Double-Slit Measurement Versus a Label
A group uses a helium-neon laser (λ = 632.8 nm) and a double-slit slide labeled d = 0.250 mm. The screen is 2.00 m from the slits. They measure the distance from the central maximum to the fifth bright fringe as 25.4 mm and the distance across ten bright-fringe spacings as 51.2 mm. The ruler has 1 mm divisions, and the screen is held by a clamp that may not be exactly parallel to the slit plane. The students compute d from each measurement and get values near 0.249 mm and 0.247 mm. One student writes, 'The label is wrong; the true slit spacing is about 0.248 mm.' Another wants to report d = 0.250 mm because that is the labeled value. The teacher asks them to evaluate the measurement uncertainty, the possible systematic errors, and the best way to report the result.
1. Calculate d from the fifth-fringe position and from the ten-spacing measurement. Estimate the uncertainty from the ruler and decide what value should be reported.
Read the worked answer
Use d = mλL/y. For the fifth fringe, y5 = 25.4 mm = 0.0254 m, so d = 5(632.8 × 10⁻⁹)(2.00)/0.0254 = 2.49 × 10⁻⁴ m = 0.249 mm. The ten-spacing measurement gives an average spacing of 51.2/10 = 5.12 mm, so d = (632.8 × 10⁻⁹)(2.00)/0.00512 = 2.47 × 10⁻⁴ m = 0.247 mm. A ruler reading uncertainty of about ±0.5 mm is 2.0% on 25.4 mm and 1.0% on 51.2 mm, so the two values differ by less than the combined uncertainty. The screen distance also has some uncertainty, but it is likely smaller. A reasonable report is d = (2.48 ± 0.01) × 10⁻⁴ m, consistent with the label.
2. Evaluate the two students' claims. What systematic errors could explain the measured values, and what is the defensible conclusion?
Read the worked answer
The claim that the label is wrong is overinterpretation: 0.247 to 0.249 mm is within about 1 to 2% of 0.250 mm, and the measured values are slightly smaller, not larger. A systematic tilt of the screen would stretch the measured fringe distance, making d appear smaller; parallax in reading the ruler and uncertainty in locating the fringe centers do the same. The slit-to-screen distance may also be slightly less than 2.00 m. Reporting the labeled value as if it were measured is also wrong, because it hides the measurement and its uncertainty. The defensible statement is that the measured spacing agrees with 0.250 mm within experimental uncertainty; the data do not establish a discrepancy. To improve, measure several orders on both sides of the central maximum, plot y_m versus m, and use the slope; check that the screen is perpendicular to the beam and measure the slit-to-screen distance from the slit plane.
3. If the actual slit separation were 0.500 mm, predict the fringe spacing and the ten-fringe span. Would the ten-fringe measurement become more or less precise with the same ruler?
Read the worked answer
If d = 0.500 mm, the fringe spacing is Δy = λL/d = (632.8 × 10⁻⁹)(2.00)/(5.00 × 10⁻⁴) = 2.53 × 10⁻³ m = 2.53 mm. Ten spacings would be 25.3 mm, about half the previous 51.2 mm. With the same ±0.5 mm ruler reading, the relative uncertainty on a 25.3 mm span is about 2%, versus about 1% for 51.2 mm, so a single ten-fringe measurement is less precise in relative terms. The group could recover precision by measuring more fringes, such as 20, or by using a traveling microscope or a finer scale, and by fitting multiple orders rather than relying on one span.