A Double Slit That Seems Mislabeled
An optics lab uses a helium-neon laser (633 nm), a double-slit slide labeled d = 0.250 mm with slit width 0.050 mm, and a screen 2.00 m from the slide. A student measures from the centre of the first bright fringe on one side of the central maximum to the centre of the tenth fringe on the same side, records 45.5 mm, and divides by 10. That gives 4.6 mm, about 9% below the value predicted by the usual formula. The student concludes the slide is mislabeled and writes 'd = 0.275 mm' in the report, recommending that the supplier be contacted. They also notice that one bright fringe around 25 mm from the centre is missing entirely, and that fringes beyond about 30 mm are noticeably dimmer than the central ones.
1. Compute the expected fringe spacing. Then test the student's conclusion against a counting error, and decide which explanation the data actually support.
Read the worked answer
Expected spacing: y = lambda L / d = 633e-9 x 2.00 / 2.50e-4 = 5.06e-3 m, or 5.06 mm.
Now examine the raw measurement rather than the derived number. The distance from the centre of fringe 1 to the centre of fringe 10 spans 9 intervals, not 10. Nine intervals at 5.06 mm is 45.6 mm. The student measured 45.5 mm. The raw length is exactly right; only the divisor is wrong. Dividing 45.5 mm by 9 gives 5.06 mm, and d = lambda L / y = 633e-9 x 2.00 / 5.06e-3 = 2.50e-4 m, which is 0.250 mm, in agreement with the label.
The mislabeling hypothesis survives arithmetic but is far less parsimonious. It requires the manufacturer to be 10% out on a parameter that is set photolithographically to better than 1%, and it happens to require an error of exactly the size produced by the fencepost mistake. It also makes a testable prediction the student did not check: if d were really 0.275 mm, every feature would shrink by the same factor, so the missing fringe would sit near 27.8 mm rather than 25 mm.
The correct report entry is that d = 0.250 mm is confirmed and the analysis method was corrected.
2. Explain the missing fringe and the dimming at large angle, and use the missing-fringe position to determine the slit width independently.
Read the worked answer
A real double slit is two finite-width slits, so the two-slit interference pattern is multiplied by a single-slit diffraction envelope. The envelope has minima where a sin(theta) = m lambda.
A fringe disappears when a bright interference maximum coincides with an envelope minimum. Interference maxima sit at y = n lambda L / d; envelope minima sit at y = lambda L / a. Equating the two for the same y gives n = d/a = 0.250/0.050 = 5. The fifth-order fringe should be absent, and it sits at 5 x 5.06 = 25.3 mm, which is exactly where the student saw the gap.
Inverting the relation gives an independent measurement of the slit width from the observed gap: a = lambda L / y = 633e-9 x 2.00 / 0.0253 = 5.0e-5 m = 0.050 mm. The slide's two labeled parameters are therefore mutually consistent, which is a second, independent confirmation that the slide is fine.
The dimming beyond about 30 mm is the same envelope falling off: past the first minimum at 25 mm the envelope has dropped to near zero, and the fringes that reappear in the second lobe are weak. The practical consequence for the lab is that fringe positions should be measured in the first lobe and near the centre, where the fringes are bright and symmetric; trying to measure the tenth fringe at 45 mm puts the measurement out where the intensity is low and the fringe centre is hard to locate.
3. The student wants a spacing measurement good to about 1%. Evaluate their measurement strategy and propose improvements, keeping the same apparatus.
Read the worked answer
The strategy of spanning many fringes is sound: measuring ten intervals divides the position-reading uncertainty by ten. The failure was in bookkeeping, not in the idea.
Three improvements are worth making with the same equipment. First, count intervals explicitly. Mark the central maximum as fringe 0 and the nth fringe as the end of n intervals, or simply count the bright spots and subtract one. A written tally on the data sheet prevents the discrepancy that caused this whole detour.
Second, measure both sides of the central maximum and average. If the laser is not exactly perpendicular to the slide, or the screen is tilted, the pattern is stretched on one side and compressed on the other; the mean of the two is much less sensitive to that misalignment than either side alone.
Third, measure L to the slit plane rather than to the front face of the slide holder, and check that the screen is parallel to the slide using a square. A 1% error in L is a 1% error in spacing, and a hand-held tape measuring 2.00 m can easily be off by 20 mm if the tape droops or if the reference surface is the holder rim.
With both sides measured and the intervals counted correctly, the spacing uncertainty drops to about the length-reading uncertainty divided by 2n, well inside 1%.