A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 626 nm (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is 290 nm and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?
nfilm = 1Your answer is incorrect.I did t=(m)(wavelengthfilm)/(2) solving for the wavelegth of filmThenI did: wavelength film = wavelength vaccum / n and my n comes out as 1.18 which is the wrong answer can anyone help??

This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then

path difference created will be 2μ t.

For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference

path diff = nλ , for minimum t , n =1

path diff = λ

2μ t. = λ

μ = λ / 2t

= 626 / 2 x 290

= 1.08