Young's Double Slit — Solving Strategy

The Setup You Need to Visualise

Young’s double-slit experiment is the cleanest demonstration of wave interference in the Class 12 syllabus. Two narrow slits, separated by distance $d$ , are illuminated by monochromatic light. A screen sits at distance $D$ behind them. On that screen we see alternating bright and dark fringes — proof that light is a wave.

The whole pattern reduces to two numbers: the slit separation $d$ and the screen distance $D$ . From these we get fringe positions, fringe widths, and the answer to almost every JEE/NEET question on this chapter.

We’ll walk through the formulas, the standard problem types, and the traps examiners love to set. The math here is high-leverage — the same logic shows up in thin-film interference, diffraction gratings, and even electron diffraction.

Key Terms & Definitions

Path difference ( $\Delta$ ): The difference in distances from the two slits to a point on the screen. For a point at distance $y$ from the central axis, $\Delta \approx yd/D$ (small-angle approximation).

Phase difference ( $\phi$ ): $\phi = (2\pi/\lambda)\Delta$ . Constructive interference at $\phi = 2n\pi$ , destructive at $\phi = (2n+1)\pi$ .

Fringe width ( $\beta$ ): Distance between consecutive bright (or dark) fringes. $\beta = \lambda D/d$ .

Order of fringe ( $n$ ): Integer labelling each maximum. $n = 0$ is the central bright fringe; $n = \pm 1, \pm 2, \ldots$ are successive bright fringes on either side.

Coherence: Both slits must emit light with a fixed phase relationship — that’s why we use one source and split it via two slits, not two independent bulbs.

The Core Formulas

$y_n^{bright} = \frac{n\lambda D}{d}, \quad n = 0, \pm 1, \pm 2, \ldots$

$y_n^{dark} = \frac{(2n+1)\lambda D}{2d}, \quad n = 0, 1, 2, \ldots$

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

$I = I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right) \approx 4I_1 \cos^2\left(\frac{\pi y d}{\lambda D}\right)$

where $I_1$ is the intensity from one slit alone.

The intensity formula is the gem most students skip. It tells you the brightness at any point — not just maxima and minima — and shows that the maxima are 4 times brighter than a single slit (constructive interference of two equal amplitudes).

Standard Problem Types

Type 1: Find Fringe Width

Given $\lambda$ , $d$ , $D$ , compute $\beta$ . Plug into $\beta = \lambda D/d$ . Done.

Most common error: unit mixing. $\lambda$ in nm, $d$ in mm, $D$ in m — convert everything to metres before plugging.

Type 2: Find Wavelength from Fringe Width

Given $\beta$ , $d$ , $D$ , find $\lambda$ .

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

Type 3: Distance Between Specific Fringes

“Distance between 3rd bright and 5th dark fringe on the same side.”

3rd bright: $y = 3\lambda D/d$ . 5th dark: $y = (2 \times 4 + 1)\lambda D/(2d) = 9\lambda D/(2d)$ .

Distance = $9\lambda D/(2d) - 3\lambda D/d = 3\lambda D/(2d) = 1.5\beta$ .

Type 4: Effect of a Slab in Front of One Slit

A glass slab of thickness $t$ and refractive index $\mu$ in front of one slit shifts the central maximum by:

$\Delta y = \frac{D(\mu - 1)t}{d}$

The shift is toward the slit with the slab (because that path now takes longer in vacuum-equivalent length).

Type 5: Change of Medium

If the whole apparatus is immersed in a medium of refractive index $n$ , the wavelength becomes $\lambda/n$ , so:

$\beta_{new} = \frac{\beta}{n}$

Solved Examples

Easy (CBSE)

Light of wavelength 600 nm illuminates two slits 0.5 mm apart. The screen is 2 m away. Find fringe width.

$\beta = \frac{600 \times 10^{-9} \times 2}{0.5 \times 10^{-3}} = 2.4 \times 10^{-3}\text{ m} = 2.4\text{ mm}$

Medium (JEE Main)

In a YDSE setup, the 4th bright fringe of light $\lambda_1$ coincides with the 5th bright fringe of light $\lambda_2 = 480$ nm. Find $\lambda_1$ .

Coincidence: $4\lambda_1 D/d = 5\lambda_2 D/d \implies \lambda_1 = 5\lambda_2/4 = 600$ nm.

Hard (JEE Advanced)

A glass slab of thickness 12 μm and refractive index 1.5 is placed in front of one slit. Slits are 1 mm apart, screen 1 m away, $\lambda = 600$ nm. Find the shift of the central maximum.

$\Delta y = \frac{D(\mu-1)t}{d} = \frac{1 \times 0.5 \times 12 \times 10^{-6}}{10^{-3}} = 6 \times 10^{-3}\text{ m} = 6\text{ mm}$

The central fringe shifts 6 mm toward the slit with the slab.

JEE Main 2024 Shift 1 had this exact pattern with slightly different numbers. The “shift toward the slab” rule is non-negotiable — many students get the magnitude right but the direction wrong.

Exam-Specific Tips

CBSE Boards

Derive $\beta = \lambda D/d$ from first principles. The 5-mark question loves the derivation.
State the conditions for sustained interference (coherent sources, equal amplitudes, monochromatic light).
Diagram with labels — slit separation, fringe positions, path difference.

JEE Main

Practice intensity questions: $I = 4I_1 \cos^2(\phi/2)$ where $\phi = (2\pi/\lambda)\Delta$ .
Slab-in-one-slit problems are frequent. Memorise the shift formula.
Refractive index of medium changes both $\lambda$ and effective $d$ if slits are in different media — read carefully.

NEET

Less calculation, more concept. Know that fringe width does not depend on slit position, only on $\lambda$ , $D$ , $d$ .
White light gives a central white fringe (all wavelengths constructively interfere there) with coloured fringes outside. Red is farthest from centre because it has the largest $\lambda$ .

Common Mistakes to Avoid

Using degrees instead of radians for $\theta$ : Phase formulas need radians. The small-angle approximation $\sin\theta \approx \tan\theta \approx y/D$ already keeps things in radians.
Confusing path difference and phase difference: $\phi = (2\pi/\lambda) \Delta$ . They differ by a factor of $2\pi/\lambda$ .
Forgetting that “nth dark fringe” starts at $n=0$ for the first dark, not $n=1$ : Some textbooks number differently. Check the convention being used.
Slab shift on the wrong side: The slab makes the path through that slit “optically longer”, so the central fringe moves toward the slab (the side that needs to “catch up”).
Treating the apparatus as in vacuum when it’s submerged: $\lambda \to \lambda/n$ , fringe width drops by factor of $n$ .

Practice Questions

Q1. Two slits 1 mm apart are illuminated by 500 nm light. Screen is at 2 m. Find fringe width.

$\beta = (500 \times 10^{-9})(2)/(10^{-3}) = 10^{-3}$ m = 1 mm.

Q2. If the slit separation is doubled, what happens to fringe width?

$\beta = \lambda D / d$ . Doubling $d$ halves $\beta$ .

Q3. A YDSE setup uses 600 nm light, $d = 0.6$ mm, $D = 1.5$ m. At what distance from the central fringe is the 5th maximum?

$y_5 = 5\lambda D/d = 5 \times 600 \times 10^{-9} \times 1.5 / (0.6 \times 10^{-3}) = 7.5 \times 10^{-3}$ m = 7.5 mm.

Q4. The intensity at the central maximum is $I_0$ . What is the intensity at a point with path difference $\lambda/4$ ?

$\phi = 2\pi(\lambda/4)/\lambda = \pi/2$ . $I = I_0 \cos^2(\pi/4) = I_0/2$ .

Q5. The whole YDSE is dipped in water ( $n = 4/3$ ). Fringe width becomes?

$\beta_{new} = \beta/n = 3\beta/4$ .

Q6. Why is interference not observed with two independent bulbs?

Independent sources have random phase relationships that vary on time scales much shorter than the eye’s response. Patterns wash out. Coherence requires phase locking, which we get by splitting one source.

Q7. In a YDSE, if light of two wavelengths 480 nm and 600 nm is used, at what minimum distance from the centre do their bright fringes coincide?

Need $n_1 \times 480 = n_2 \times 600 \implies n_1/n_2 = 5/4$ . Minimum: $n_1 = 5, n_2 = 4$ . So at $5\lambda_1 D/d = 5(480 \times 10^{-9})D/d$ . Plug in $D, d$ values for numerical answer.

Q8. A thin film of $\mu = 1.4$ and thickness 6 μm is placed before one slit. By how many fringes does the pattern shift? ( $\lambda = 600$ nm.)

Shift in path = $(\mu - 1)t = 0.4 \times 6 \times 10^{-6} = 2.4 \times 10^{-6}$ m. Number of fringes shifted = $2.4 \times 10^{-6} / 600 \times 10^{-9} = 4$ fringes.

FAQs

What if the two slits have unequal widths? The amplitudes at the screen are unequal, so the minima aren’t completely dark. Visibility drops but the fringe spacing stays $\lambda D/d$ .

Why is the central fringe always bright? At the central point, both paths are equal — zero path difference, so constructive interference for all wavelengths.

What changes if we use white light? Central fringe is white, surrounded by coloured fringes. Higher orders blur into a smear because different wavelengths place fringes at different positions.

Does fringe width depend on slit width (not slit separation)? No. Slit width affects single-slit diffraction envelope, which modulates the YDSE pattern. The fringe spacing depends only on slit separation $d$ .

Why does the formula $\beta = \lambda D/d$ assume small angles? It uses $\sin\theta \approx \tan\theta \approx y/D$ . For large angles (or very wide screens), use $d\sin\theta = n\lambda$ exactly.

What’s the relationship to the diffraction grating? A grating is YDSE with thousands of slits. Maxima are sharper but at the same angles as YDSE: $d\sin\theta = n\lambda$ .