Young's Double-Slit Experiment

Why Two Lamps Fail, and Young's Fix

Illuminate two pinholes with two separate sodium lamps and you see… nothing — no fringes. An ordinary source emits light in bursts with abrupt phase changes every ~$10^{-10}$ s, so two independent sources have no fixed phase relation: incoherent, intensities just add.

Young's 1801 trick: a single source S illuminates one pinhole, whose light then falls on two closely spaced pinholes S$_1$ and S$_2$. Because S$_1$ and S$_2$ are fed from the same original wavefront, any phase jiggle in S appears identically in both — they are locked in phase, i.e. coherent. Spherical waves from S$_1$ and S$_2$ overlap on the screen and produce a steady pattern of bright and dark fringes.

The Fringe Formulas

Slit separation d, screen distance D ($D \gg d$), point P at distance x from the centre. The path difference is $\Delta = \frac{xd}{D}$.

Bright fringes ($\Delta = n\lambda$): $$\boxed{x_n = \frac{n\lambda D}{d}, \quad n = 0, \pm1, \pm2, ...}$$

Dark fringes ($\Delta = (n+\tfrac12)\lambda$): $$\boxed{x_n = \left(n+\tfrac12\right)\frac{\lambda D}{d}}$$

Consecutive bright (or dark) fringes are equally spaced by the fringe width:

$$\boxed{\beta = \frac{\lambda D}{d}}$$

This is the chapter's flagship formula — the one that let Young measure $\lambda$ (rearrange to $\lambda = \beta d/D$).

The variation rules (every exam loves these):

• $\beta \propto \lambda$: red fringes wider than blue.
• $\beta \propto D$: move the screen back, fringes spread.
• $\beta \propto 1/d$: closer slits, wider fringes.
• Immerse the apparatus in water (n): $\lambda \to \lambda/n$, so $\beta \to \beta/n$ — fringes shrink.

[NEET Important] Central fringe (n = 0) is bright and wavelength-independent — in white light it is white, flanked by coloured fringes. [JEE Tip] Angular fringe width $\theta = \beta/D = \lambda/d$ — independent of D.

Solved Examples

Example 1: Fringe width [NEET Numerical]

In a double-slit setup, d = 0.5 mm, D = 1.0 m, $\lambda$ = 600 nm. Find the fringe width.

Solution:

1. $\beta = \frac{\lambda D}{d} = \frac{600\times10^{-9}\times1.0}{0.5\times10^{-3}}$.
2. $\beta = 1.2\times10^{-3}$ m = 1.2 mm.
3. Visible, measurable bands — the experiment's whole point.

Example 2: Measuring the wavelength [JEE Numerical]

Fringes 0.9 mm wide are seen with d = 0.4 mm and D = 0.6 m. Find $\lambda$.

Solution:

1. $\lambda = \frac{\beta d}{D} = \frac{0.9\times10^{-3}\times0.4\times10^{-3}}{0.6}$.
2. $\lambda = 6\times10^{-7}$ m = 600 nm.
3. Exactly how Young first pinned down the wavelength of light.

Example 3: Position of a fringe [NEET Numerical]

For $\beta$ = 1.2 mm, where is the 3rd bright fringe and the 2nd dark fringe from the centre?

Solution:

1. 3rd bright: $x = 3\beta = 3.6$ mm.
2. 2nd dark: $x = (1 + \tfrac12)\beta = 1.5\beta = 1.8$ mm (n = 1 gives the 2nd dark).
3. Bright at integer $\beta$, dark at half-integer — fingers off by one cost marks here.

Example 4: Immersion in water [JEE Numerical]

The fringe width is 1.2 mm in air. What does it become when the whole apparatus is dipped in water (n = 1.33)?

Solution:

1. In water $\lambda \to \lambda/n$, so $\beta \to \beta/n$.
2. $\beta_w = \frac{1.2}{1.33} = 0.90$ mm.
3. Denser medium, shorter wavelength, tighter fringes.

Example 5: Changing the slit separation [NEET Numerical]

If the slit separation d is doubled (everything else fixed), what happens to the fringe width?

Solution:

1. $\beta \propto 1/d$.
2. Doubling d halves $\beta$ — fringes crowd together.
3. (Bring the slits closer to spread fringes out — the opposite manoeuvre.)

Example 6: Two wavelengths, coincident fringes [JEE Numerical]

Light of 480 nm and 600 nm illuminates a double slit together. At what order do their bright fringes first coincide?

Solution:

1. Coincidence: $n_1\lambda_1 = n_2\lambda_2$, i.e. $\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$.
2. Smallest integers: $n_1 = 5$ (for 480 nm), $n_2 = 4$ (for 600 nm).
3. The 5th fringe of 480 nm overlaps the 4th of 600 nm — a standard two-colour overlap problem.

Example 7: Number of fringes in a width [JEE Numerical]

How many bright fringes fit in a 1.5 cm wide region of the screen if $\beta$ = 1.2 mm?

Solution:

1. Fringes per region $\approx \frac{15\text{ mm}}{1.2\text{ mm}} = 12.5$.
2. So about 12 fringes span the region (counting bright fringes).
3. Wider screens or wider fringes hold fewer per centimetre — the geometry on a leash.

Example 8: Why the central fringe is white

In white-light Young's fringes, why is the central fringe white and the rest coloured?

Solution:

1. At the centre (n = 0) the path difference is zero for every wavelength — all colours interfere constructively: white.
2. Away from centre, $x_n = n\lambda D/d$ depends on $\lambda$: each colour's fringe sits at a slightly different place.
3. So flanking fringes spread into spectra; only the centre is achromatic. A classic reasoning question.

Example 9: Angular fringe width [JEE Numerical]

For d = 0.5 mm and $\lambda$ = 600 nm, find the angular fringe width, and confirm it is independent of D.

Solution:

1. $\theta = \frac{\beta}{D} = \frac{\lambda}{d} = \frac{600\times10^{-9}}{0.5\times10^{-3}}$.
2. $\theta = 1.2\times10^{-3}$ rad.
3. D cancels — the angular spacing is fixed by $\lambda/d$ alone, which is why it survives moving the screen.

Example 10: Shift on covering one slit with a thin sheet [JEE pattern]

Qualitatively, what happens to the fringe pattern if a thin transparent sheet is placed over one slit?

Solution:

1. The sheet adds extra optical path $(n-1)t$ to one beam, introducing a constant path difference.
2. The whole fringe pattern shifts sideways (towards the covered slit) by $\frac{D(n-1)t}{d}$ — the fringe width $\beta$ is unchanged.
3. (Beyond the rationalized core but a JEE staple: shift counts fringes, $\frac{(n-1)t}{\lambda}$ of them.)