Why Does Oil Appear Red on Water at Certain Thicknesses?

[GreenDinos]

A thin film of oil (n = 1.46 is spread over a puddle of water (n = 1.33). In a region where the film looks red from directly above ( = 626 nm), what is the minimum possible thickness of the film?

I thought this was simple but it says my answer is wrong. Ok the light goes from a smaller to larger n value when it goes from air to the oil, so it shifts 180 deg. The light that goes through the oil reflects off the water and doesn't shift so now the waves have to be lined up so they're not destructive. The minimum thinkness that would cause the wave reflecting off the water to match up constructively with the first reflection off the oil should cause the wave to shift by 1/2 wavelength- so the minimum thickness should be wavelength/4 ...but it's not. I even found an equation for this in the book
2t = m wavelength
and since we have to shift it by .5 it would be
2t = (m+.5) wavelength

but I'm doing something wrong.
thanks

 

[Bystander]

Wavelength in air is related to wavelength in oil in what way?

 

[wais]

for any help!

It seems like you have the right idea, but there may be some error in your calculations. Let's break it down step by step to see where the issue may be.

First, let's define some variables:
- t = thickness of the oil film
- n1 = refractive index of air (1.00)
- n2 = refractive index of oil (1.46)
- n3 = refractive index of water (1.33)
- λ = wavelength of light (626 nm)

Now, we can use the equation you mentioned: 2t = (m + 1/2)λ, where m is the order of the interference.

We know that the light is first traveling through air, then through the oil, and finally reflecting off the water. So the total phase shift would be:
Δφ = (n1t + n2t) + (n2t + n3t)
We can simplify this to:
Δφ = (n1 + n2)t + (n2 + n3)t
And since we want the phase shift to be 180 degrees, we can set Δφ = π.

Now, plugging in the values for n1, n2, and n3, we get:
π = (2.46)t + (2.79)t
π = 5.25t
t = π/5.25 = 0.599λ

So the minimum thickness of the oil film for the red light to be visible would be 0.599λ, which is approximately 374 nm. This is slightly different from your answer of λ/4, which would be 156.5 nm. It's possible that you made a calculation error or used slightly different values for the refractive indices.

I hope this helps clarify the concept and find the correct answer. Keep in mind that the equation 2t = (m + 1/2)λ is a simplified version and may not always give the exact answer. In some cases, a more complex formula may be needed to account for multiple reflections and phase shifts.