A Double-Slit Result That Does Not Match the Label
In a dark room, a learner uses a laser of wavelength 633 nm, a double slit with nominal separation 0.250 mm, and a screen 2.000 m from the slits. She measures the distance from the centre of the first bright fringe to the centre of the eleventh bright fringe as 48.0 mm using a 300 mm ruler with 1 mm divisions, held against the screen. The expected ten-fringe separation from the marked slit spacing is 50.6 mm. Her calculated wavelength is 600 nm. She writes: 'The wavelength is 600 nm, so the slit label is wrong.' A second learner says the 5% difference is just experimental error and no conclusion can be drawn. The teacher asks for a judgement based on the uncertainties, not on opinion.
1. Reproduce the expected fringe spacing and the wavelength calculation.
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Expected fringe spacing x = lambda L/d = (633 x 10^-9 x 2.000)/(0.250 x 10^-3) = 5.064 x 10^-3 m = 5.06 mm. Ten spaces = 50.6 mm. From the measurement, x = 48.0/10 = 4.80 mm. Then lambda = xd/L = (4.80 x 10^-3 x 0.250 x 10^-3)/2.000 = 6.00 x 10^-7 m = 600 nm. The calculation is consistent; the question is whether the 5.2% difference is significant.
2. Estimate the uncertainty in the measured fringe separation and in the slit separation. Which is likely to dominate, and does the 5% discrepancy require the label to be wrong?
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The ruler contributes at least plus or minus 0.5 mm over 48 mm = 1.0%, and identifying the centre of each fringe adds perhaps plus or minus 0.5 mm at each end, giving plus or minus 1 mm/48 mm, about 2%. The screen distance is known to about 0.05%. The slit separation is only a nominal label; a typical tolerance of plus or minus 5% would give plus or minus 2.5 mm on the 50.6 mm prediction, which covers the measured 48.0 mm. So the dominant uncertainty is likely d, not the ruler. The 5% difference is therefore not strong evidence that the label is wrong, and the second learner is too strong in saying no conclusion can be drawn: the result is consistent with 633 nm within the plausible slit tolerance, but it cannot confirm it.
3. Describe one modification that would reduce the measurement uncertainty and explain why it works.
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Measure the distance across 20 or 30 fringes rather than 10. If the total distance is about 100 to 150 mm, a plus or minus 0.5 mm ruler uncertainty is 0.3 to 0.5%, and fringe-centre identification is also a smaller fraction. Alternatively use a travelling microscope or a vernier scale fixed to the screen, and measure L with a steel tape. Repeating at several screen distances and plotting x against L is stronger still: the slope is lambda/d, so an unknown d still prevents an absolute lambda, but the linearity tests the theory and the precision of x improves.