Light of wavelength 601 nm (in vacuum) is incident perpendicularly on a soap fil

Question

Light of wavelength 601 nm (in vacuum) is incident perpendicularly on a soap film (assume n = 1.30) suspended in the air. What are the two smallest nonzero film thicknesses (in nm) for which the reflected light undergoes constructive interference?
nm (smallest thickness)
nm (next smallest thickness)

Explanation / Answer

At the first surface of the soap film, there is a 180 degree phase shift of the reflected light (because nsoap>nair). At the bottom surface, the reflected light does not have a phase shift, so the reflected waves that need to interfere constructively need to travel a distance equal to 1/2 a wavelength (and 3/2 a wavelength for the second smallest thickness). One more thing, the wavelength of the light in the film is 622/1.34, don't forget that! So (1/2)(622/1.34) = 2T (2T is 2* the thickness of the film, or the difference of distance travelled by the two reflected wave fronts) Solve for T to get the minimum thickness, then replace the (1/2) with (3/2) and solve for T to get the second smallest thickness

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