A scientist notices that an oil slick floating on water when viewed from above h

Question

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 750nm (in air). The index of refraction of water is 1.33
a)The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
Express your answer in nanometers to three significant figures. in nm
b)Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
Express your answer in nanometers to three significant figures. in nm

Explanation / Answer

Part A) This problem involves determining whether the refracted light rays are in phase or out of phase with each other.

n=1 (air)
________________
n=1.2 (oil)
_______________
n=1.33 (water)

Light that is incident upon the oil from the air goes from a lower index of refraction to a higher index of refraction. The light incident upon the water from the oil also goes from a lower index of refraction to a higher index of refraction. when light from a lower index of refraction to a higher index of refraction it becomes out of phase but since both refracted rays are out of phase they are actually in phase with each other.

For refracted rays that are in phase use eqtn: 2nt=m

plug in your values:

2(1.2)t=(1)(750x10-9m)

solve for t to get the thickness of the film

t= 3.125x10-7m or 313 nm

Part B) same basic setup only the thin film now has index of refraction of 1.5. The first refracted ray from air to oil is still out of phase but the second refracted ray from oil to water is not out of phase because it is going from a higher index (oil 1.5) to a lower index (water 1.33). This changes the equation we must use because now the two refracted rays are out of phase with each other. Use m=0 for minimum thickness

Eqtn: 2nt=(m+1/2)

2(1.5)t=(0+.5)(750x10-9m)

solve for t= 1.25x10-7m = 125 nm

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