The refraction of perpendicularly incident white light by a soap film in air has an interface maximum at 600nm and a minimum at 450nm, with no minimum in between. If n=1.33 for the film, what is the film thickness, assumed uniform?
For interference maxima by a soap film of thickness t is given as,
2*n*t = (k+(1/2))*lambda_1 ...(1)
where, n= refrective index of film = 1.33
lambda_1 = wavelegth for which there is maxima = 600 nm
k = an integer
For minima with a ...
The solution works through the problem step by step to clearly arrive at the answer detailing the necessary thickness of soap film to produce the described ray of white light.