# Index of refraction and oil/water

Tags: index, oil or water, refraction
 P: 150 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? _________________________________ I know that index of refraction = c/wavelength where c is a velocity.... Can anyone PLEASE give me a hint as to how I would approach this Question????
 Sci Advisor HW Helper P: 8,953 Have a look at http://hyperphysics.phy-astr.gsu.edu...t/oilfilm.html
 P: 150 i get an equation that is irrelevant to the problem.. 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??????? WOW this question is CONFUSING
 Sci Advisor HW Helper P: 8,953 Index of refraction and oil/water The minimum thickness is that needed to give constructive interference - ie a whole number of wavelengths in the ray passing through the oil and back again. But there is an extra complication that there is a phase change on the oil/water reflection. So the equation 2ndcos(beta) = (m-0.5) lambda has a minimal value for m=1 You can have a layer of oil thicker than this as long as it is a whole number of wavelengths thicker. beta you can get from the incident angle and the refractive index of the oil.
 P: 150 what is this 2nd in front of the cos? beta i cannot solve for because i do not know what this incident angle is...it wasn't given in the prob. I just use the equation 2ndcos(beta) = (m-0.5) lambda to solve for lambda to get the answer to my question?...
 Sci Advisor HW Helper P: 8,953 2nd is 2 * n (refractive index of oil) * d (thickness of oil) Sorry misread your question, I thought you were given the angle but not 'n' Cos beta has a maximum value of 1 so you can get the minimum value of 'd'
P: 150
 Quote by mgb_phys 2nd is 2 * n (refractive index of oil) * d (thickness of oil) Sorry misread your question, I thought you were given the angle but not 'n' Cos beta has a maximum value of 1 so you can get the minimum value of 'd'
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness t be now?

Do I use the same method but instead change the refractive index?
 Sci Advisor HW Helper P: 8,953 Nearly, the rules change if the first medium is higher than the second - there is no phase change so there is no 1/2 in the equation. Optics isn't always easy ;-( The other pages on the link I gave you explains it further.
P: 150
 Quote by mgb_phys Nearly, the rules change if the first medium is higher than the second - there is no phase change so there is no 1/2 in the equation. Optics isn't always easy ;-( The other pages on the link I gave you explains it further.
Ok, so when light reflects off a surface with a higher index of refraction, it gains an extra shift of half of a wavelength. Therefore: 2ndcosbeta=(m)lamda?
is cosbeta = 1 still?

then why am i not getting the right answer when i solve for d??
 P: 150 My answer would be correct if both reflected rays had pi phase shifts... but they don't When light reflects off a surface with a higher index of refraction, it gains an extra shift of half of a wavelength. But if two beams reflect, they will both get a half-wavelength shift, canceling out that effect. Also, reflection off a surface with a lower index of refraction yields no phase shift. So I don't know why I would be getting a wrong answer using that method you described.
 P: 150 no one?
 P: 6 Isn't it just lamda(c) = 2nd/m, m=1 for thinnest therefore d = 312.5nm
 P: 3 jaded im pretty sure you mixed up the two equations for both parts. For the first question i used the 2ndcosB = mlambda equation which is basically the wavelength given divided by twice the index of refraction of oil Then for the second part I used the equation 2ndcosB = (m*1/2)lamba so basically half the lambda given divided by twice the index of refraction of oil I got the answers right for both of them

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