Thin film interference on water

[Taniaz]

Homework Statement

An oil tanker spills a large amount of oil (n =1.47) into the sea (n = 1.33).

a) If you look down onto the oil spill from overhead, what predominant wavelength of light do you see at a point where the oil is 390 nm thick? What color is the light?
b) In the water transmitted under the slick, what visible wavelength (as measured in air) is predominant in the transmitted light at the same place in the slick as in part a?
c) If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelengths would there be constructive interference in the light that reflects back downward?

Homework Equations

If anyone of the waves has a half cycle phase shift then for constructive interference 2t=(m+1/2)(wavelength).

The Attempt at a Solution

I've done parts a and b and I know the equation for constructive interference will be the same as above as one of the waves going from water will have a half cycle phase shift but I'm not sure what wavelength they are asking for?Do they want the wavelength as seen in air?water?

In parts a and b I calculated the wavelength in air by substituting
wavelength in oil = wavelength in air / index of refraction of oil.

 

[Charles Link]

What did you get for parts $a$ and $b$? I would be glad to check your calculations if you showed them.

 

[Taniaz]

Hi Chales, thank you for your reply.

I got the wavelength in air for part a to be around 458 nm. I used the constructive interference equation when one of the reflected waves has a half cycle phase shift 2t=(m+ 1/2)(lamda). I substituted lambda in oil = lambda in air / n using the equation (n1)(lambda 1)=(n2)(lambda 2)

For part b I got 573 nm and I used the destructive interference equation when one of the reflected waves has a half cycle phase shift so 2t=m(lambda). Because the question was asking for the wavelength in the visible range, only m=2 worked for both cases.

For part c, I'm not which wavelength the question is asking for, the wavelength in air or water? I know the equation will still be as in part a because one the reflected waves in water has a half cycle phase shift.

 

[Charles Link]

I agree with your answers to "a" and "b" . (And for part "a", it is the top surface that has the $\pi$ phase shift for the reflected wave). $\\$ For "c" the wavelength he sees will be the wavelength in air. Even though he is under water, what he sees is related to the wavelength in air. They want the wavelength in air. He sees the light that reaches him. $\\$ One way of looking at it: The light that an observer sees is related to the frequency of the light. For an observer is really immaterial what medium the light travels through before it reaches him. The observer will see the wavelength that the light has in air. The medium doesn't change the frequency.