Fringe separation is the distance between adjacent bright fringes (or adjacent dark fringes) on the screen in a double-slit interference pattern. For small angles it equals λL/d, so it increases with wavelength λ and screen distance L, and decreases with slit separation d.
Fringe separation is how far apart the bright bands (or dark bands) sit on the screen when monochromatic light passes through a double slit. The pattern exists because waves from the two slits travel different distances to each point on the screen. Where the path length difference is a whole number of wavelengths, the waves arrive in step and you get a bright fringe. Where it's a half-integer number of wavelengths, they cancel and you get a dark fringe.
The maxima are located by d sin θ = mλ, where d is the slit separation and m is the order number. For the small angles you almost always deal with on the AP exam, sin θ ≈ tan θ = y/L, which gives the position of each bright fringe as y = mλL/d. The spacing between neighboring fringes is therefore Δy = λL/d. Read that formula like a story. Longer wavelength spreads the pattern out, a farther screen spreads it out, and slits that are closer together spread it out too. That last one trips people up because it feels backwards, but it falls straight out of the math.
Fringe separation lives in Topic 14.8 (Double-Slit Interference and Diffraction Gratings) in Unit 14: Waves, Sound, and Physical Optics. It directly supports learning objective 14.8.A, which asks you to describe the pattern produced when a wave passes through multiple openings. The essential knowledge for 14.8.A spells out the logic chain you need. Interference alone from two slits produces uniformly spaced maxima, the amount of interference depends on the path length difference ΔD between the two wavefronts, and ΔD is set by the slit separation d and the angle to the screen. Fringe separation is the measurable, on-screen consequence of all that wave reasoning, which is exactly why exam questions love it. It turns 'do you understand interference?' into a concrete prediction you can calculate or rank.
Keep studying AP® Physics 2 Unit 14
d sin θ = mλ (Unit 14)
This is the parent equation. Fringe separation is what you get when you apply the small-angle approximation to it, turning angles into screen positions. If you can derive Δy = λL/d from d sin θ = mλ, you understand the whole topic, not just a memorized formula.
Path length difference (Unit 14)
Path length difference is the cause; fringe separation is the effect. Each bright fringe marks a spot where the path difference ΔD jumps by exactly one more wavelength, so the fringe spacing on the screen is really a map of where ΔD hits 0, λ, 2λ, and so on.
Wavelength in transparent media (Unit 14)
When light enters a medium with index of refraction n, its wavelength shrinks to λ/n. Submerge a double-slit setup in water or oil and the fringe separation shrinks by the same factor, since Δy = λL/d depends directly on λ. This connects interference to refraction, and it's a favorite twist on exam questions.
Central bright fringe (Unit 14)
The central bright fringe (m = 0) is the anchor point where the path difference is zero, so it sits in the same place no matter the wavelength. Every other fringe is measured outward from it in steps of Δy, which is why y₁, the distance to the first-order maximum, equals one fringe separation.
Fringe separation shows up mostly as 'predict and justify' questions. A classic stem gives you a change to the setup and asks how the spacing responds. For example, a student claims moving the screen farther away makes the fringes closer together, and you have to evaluate the claim (it's wrong, since Δy = λL/d means doubling L doubles the spacing, so moving the screen from L to 2L doubles y₁). Another favorite is submerging the whole apparatus in a liquid with n = 1.5, which shrinks the wavelength to λ/1.5 and the fringe separation to 2y/3. The College Board has also framed interference in transparent media on short-answer questions, like the 2022 exam's investigation of electromagnetic waves in a transparent block. Your job is rarely just to plug numbers. You need to identify which variable changed (λ, L, or d), state the proportionality, and justify it from interference reasoning, ideally tying the answer back to path length difference rather than just citing the equation.
Slit separation d is the distance between the two openings in the barrier. Fringe separation Δy is the distance between bright bands on the screen. They're both 'separations,' but they live in different places and pull in opposite directions. Because Δy = λL/d, making the slits closer together (smaller d) makes the fringes farther apart (bigger Δy). If you swap the two on a ranking question, you'll get the exact opposite of the right answer.
Fringe separation is the distance between adjacent bright fringes (or adjacent dark fringes) on the screen, and in the small-angle approximation it equals Δy = λL/d.
Fringe separation increases with longer wavelength and a farther screen, but decreases when the slits are moved farther apart.
Bright fringes occur where the path length difference between the two slits equals a whole number of wavelengths (d sin θ = mλ), and fringe separation is the on-screen result of that condition.
Doubling the screen distance L doubles the fringe separation, so a claim that moving the screen back squeezes the fringes together is wrong.
Submerging the apparatus in a medium with index n shrinks the wavelength to λ/n, so the fringe separation shrinks by the same factor (n = 1.5 turns y into 2y/3).
Interference alone from two slits gives uniformly spaced maxima, so the spacing between any two neighboring bright fringes is the same.
It's the distance between adjacent bright fringes (or adjacent dark fringes) in a double-slit interference pattern. In the small-angle approximation, Δy = λL/d, where λ is the wavelength, L is the slit-to-screen distance, and d is the slit separation.
No, it's the opposite. Since Δy = λL/d, fringe separation is directly proportional to screen distance, so moving the screen from L to 2L doubles the spacing between fringes.
Slit separation d is the gap between the two slits in the barrier; fringe separation Δy is the gap between bright bands on the screen. They're inversely related, so smaller d means larger Δy.
It shrinks. In a medium with index of refraction n, the wavelength becomes λ/n, so the fringe separation becomes Δy/n. With n = 1.5, a spacing of y drops to 2y/3.
Bright fringes occur where the path length difference between the two slits is a whole number of wavelengths. A longer wavelength means you have to move farther along the screen before the path difference grows by one more λ, so the fringes spread out. That's why red light produces wider spacing than blue light.
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