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CED objectives

Double-Slit Interference and Diffraction Gratings

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletFringe Bench · read the path difference in wavelengths rather than nanometres, number the centre zero, and watch fifty slits sharpen what two slits blur

Interference depends on the path length difference measured in wavelengths: whole numbers give constructive interference and odd half-integers give destructive, so the raw difference in metres settles nothing until it is divided by $\lambda$. For two slits separated by $d$, $d\sin\theta = m\lambda$ locates the maxima with the central one at $m = 0$, and the linear form $y = m\lambda L/d$ follows only while the angle is small. A grating obeys the same condition, so more slits leave the maxima where they were and make them far narrower, and because $\theta$ rises with $\lambda$ the red end of each order sits farthest from the centre.

Six errors dominate. Deciding constructive or destructive from the raw path difference rather than from its ratio to the wavelength. Using $y = m\lambda L/d$ at angles where the small-angle approximation has already failed, which gratings routinely produce. Counting the central bright fringe as the first order, which shifts every position by a fixed ratio. Swapping the slit separation with the screen distance, or reading $d$ as the width of one slit. Treating a grating as a double slit with extra openings, so adding slits is expected to smear the pattern. And putting violet at the outside of a grating spectrum because a prism bends violet the most.

label the diagram before substituting anything d micrometres slit WIDTH: sets the envelope L, a metre or two m = 0 m = 1 m = 2 m = 1 m = 2 the centre is m = 0. Counting it as 1 shifts every position by a fixed ratio
Three different lengths appear in this diagram and only two of them are in the fringe formula. Size-checking the answer catches a swap immediately.
the two devices order the colours OPPOSITE ways white red violet GRATING: sinθ = mλ/d θ rises with λ, so RED is farthest out and the centre stays WHITE: zero path difference for every λ red violet PRISM: n is larger for violet so VIOLET is deviated most: a different mechanism
Both devices spread white light and neither result follows from the other. The grating's ordering comes straight out of solving its own condition for the angle.

The work

3 ways in · any order
Lesson
Double-Slit Interference and Diffraction Gratings

Compares path difference with the wavelength before deciding anything, restricts the small-angle form to small angles, labels the orders from zero, keeps the slit separation apart from the screen distance, and orders a grating's colours from its own condition.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: judging interference from a raw path difference, using the linear fringe formula at large angles, counting the central fringe as order one, swapping d and L, expecting extra slits to blur a grating, and putting violet at the outer edge. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions