Optics - destructive interference

[ChuckB]

Homework Statement

White light (most intense at 550nm wevelength) shines on a layer of water (n=1.33) setting on a sheet of plastic film (n=1.25). What minimum thickness of water will produce destructive interference on the surface?[/B]

Homework Equations

a) phase reversal of light reflecting off an optically denser medium
b) n=c/v (?)[/B]

The Attempt at a Solution

I am absolutely stumped on this problem. 1) The light reflecting off the top of the water layer will be, as I understand it, phase reversed by the reflection. 2) The light will also slow as it enters the water such that v = c / Nwater, and then reflect back through the water to interfere destructively with the light in 1. This seems to tell me that if the water is one molecule thick, there would be destructive interference due to the phase reversal? As you may be able to tell, I'm just flat stumped. Any help greatly appreciated.[/B]

 

[blue_leaf77]

will produce destructive interference

It doesn't ask for the constructive interference?

 

[ChuckB]

I double checked, it asks for that minimum thickness of water that will result in destructive interference.

 

[blue_leaf77]

Ok, then what you need is the expression of the optical path difference between the rays reflected from the upper surface of the water (air-water interface) and those reflected from the bottom surface (water-plastic interface). Since it's not specified, you can assume normal incidence of the light. See the picture below,

If the thickness is ##d##, what is the distance traveled by the rays which is reflected from the bottom surface?

 

[ChuckB]

That would be 2d. So d should be half the wavelength of light in water (n=1.33). Or am I oversimplifying?

 

[Physics-Tutor]

Although, this is probably not what is asked from you in the exercise you mention, you are correct.

A soap bubble film has a similar construction similar to your problem: low n (air) / High n (soap) / Low n (air). As you may suspect, the iridescent colors of a bubble are due to thin film interference. With time, the thickness of the bubble wall will decreases. When the film reaches a thickness such that the bubble is about to pop, notice the change in color: the bubble becomes transparent - all reflected (visible) light is cancelled.

However, I suspect that what is asked from you is to find the a non-negligible thickness.
First, determine the phase difference between incident beam and beam reflected at the air / water interface: Φ₁
Second , determine the phase difference between incident beam and beam reflected at the water / plastic interface: Φ₂
equate the difference (Φ₂ - Φ₁) to the phase difference condition for destructive interference, and rearrange the equation so you can calculate the minimum thickness.

 

[blue_leaf77]

That would be 2d. So d should be half the wavelength of light in water (n=1.33). Or am I oversimplifying?

Right, it's 2d. So what is the phase accumulated by this ray when it emerges at the upper surface? Having found this, find the phase accumulated by the ray reflected at the upper surface, then take the phase difference between the phases of the two rays.

However, I suspect that what is asked from you is to find the a non-negligible thickness.

That might be the case, but I will let the OP derive the general relation between the thickness and phase difference to allow him to know why a negligible thickness can occur as the answer in this problem.