Thin film light interference question

[ia22]

Homework Statement

A uniform film of oil (n=1.31) is floating on water. When sunlight in air is incident normally on the film, an observer finds that the reflected light has a maximum at wavelength (lambda) = 450nm and a minimum at wavelength (lambda) = 600nm. What is the thickness of the oil film?

Homework Equations

The light equations for constructive and destructive interference. 2t = (m + 1/2) lambda / n and 2t = m * (lambda / n)

The Attempt at a Solution

I am having difficulties knowing where to start. We haven't had a question like this before.

 

[buffordboy23]

Welcome to PF!

You may need to be careful with the application of those equations. Under what scenario were they derived? a thin-film suspended in air?

Here's why: Let n2 be the index of refraction of the thin-film and let n1 and n3 be the indexes for other media, which may or may not be the same. Now suppose that the equations that you have quoted were derived for a thin-film suspended in air. Then n2 > n1 and n2 > n3. Now compare this with your scenario in the problem, where n1 is the air, n2 is the thin-film of oil, and n3 is water. According to my physics text, the index of refraction of water is 1.33. Thus, for this scenario, n2 > n1 and n3 > n2 and the scenarios are not equivalent. Recall that reflection off a higher-index results in a phase shift of 0.5 wavelength.

You may need to derive the equations yourself. My hunch is that you just need to reverse the equations, but I'll let you verify that on your own. Just let me know if you need further assistance.

 

[nlightner]

I can understand your initial confusion and difficulty in knowing where to start with this problem. However, let's break it down step by step and use some basic principles of thin film light interference to find a solution.

First, let's start with the fact that the reflected light has a maximum at 450nm and a minimum at 600nm. This means that the thickness of the oil film is causing constructive and destructive interference of the incident light at these two wavelengths. We can use the equations for constructive and destructive interference to determine the thickness of the film.

For constructive interference, we can use the equation 2t = (m + 1/2) lambda / n, where t is the thickness of the film, m is the order of the interference (in this case, m=0 since we are looking at the first maximum), lambda is the wavelength, and n is the refractive index of the film. Plugging in the values given in the problem, we get:

2t = (0 + 1/2) * 450nm / 1.31

Simplifying, we get:

2t = 171.37nm

Dividing both sides by 2, we get:

t = 85.69nm

This is the thickness of the oil film at which constructive interference occurs for light with a wavelength of 450nm.

Similarly, for destructive interference, we can use the equation 2t = m * (lambda / n). Using the same values as before, we get:

2t = m * 600nm / 1.31

Since we are looking at the first minimum, m=1. Plugging in this value, we get:

2t = 1 * 600nm / 1.31

Simplifying, we get:

2t = 458.02nm

Dividing both sides by 2, we get:

t = 229.01nm

This is the thickness of the oil film at which destructive interference occurs for light with a wavelength of 600nm.

Now, since we know that the thickness of the film is causing both constructive and destructive interference at these two wavelengths, we can assume that the thickness of the film lies somewhere between these two values. Therefore, we can take the average of these two values to get an estimate for the thickness of the film:

t = (85.69nm + 229.01nm) /