# Thin film light interference question

## 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.

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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.