# Young's Double Slit Experiment

> Young's double slit experiment for JEE Physics; fringe width, path difference, intensity variation, effect of medium and thin sheets.

- Canonical URL: https://prepelephant.com/topics/jee/physics/youngs-double-slit-experiment
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Young's Double Slit Experiment", PrepElephant, https://prepelephant.com/topics/jee/physics/youngs-double-slit-experiment

## 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.