Question

A soap bubble is floating in the air. The refractive index of the bubble is n = 1.33. The width of the wall is 115 nm. The light that hits the bubble is reflected on the outer and inner surface and both interfere once they leave. The difference in phase for both for both lights will be wt, where t is the time it takes for the light to go and return through the soap layer (t = d / (c / n)).

a) Draw a picture showing how interference occurs when light is reflected from thin films.

b) The light that most reflects towards the front of the bubble is that where the phase is 2pi. Find the formula for the frequency of light that is most feflected from the front. (At other angles this changes, since d = t / cos (0)). What is this value for the bubble?

c) What happens in other angles? What do the colors of the light reflected by the bubble look like?

Answer 1

When the light reflects from the upper and lower layer of the film, the path difference between them is:

Since the light is incident normally here,

For the case where the reflected light is maximum, the path difference between the light reflected from upper and lower surface must be integral multiples of the wavelength.

where m = 1,2,3,...

substitute frequency in place of the wavelength to get,

=>

for 1st order reflection, (m = 1) the phase difference is . Since this is the phase difference given, the reflected order is m = 1.

therefore,

=>

=> f = 9.807 x 10¹⁴ Hz = 980.71 THz

This is the frequency that is most reflected from the film.

c] At other angles, the path difference changes and so, the frequency reflected will be larger than this (since maximum of cosine function is 1). With increasing angle, different colors will be reflected thereby creating a spectrum of colors on the film.