This was part of an AP problem assigned to my class regardingsnell\'s law and th

Question

This was part of an AP problem assigned to my class regardingsnell's law and the index of refraction. A student is asked to determine the thickness of a film of oil(n = 1.42) on the surface of water (n=1.33). Light from a variablewavelength source is incident vertically onto the oil film.The student measures a maximum in the intensity of the reflectedlight when the incident light has a wavelength of 600 nm. 1. At which of the two interfaces does the light undergo a 180degrees phase change on reflection? a. The air-oil interface only b. Both interfaces c. The oil-water interface only d. Neither interface 2. Calculate the minimum possible thickness of the oilfilm. This was part of an AP problem assigned to my class regardingsnell's law and the index of refraction. A student is asked to determine the thickness of a film of oil(n = 1.42) on the surface of water (n=1.33). Light from a variablewavelength source is incident vertically onto the oil film.The student measures a maximum in the intensity of the reflectedlight when the incident light has a wavelength of 600 nm. 1. At which of the two interfaces does the light undergo a 180degrees phase change on reflection? a. The air-oil interface only b. Both interfaces c. The oil-water interface only d. Neither interface 2. Calculate the minimum possible thickness of the oilfilm.

Explanation / Answer

given index of refraction of a film of oil n = 1.42 wavelength of light = 600nm 1) The two interfaces at which the light undergo 180degrees phase change on reflection is the air-oil interface. (since the light ray undergo reflection at the surfacebacked by oil which the rarer medium ) 2) here we have the condition for brightness as       2nt = (m+1/2)* for minimum thickness m = 0 so    2nt = /2 or     t = /4n        = 600nm/4*1.42        = 105.6 nm

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