Please thoroughly explain step by step of just part c and th

Please thoroughly explain step by step of just part c, and thanks in advance!

Light of wavelength 480 nm is incident normally on a film of water 10\'4 cm thick. The index of refraction of water is 1.33. What is the wavelength of the light in the water? 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.) 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 after it has traveled this distance?

Solution

a) Wavelength of light in the water _water=/n

                     _water=480/1.33=360.9=361 nm

b) Number of wavelengths N=2t/ _water = 2X10^-4x10^-2m/361X10^-9

                                                                   = 5.54

c) Light is travelled from low refractive index medium (n=1)to high refractive index medium (n=1.33) Hence reflected wave is out of phase

Phase difference = -+2 X2t/ _water

                                                = -+ ( X11.08)=10.08

Subtracting this with nearest multiples of (ie 10)

=10.08-10=0.08=0.25 Rad