Double Slit, an Out-of-Spec Laser, and a Hidden Distance Error
An inexpensive diode laser is labeled 650 nm. A student sets up a double-slit slide whose package states d = 0.250 mm. The screen is a whiteboard 1.500 m from the front face of the slit holder, measured with a steel tape. Using a ruler with 1 mm graduations, the student measures the distance from the center of the central maximum to the center of the 10th bright fringe on one side as 37.0 mm, with the fringe centers locatable to about ±0.5 mm; the fringes are clean and evenly spaced. Computing λ = Δy·d/L, the student obtains 617 nm and reports that the laser is out of specification. A second student points out that the slit apertures sit about 2.0 cm behind the front face of the holder, and that the slide's value of d is specified only to ±4%. The instructor asks the pair to decide which quantity is actually responsible for the disagreement.
1. Reproduce the student's calculation and build an uncertainty budget that shows how much of the 650 nm versus 617 nm disagreement each measurement can explain.
Read the worked answer
Ten spacings span 37.0 mm, so Δy = 3.70 mm and λ = (3.70 × 10⁻³ m)(2.50 × 10⁻⁴ m)/(1.500 m) = 6.17 × 10⁻⁷ m = 617 nm.
Uncertainty budget: locating the 10th fringe to ±0.5 mm out of 37.0 mm contributes 1.4%; the tape over 1.5 m contributes about 0.3%; the stated slit separation contributes ±4%. Adding in quadrature gives about 4.2%, so 617 nm really represents roughly 591–643 nm before any correction. The 650 nm label is about 5% away, barely outside one combined standard uncertainty, and nowhere near enough to prove the laser is bad. The dominant term is d, which the manufacturer specifies only to ±4%: a true d of 0.260 mm alone would give 642 nm. A single measurement cannot separate λ from d, so the student's conclusion overreaches the data even though the arithmetic is correct.
2. Which geometric correction can be made immediately, and what independent measurement would settle the d-versus-λ ambiguity?
Read the worked answer
The slit plane is about 2.0 cm behind the face from which L was taped, so L is really about 1.480 m. Recomputing: λ = (3.70 × 10⁻³)(2.50 × 10⁻⁴)/1.480 = 6.25 × 10⁻⁷ m = 625 nm, a 1.3% correction toward the label. The residual is now about 4%, exactly the tolerance on d, so the evidence still does not justify declaring the laser out of specification.
To break the degeneracy, measure the wavelength independently of the slide. Use a ruled diffraction grating of known line density, for example 600 lines/mm, and apply d sinθ = mλ to the same laser. Alternatively, illuminate the slide with a known 632.8 nm helium-neon line and solve for d. Either procedure anchors one of the two coupled quantities. A useful cross-check is the single-slit envelope of the pattern, which depends on the slit width and can reveal whether the slide matches its specification at all.
3. Predict what happens to the pattern if the slits and screen are immersed in water (n = 1.333), and explain what the student would wrongly conclude from the change.
Read the worked answer
Frequency is set by the source and does not change, but the speed and wavelength in water fall by the factor n, so λ_water = 650/1.333 = 488 nm. Every fringe spacing shrinks by the same factor: Δy = 3.70/1.333 = 2.78 mm, and the 10-fringe span becomes about 27.8 mm. The geometry d sinθ = mλ is unchanged; only λ differs.
If the student did not account for the medium, the reduced spacing would look like a shorter wavelength and would appear to confirm the earlier claim that the laser is out of specification, when in fact the apparatus is behaving exactly as predicted. Two practical cautions follow. The water must actually fill the space the light travels through, so a flat-sided tank is required. And the tank walls must be parallel, because a wedged or curved wall refracts the beams and changes the effective distance to the screen, corrupting L in a way that no fringe-spacing formula will reveal.