Course Timer Notes Evaluate FeedbackPrint A thin film of soap with n= 1.40 hangi

Question

Course Timer Notes Evaluate FeedbackPrint A thin film of soap with n= 1.40 hanging in the air ran dominantly red light with = 664 nm. What is minimum thickness of the film? 1.186×102 nm You are correct. Your receipt no. is 160-412 o Previous Tries Now this film is on a sheet of glass, with n 1.50, what is the wavelength of the light in air that will now be predominantly reflected 3.557 10-7 nm What changed compared to previous problem? What is the requirement for a maximum for the pathlength difference now? Submit Answer Incorrect. Tries 5/10 Previous Tries

Explanation / Answer

given
n = 1.4
reflects doninantly lambda = 664 nm
thickness = t
hence
the light moving from air to soap film undergoes a phase shift while light going from soap film to air does not
hence
n*2*t = (2n - 1)lambda/2
for minimum thickness
2nt = lambda/2
t = 118.5714 nm

on a sheet of glass, ng = 1.5
wavelength of ight that now willbe reflected = lambda
now, in this case, the second light leaving soap to glass also has a phase shift
hence
n*2*t = m*lambda
lambda = 331.99992 nm for m = 1

the phase shift from the soap glass interface is what changed now. the requirement for path difference is now that it should be an integral multiple of wavelength

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