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 distance $D = 1$ m, and light of wavelength $\lambda = 600$ nm is used. Find the fringe width. What happens to the fringe width if we: (a) double the slit separation, (b) use blue light instead, (c) move the screen closer?

Solution — Step by Step

The fringe width (distance between consecutive bright or dark fringes) is:

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

\beta = \frac{600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = \frac{6 \times 10^{-7}}{5 \times 10^{-4}} = 1.2 \times 10^{-3} \text{ m} = \mathbf{1.2 \text{ mm}}

From $\beta = \lambda D / d$ :

(a) Double slit separation ( $d \to 2d$ ): $\beta' = \lambda D / 2d = \beta/2 = \mathbf{0.6 \text{ mm}}$ . Fringe width halves.

(b) Use blue light ( $\lambda_{blue} \approx 450$ nm instead of 600 nm): $\beta' = (450/600)\beta = 0.75 \times 1.2 = \mathbf{0.9 \text{ mm}}$ . Fringe width decreases because blue has shorter wavelength.

(c) Move screen closer (say $D \to D/2$ ): $\beta' = \lambda(D/2)/d = \beta/2 = \mathbf{0.6 \text{ mm}}$ . Fringe width halves.

$\beta \propto \lambda$ , $\beta \propto D$ , $\beta \propto 1/d$ . Increase wavelength or screen distance → fringes spread out. Increase slit separation → fringes squeeze together.

Why This Works

graph TD
    A["YDSE Parameter Changes"] --> B["Increase λ"]
    A --> C["Increase D"]
    A --> D["Increase d"]
    A --> E["Immerse in water"]
    B --> F["β increases, fringes spread"]
    C --> G["β increases, fringes spread"]
    D --> H["β decreases, fringes squeeze"]
    E --> I["λ decreases to λ/n, β decreases"]
    A --> J["Use white light"]
    J --> K["Central fringe is white, higher fringes show colours"]

The derivation comes from path difference analysis. Two slits separated by $d$ emit coherent waves. At a point on the screen at distance $y$ from the centre, the path difference is approximately $\Delta = yd/D$ (for small angles). Bright fringes occur when $\Delta = n\lambda$ , giving $y_n = n\lambda D/d$ . The spacing between consecutive fringes is $\beta = y_{n+1} - y_n = \lambda D/d$ .

This is one of the most beautiful experiments in physics. Young performed it in 1801 and settled the wave-particle debate for light — at least for the next century until Einstein brought photons back.

Alternative Method

For “what happens when” questions in MCQs, just write $\beta = \lambda D/d$ and check proportionality. You do not need to recalculate — just use ratios:

\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \times \frac{D_2}{D_1} \times \frac{d_1}{d_2}

This ratio method works for any combination of parameter changes and saves significant time in competitive exams.

Common Mistake

Forgetting that immersing the setup in water changes wavelength. When the YDSE apparatus is submerged in a medium of refractive index $n$ , the wavelength becomes $\lambda/n$ but $D$ and $d$ remain unchanged. So $\beta_{water} = \lambda D/(nd)$ — fringe width decreases by factor $n$ . Students often leave the fringe width unchanged, forgetting that wavelength changes in a medium. This has appeared in JEE Main 2023 Shift 1 and NEET 2022.

Fringe width: $\beta = \frac{\lambda D}{d}$

Bright fringe position: $y_n = \frac{n\lambda D}{d}$ , $n = 0, 1, 2, ...$

Dark fringe position: $y_n = \frac{(2n-1)\lambda D}{2d}$ , $n = 1, 2, 3, ...$

Path difference: $\Delta = \frac{yd}{D}$ (for small angles)

In medium of refractive index $n$ : $\beta' = \frac{\beta}{n}$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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