Calculating Minimum Thickness of Oil Slick Using Refractive Indices

[jaded18]

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.

The index of refraction of the oil is 1.20. What is the minimum thickness t of the oil slick at that spot?
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All right, I know that light reflects off the two different surfaces of the film (air-oil interface and oil-air interface).
I also know that this phase difference one talks about is between two waves and when the difference is a full wavelength then the waves interfere constructively and if the difference is half a wavelength then the waves interfere destructively...
The light that that reflects off the oil-water interface has to pass through the oil slick where it will have a different wavelength. & The total extra distance it travels is twice the thickness of the slick.

How do I put this info together to solve the prob??

So I found this equation 2 * n (refractive index of oil) * d (thickness of oil) cos beta= (m-0.5) lambda where Cos beta has a maximum value of 1 so you can get the minimum value of 'd'. But when I solve fore d, I don't get the right answer...

I really want to understand this, so please tell me if I am approaching this wrong.

 

[Nino MMP]

The correct way to solve the problem is to use Snell's Law. Snell's Law states that n1 * sin(theta1) = n2 * sin (theta2), where n1 and n2 are the indices of refraction of the two media, and theta1 and theta2 are the angles of incidence and refraction respectively. Using this equation, you can solve for the angle theta2 (the angle of refraction) in terms of theta1 (the angle of incidence). You can then use the law of sines to solve for the ratio of the thickness of the oil slick (t) to the distance from the surface of the water (d). When you substitute in the given values for n1 and n2, you get t/d = 0.9. Finally, you can solve for the thickness of the oil slick by multiplying both sides of the equation by d and solving for t. In this case, t = 0.9d = 0.9*750 nm = 675 nm.

 

[Blueshift5]

First of all, great job on recognizing the different surfaces and the concept of phase difference in interference! You are on the right track.

To solve this problem, we can use the equation you mentioned: 2 * n * d * cos beta = (m-0.5) * lambda. However, there are a few things we need to consider.

First, we need to convert the wavelength in air (750 nm) to the wavelength in oil. Since we know that the index of refraction of oil is 1.20, we can use the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the medium. So the wavelength in oil would be 750 nm * (1.33/1.20) = 830 nm.

Next, we need to find the angle beta. This angle is the angle of incidence of the light ray on the oil-water interface. We can use Snell's law to find this angle: n1 * sin theta1 = n2 * sin theta2, where n1 and n2 are the indices of refraction of the two media and theta1 and theta2 are the angles of incidence and refraction, respectively. In this case, n1 = 1.33 (index of refraction of water) and n2 = 1.20 (index of refraction of oil). We also know that theta1 = 0 degrees (since the light is coming straight down onto the oil slick) and we can solve for theta2, which turns out to be 39.7 degrees.

Now we can plug in all the values into our equation: 2 * 1.20 * d * cos 39.7 = (m-0.5) * 830. We want to find the minimum thickness, so we can use the smallest value for m, which is 1. So the equation becomes: 2 * 1.20 * d * cos 39.7 = 0.5 * 830. Solving for d, we get d = 0.55 micrometers. This is the minimum thickness of the oil slick at that particular spot.

I hope this helps and clarifies any confusion you had. Keep up the good work!