1. The problem statement, all variables and given/known data Light of wavelength 418 nm is incident normally on a film of water 1.0 μm thick. The index of refraction of water is 1.33. (a) What is the wavelength of the light in the water? (b) How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (Do not round your answer to a whole number.) (c) What is the phase difference between the wave reflected from the top of the air-water interface and the one reflected from the bottom of the water-air interface in the region where the two reflected waves superpose? 2. Relevant equations 3. The attempt at a solution my attemp at a solution: (a)lambda"water" = lambda/n = 445nm (b)N = 2t/ lambda"water" = 4.5 (c) Light is going from a medium of smaller index of refraction (air n = 1) to a medium with higher one (water n = 1.33) so the reflected wave is out of phase by π respect to the transmitted wave right? So the phase diff = π + 2π(2t/lambda"water") = -π +2π(4.5) = ??? My answers i have tried: 25.1 rad, 3.14rad, 31.4 rad, 1 rad, 4.72 rad, 0 rad 31.42rad 1 rad Im really stumped...pls help!!!!!!!!!!!!!!!!!!!!!!!