A soap film is 129 nm thick. The film is in air and illuminated by white light at normal incidence as shown in the figure below. (In the figure the rays have been drawn at an angle to show the multiple reflections and transmissions in the film.) Assume the index of refraction of the soap film is the same as the index of refraction of water n = 1.33. The film is viewed from above and below by a video system which is sensitive to wavelengths from 200 to 1100 nm.
1/2 + 2d/lambda = m
1/2 + 2d/lambda = m + 1/2
The Attempt at a Solution
a) When viewed in reflection, what is the longest visible wavelength that will appear enhanced?
b) When viewed in reflection, what is the longest visible wavelength that will have the least intensity?
c) When viewed in transmission, what is the longest visible wavelength that will appear enhanced?
a) Wouldn't this be 1/2 + 2d/lambda = m and solving for lambda where it's inbetween 200-1100 nm?
b) Got this, ended up just being 2nt/m = lambda
c) Not sure on this one, the help provides a link to some slides but the link is broken and I don't think our professor really cares.. Thanks!