Question

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nmnm (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 ttt of the oil slick at that spot?
b. Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
c. Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver?

Answer 1

Answer:

(a) 313nm

(b) 250nm

(c) 301nm

Explanation:

This problem involves the concept of thin-film interference. The scientist observes constructive interference. In this case m = 1.

(a) To find the thickness of the oil at that spot we need to first calculate the wavelength in the oil.

       λ = λo/n = 750/ 1.20 = 625nm

λ = wavelength in the material of interest into which light is entering (the oil)

λo = wavelength in the material from which light is coming in or entering.

n = refractive index of the material into which the light is entering.

Light travels from air into the oil. So the refractive index used above is that of the oil and not air.

The formula relating the thickness t to the wavelength is 2t = mλ

m = 1 so,

    2t = λ = 625nm

     t = 625/2 = 313nm

(b) Supposing the refractive index of the water is now n = 1.50

The wavelength in the oil λ = 750/1.50 = 500nm

 The thickness t = λ/2 = 500/2 = 250nm

(c) If the oil has a thickness of 200nm then the wavelength λ = 2t = 2 ×200nm = 400nm

The wavelength in the oil is 400nm. In order to find the wavelength in water we set this wavelength to λo = 400nm. So the wavelength in water seen by the diver is given by the formula

       λ = λo/n

n for water is 1.33

      λ = 400/1.33 = 300.8nm ≈ 301nm.