Young's double slit — derivation, fringe width, what happens when parameters change

Question

In Young’s double slit experiment, the slit separation is $d = 0.5$ mm, the screen is at $D = 1$ m, and the wavelength is $\lambda = 500$ nm. Find the fringe width. What happens if we (a) double the wavelength, (b) halve the slit separation, (c) use white light?

(JEE Main & NEET — tested every single year)

Solution — Step by Step

Two slits separated by $d$ produce waves that reach a point $P$ on the screen with a path difference:

\Delta = d\sin\theta \approx \frac{dy}{D}

(for small angles, where $y$ is the distance from the central maximum)

Bright fringes: $\Delta = n\lambda$ , so $y_n = \dfrac{n\lambda D}{d}$

Fringe width = distance between consecutive bright fringes:

\beta = y_{n+1} - y_n = \frac{\lambda D}{d}

\beta = \frac{\lambda D}{d} = \frac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = \frac{5 \times 10^{-7}}{5 \times 10^{-4}} = \mathbf{1 \times 10^{-3} \text{ m} = 1 \text{ mm}}

(a) Double the wavelength ( $\lambda \to 2\lambda$ ): $\beta \to 2\beta = 2$ mm. Fringes spread out.

(b) Halve the slit separation ( $d \to d/2$ ): $\beta \to 2\beta = 2$ mm. Same effect — fringes spread out.

(c) White light: Each colour has a different $\lambda$ , so each has a different $\beta$ . The central bright fringe is white (all colours overlap at $y = 0$ ). Away from centre, colours separate — you see coloured fringes with violet closest to centre (smallest $\lambda$ , smallest $\beta$ ) and red farthest (largest $\lambda$ , largest $\beta$ ). After a few fringes, colours overlap and the pattern becomes washed out.

Why This Works

The double slit creates two coherent sources. At any point on the screen, the two waves arrive with a fixed path difference that depends on geometry. When this path difference equals a whole number of wavelengths, we get constructive interference (bright). When it’s a half-integer, destructive (dark).

graph TD
    A["YDSE Parameter Change"] --> B{"What changed?"}
    B -->|"λ increases"| C["β increases<br/>Fringes spread"]
    B -->|"d increases"| D["β decreases<br/>Fringes narrow"]
    B -->|"D increases"| E["β increases<br/>Fringes spread"]
    B -->|"Immersed in medium<br/>(refractive index n)"| F["λ_eff = λ/n<br/>β decreases"]
    B -->|"One slit covered<br/>with thin film"| G["Central fringe<br/>shifts toward film"]
    B -->|"White light used"| H["Coloured fringes<br/>Central max is white"]

Alternative Method — Intensity Formula

For a more complete analysis, the intensity at any point is:

I = 4I_0 \cos^2\left(\frac{\pi d y}{\lambda D}\right)

Maximum ( $4I_0$ ) when $\cos^2 = 1$ (bright fringe), minimum ( $0$ ) when $\cos^2 = 0$ (dark fringe). This also shows the fringes are equally spaced (period of $\cos^2$ is constant).

For JEE numericals: if one slit is covered with a thin transparent film of thickness $t$ and refractive index $\mu$ , the extra path = $(\mu - 1)t$ . The central fringe shifts by $\Delta y = (\mu - 1)t \times D/d$ . This is a very common JEE Main numerical.

Common Mistake

When the apparatus is immersed in water (refractive index $n$ ), students forget to use the modified wavelength $\lambda' = \lambda/n$ . The fringe width becomes $\beta' = \lambda D/(nd)$ , which is smaller — fringes get closer together. Using the vacuum wavelength gives the wrong answer. NEET 2022 had this exact scenario.

Question

Solution — Step by Step

Why This Works

Alternative Method — Intensity Formula

Common Mistake

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