A wedge-shaped soap film is floating in air. The index of refraction of the film

Question

A wedge-shaped soap film is floating in air. The index of refraction of the film is that of water (1.33). The thickness at the left end of the film is essentially zero. The film is viewed in reflected light with near-normal incidence. Does the left edge of the film appear dark or bright? Explain why. The first interference maximum for violet light (wavelength in air = 450 nm) is observed 3.25 cm to the right of the left edge of the film. What is the thickness of the film at this point? What is the angle of the wedge? What is the thickness of the film where the first bright fringe occurs for red light (wavelength in air = 650 nm)? How far to the right of the left edge does this fringe occur?

Explanation / Answer

a) Left edge of the film appears dark . Because , minima will be formed at that place . This is due to phase difference of pi between the two interference waves   .

b)    Here, 2t = (m + 1/2)/n * lambda

=>   2t = (1 + 1/2)/1.33 * 450

=> t = 253.76 nm           ------------------>   thickness of film

c)    angle of wedge   =   tan-1(253.76 * 10-9/0.0325)

                                  =   4.47 * 10-4   degree

d)    2t = (m + 1/2)/n * lambda

=>   2t = (1 + 1/2)/1.33 * 650

=> t = 366.54 nm           ------------------>   thickness of film

e)    To the right of left edge this occur =    (366.54 * 10-9)/tan(4.47 * 10-4)

                                                            = 0.04698 m

                                                           = 4.698 cm

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