# Thin Films

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33

Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectrometer. What is the longest wavelength of the light in water that is transmitted most easily to the diver?

air n = 1
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oil n = 1.5
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water n = 1.33
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t = 200 nm, there is 1 phase change in this problem and we want constructive interference

2t = (m + .5)(lambda of oil / n of oil)

When I solve for lambda of oil I get 1200 nm.

Then I use (lambda of oil)(n of oil) = (lambda of water)(n of water)

When I solved for the wavelength in water I got 1353 nm. The correct answer is 451 nm, and I am not sure what I am doing wrong.

## The Attempt at a Solution

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I now see I made the following mistake:

2t = (m + .5)(lambda of air / n of film) and let m = 0

(lambda of air)(n of air) = (lambda of water)(n of water)

After solving for lambda of water, I am still getting the wrong answer.

One of the hints given is the following: "This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energy ensures that maximum transmission and minimum reflection occur at the same time (i.e., if the energy did not reflect, then it must have been transmitted to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the wavelength of maximum transmission. In some cases, working the problem one way may be substantially easier, so you should keep both approaches in mind."

Does anyone have any input?