Please give clear answers to A B and C, thank you! Please do not plagiarize this
ID: 1500081 • Letter: P
Question
Please give clear answers to A B and C, thank you!
Please do not plagiarize this from another Chegg answer
water in air s held vertically and viewed in reflected light. The film has an index of refraction n 1.36. refraction n = 1.36. a) b) Explain why the film appears black at the top The light reflected perpendicular to the film at a certain point is missing the wavelengths 504 nm and 630.0 nm. No wavelengths between these two are missing. What is the thickness of the film at that point? c) What other visible wavelengths are missing, if any?Explanation / Answer
(a) The speed of light in the film is slower than in air. Therefore ray 1, which reflects from a slower medium (the film), is inverted; ray 2, which reflects from a faster medium (air), is not inverted. There is a relative phase difference of 180 deg between the two regardless of wavelength Due to gravity, the film is thineest at the top and thickest at the bottom. Ray 2 has a phase shift copared to ray 1 due to the extra distance traveled in the film. The only way to presrve destrutive interference for all the wavelengths is if the top of the film is thin compared to the wavelengths of visible light; then the phase change of ray 2 due to the extra path traveled is negligibly small.
(b) For light reflected perpendicular to the film (nprmal incidence), reflected ray 2 travels an extra distance 2r compared to ray 1, which introduces a phase differences between them. Since there is already a relative phase difference of 180 deg due to reflection, the path difference 2t must be an integral number of wavelngths to preserve destructive interference:
2t = m*lamda = m*lamda0/n
Suppose lamda0,m =630.0 nm is the vacuum wave length for which the path difference is m*lamda for a certain value of m. Since there are no missing wavelengths between the two, lamda0,(m+1) =504 nm must be the vacuum wavelength for which the path difference is m+1 times the wavelength in the film.
2nt = m* lamda0,m = (m+1) lamda0,(m+1)
we can solve m :
m*630.0 nm =(m+1)*504 nm =mx504 nm + 504nm
m*126 nm = 504 nm
m =4.0
then the thickness is
t = m*lamda0/2n = 4*630 nm/ 2*1.36 =926.47 nm
(c) we already know the missing wavelenngth for m=4, and m=5. Let's check other values of m
2nt = 2*1.36*926.47 = 2520 nm
for m =3,
lamda0 = 2nt/m = 2520/3 = 840 nm
for m=6
lamda0 = 2nt/m = 2520/6 = 420 nm
for m=7
lamda0 = 2nt/m = 2520/7 = 360 nm
360 nm is UV. Thus the only other missing visible wavelength is 420 nm.