Young's Double Slit Experiment
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Direct answer
In Young's double slit experiment two coherent slits illuminate a screen, and the intensity at each point depends on the path difference Delta = d sin(theta), approximately d y/D for slit separation d and screen distance D: bright fringes where Delta = n lambda, dark where Delta = (2n + 1) lambda/2. The fringes are equally spaced with width beta = lambda D/d — the single most examined result of wave optics.
What you must remember
- Path difference d y/D at position y; phase difference phi = 2 pi Delta/lambda.
- Fringe width beta = lambda D/d, with angular width lambda/d; the nth bright fringe lies at n beta, the nth dark fringe at (n + 1/2) beta from the centre.
- Intensity: I = I1 + I2 + 2 sqrt(I1 I2) cos(phi); equal slits give I = 4 I0 cos^2(phi/2) — bright fringes four times a single slit's intensity.
- Unequal amplitudes a1, a2: Imax/Imin = (a1 + a2)^2/(a1 − a2)^2; only equal-intensity slits give perfect darkness.
- Immersion in a medium of refractive index n shortens the wavelength, shrinking beta to lambda D/(d n) — fringes crowd inward.
- A thin sheet (thickness t, refractive index mu) over one slit shifts the pattern by (mu − 1) t D/d toward that slit; the number of fringes shifted is (mu − 1) t/lambda.
- Sustained fringes need coherence — fixed phase difference; closing one slit erases fringes, and white light yields only a few coloured fringes around a white centre.
Common confusion
The recurring confusion is geometric versus optical path: interference is decided by optical path (geometric path × refractive index), which is why a sheet over one slit shifts fringes while symmetric air-path changes do not. Students also miscount — the central maximum (n = 0) is bright, so the third bright fringe sits at 3 beta while the third dark depends on where counting begins. And the fourfold intensity jump holds only for equal slits; unequal slits raise the minima above zero.
Exam-focused takeaway
JEE Main tests beta arithmetic, sheet shifts, intensity ratios and immersion as numerical-value questions — reliable marks from two formulas. JEE Advanced prefers the variants: Lloyd's mirror (one reflection adds a pi flip, swapping bright and dark), slits of unequal width through Imax/Imin, fringes visible on a finite screen, and matter-wave interference in modern-physics crossover. Anchor every question on the optical path difference and convert to phase before touching intensity.
Frequently asked questions
Why is the central fringe bright, and white in white light?
Both paths are equal at the centre, so every colour interferes constructively there; off-centre, colours reinforce at different positions, spreading into a spectrum around the white centre.
What changes when the apparatus is immersed in water?
The wavelength falls to lambda/n, so the fringe width becomes lambda D/(d n) — fringes move closer and more appear on a given screen.
Why does covering one slit destroy the pattern?
Interference needs two coherent waves crossing at each point; a single slit gives only broad diffraction-brightened illumination with no fringes.
What does a thin sheet over one slit do?
It adds optical path (mu − 1) t on that side, shifting every fringe by (mu − 1) t D/d toward the covered slit without changing the fringe width.
What is Imax/Imin for unequal slits?
(a1 + a2)^2/(a1 − a2)^2; contrast weakens as the amplitudes diverge, since the weaker wave cannot fully cancel the stronger.
Why must the sources be coherent?
Only a fixed phase difference yields stationary fringes; a drifting phase averages bright and dark within the detector's response and washes the pattern out.