Thin Film Over One Slit: Fringe Shift and Loss of Coherence
In a Young's double-slit experiment, a student uses sodium light of wavelength 589 nm, slit separation 0.50 mm and screen distance 1.20 m. The fringe width measured with a travelling microscope is 1.41 mm. She then places a thin sheet of transparent film of thickness 12.4 μm and refractive index 1.50 over the upper slit. The central point of the screen, originally bright, is now dark. She also notices that when she replaces the narrow source slit with a wide slit, the fringes disappear, although each individual slit still produces a broad diffraction pattern. The apparatus is in air, the film covers only the upper slit, and the source remains monochromatic.
1. Calculate the fringe width and the extra optical path introduced by the film. Show quantitatively why the centre becomes dark.
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Fringe width β = λD/d = (589 × 10⁻⁹)(1.20)/(0.50 × 10⁻³) = 1.41 × 10⁻³ m, matching the measurement. The film adds an optical path (n − 1)t to the upper beam: (n − 1)t = (1.50 − 1)(12.4 × 10⁻⁶) = 6.20 × 10⁻⁶ m. In wavelengths this is 6.20 × 10⁻⁶ / 589 × 10⁻⁹ = 10.5 wavelengths. The fringe pattern shifts by (n − 1)tD/d = (6.20 × 10⁻⁶)(1.20)/(0.50 × 10⁻³) = 1.48 × 10⁻² m = 14.8 mm, which is 10.5 fringe widths. At the original centre the two beams now differ by 10.5λ. An odd half-integer number of wavelengths gives destructive interference, so the centre is dark.
2. The student wants the centre to be bright again by changing only the film thickness. Find the smallest thickness change that achieves this and state whether the film should be thicker or thinner.
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The extra path is currently 10.5λ. The nearest bright conditions correspond to 10λ or 11λ, so the change needed is 0.5λ in optical path. Δ[(n − 1)t] = λ/2, so Δt = λ/[2(n − 1)] = (589 × 10⁻⁹)/(2 × 0.50) = 5.89 × 10⁻⁷ m, about 0.59 μm. Reducing the thickness by 0.59 μm makes the extra path 10λ, giving constructive interference. Increasing it by 0.59 μm makes the extra path 11λ, also constructive. The smallest change is therefore about 0.59 μm, and either a slightly thinner or a slightly thicker film works; the film should not be changed by a full wavelength in optical path, which would return the centre to dark.
3. Explain why replacing the narrow source slit with a wide slit makes the fringes disappear, even though the double slit is unchanged. Would a laser source necessarily solve the problem?
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A wide source slit behaves as many independent, mutually incoherent line sources. Each source point produces its own double-slit fringe pattern, but these patterns are laterally shifted. When the slit is wide enough that the shift between the extreme patterns is comparable with half a fringe width, bright fringes of one pattern fall on dark fringes of another and the visibility falls to zero. The double slit still diffracts light, so a broad pattern remains, but the fine interference fringes are washed out. A laser can help because it is highly spatially coherent, so a narrow laser beam can produce fringes without a narrow source slit. However, a wide laser beam containing many transverse modes can also reduce coherence, and the double slit still needs to be illuminated coherently across both slits. Coherence, not brightness alone, is the deciding factor.