any other energy must be going into the water (ignoring any absorbed within the kerosene itself).
 
  • #3
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It's just conservation of energy. If the energy doesn't come back, it must go forward. At the first interface, that means it's in the refracted beam. So the maximum energy refracted into the kerosene would be when reflection is at a minimum..

I do not think it's that.
I am afraid that somehow I dragged you in my doubt. Apologies. But by replying to your comment I think I cleared my doubt.

I think my confusion arises from two different meaning of "maximum": the first has to do with Fresnel equations and conservation of energy, the second has to do with constructive interference.

What you (and the textbook solution) seem to suggest is that it is a matter of energy: whatever light is not reflected is refracted (assuming no absorption).
But Fresnel equations depend on the angle, that here is fixed (and very small). So the ratio of transmitted/incident light is fixed, for a "single ray". Transmitted light is what it is.

Here "maximum" and "minimum" refer to constructive and destructive interference, so it cannot be a matter of a "single ray".
Even if the reflected intensity is maximum, the thickness of the film could be such that the reflected light cannot be seen due to destructive interference.

In the end, the textbook is not interested in assessing the amount of transmitted light, but only the wavelength attaining the maximum intensity (considering interference effects).

I think the textbook solution is correct, and probably works in any situation (any combination of refractive indices involved). But it's not a matter of energy, as it suggests (IMO).
 

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