Two-Slit Pattern with a Phase-Shifting Film
In a double-slit experiment, slit separation d = 0.20 mm, screen distance D = 1.5 m, and light of wavelength lambda = 600 nm in vacuum. A thin film of index n = 1.50 and thickness t = 200 nm is placed over one slit only. A student predicts the central maximum will simply shift to a new angle and that the fringe spacing will change because 'the wavelength inside the film is different.' She measures the fringe spacing on the screen and finds it unchanged from the no-film case, and she is confused. The film covers only one slit and is thin compared to D. Assume the film does not absorb light.
1. Determine the phase shift introduced by the film and predict the new position of the central maximum.
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The film adds an optical path difference (n-1)t = (1.50-1)(200 nm) = 100 nm to the light through the covered slit. The corresponding phase shift is delta_phi = 2pi*(n-1)t/lambda = 2pi100/600 = pi/3 radians (60 degrees). The central maximum (zero total phase difference) now occurs where the geometric path difference compensates: dsin(theta) = -(n-1)t = -100 nm, so sin(theta) = -100e-9/0.20e-3 = -5.0e-4, giving theta ≈ -5.0e-4 rad and a screen displacement y = Dtan(theta) ≈ 1.5*(-5.0e-4) = -0.75 mm. So the central maximum shifts by about 0.75 mm toward the uncovered-slit side (sign depends on convention), not by a full fringe.
2. Explain why the fringe spacing is unchanged despite the film, addressing the student's reasoning.
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Fringe spacing is set by the phase difference between adjacent maxima, which depends on the rate of change of geometric path difference with angle, dsin(theta), not on a constant offset. The film adds the same constant phase (n-1)tk to every point on the screen, so it shifts the whole pattern but does not change the separation between maxima. The student's error is conflating 'wavelength inside the film is different' with a change in the interference condition on the screen. The wavelength in the film affects the accumulated phase through the film (already captured by the optical path), but after the light exits the slit region it propagates in air at lambda, and the spacing remains Delta_y = lambdaD/d = (600e-9)(1.5)/(0.20e-3) = 4.5 mm. The measurement is consistent with theory; only the offset changed.
3. If the student instead used white light, describe qualitatively how the pattern would differ and why.
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With white light each wavelength produces its own interference pattern with its own fringe spacing (proportional to lambda) and its own film-induced offset (proportional to lambda as well, since delta_phi = 2pi(n-1)t/lambda varies with lambda). The central region would show a white-ish fringe only where all wavelengths constructively interfere; away from it, colors separate because maxima for different lambda occur at different angles. The film's dispersive phase shift (n also varies with lambda in real materials) would further smear the pattern. The net effect is a colored, washed-out fringe pattern rather than the clean monochromatic fringes, and the 'central' white fringe position would be set by the wavelength-dependent offset, so the pattern would not simply be a shifted version of the monochromatic one.