Interference of light and Young's experiment

Hold a single hair up to a bright streetlamp at night and squint past it — you will see faint, evenly-spaced bands of light and dark fringing its shadow. That flicker is not a trick of the eye. It is light proving, in the simplest possible setup, that it is a wave. In 1801 Thomas Young shone a single beam through two narrow slits cut side by side in an opaque screen and let the two emerging beams overlap on a wall beyond. Instead of two blurred patches of brightness, he saw a neat ladder of bright and dark stripes — fringes — repeating with almost machine-like regularity. No particle picture of light, however clever, predicts that pattern. Two streams of bullets fired through two slits pile up in two heaps, not a striped ladder. Only waves, arriving in step at some points and out of step at others, can build a pattern like that. This is Young's double-slit experiment (YDSE), and it is the single most important demonstration in the Wave Optics chapter. The idea is almost stubbornly simple: split one wave into two coherent copies, let them walk slightly different distances to reach a screen, and read off where they land in phase and where they land out of phase. The physics is compact enough to fit in one formula, yet rich enough to explain everything from the shimmer of a soap bubble to why a laser pointer makes crisp, colourful bands. This article builds the physical picture of coherent addition producing bright and dark fringes, hands you the exam-ready fringe-width formula $\beta = \lambda D/d$ , walks through the reasoning that connects geometry to that formula, and then drills it with a board-style and a competitive-style worked problem — reusing and extending the two numerical examples already anchored in this topic's core notes.

Key takeaways

Two narrow slits lit by one coherent source act as two coherent secondary sources, and their overlapping light produces alternating bright and dark fringes on a screen.
A bright fringe forms where the path difference $\Delta = S_2P - S_1P$ is a whole number of wavelengths; a dark fringe forms where it is an odd multiple of half a wavelength.
The exam-ready result is the fringe width: $\beta = \dfrac{\lambda D}{d}$ — the syllabus asks you to know and apply this formula, not reproduce its geometric derivation.
All fringes are equally spaced by $\beta$ , except that the central fringe at $n = 0$ sits exactly midway between the slits' perpendicular bisector and is always bright.
$\beta$ grows with wavelength $\lambda$ and screen distance $D$ , and shrinks as the slit separation $d$ is widened — so red fringes are wider than blue, and closer slits give a coarser, easier-to-see pattern.
Common exam slip-ups: swapping $d$ and $D$ , forgetting the central fringe is bright, and trying to re-derive $\beta$ from scratch when the formula alone is what is asked.
Key idea: two coherent slits create a path difference $\Delta$ that alternates between constructive ( $\Delta = n\lambda$ ) and destructive ( $\Delta = (n+\tfrac12)\lambda$ ) interference across the screen, producing equally spaced fringes of width $\beta = \lambda D/d$ .

The core insight

Coherent light from two narrow slits arrives at each screen point with a path difference $\Delta$ that is a whole number of wavelengths for bright fringes and a half-integer number for dark fringes, giving equally spaced fringes of width $\beta = \lambda D/d$ .

Where it fits in the syllabus

This is a central, board-examined topic of the CBSE Class 12 Wave Optics chapter, sitting right after coherent and incoherent addition of waves and just before diffraction. The syllabus is explicit and narrow here: you must know the YDSE setup, the conditions for coherent sources, and the final expression for fringe width, $\beta = \lambda D/d$ , with no derivation required — the geometric path-difference derivation of that formula is explicitly kept out of the prescribed scope, so exams will test your ability to use the formula, not reproduce where it comes from. This article respects that boundary: the derivation block below is offered only as background reasoning for a curious reader, clearly separated from what the board actually expects you to write in an answer.

The physical picture

In Young's experiment, a single monochromatic source illuminates two narrow, closely spaced slits $S_1$ and $S_2$ , separated by a small distance $d$ . By Huygens' principle, each slit acts as a fresh point source of secondary wavelets, and because both wavelets originate from the same primary wavefront, they are automatically coherent — they keep a constant phase relationship and can sustain a stable interference pattern instead of a pattern that flickers and washes out. A screen is placed a distance $D$ from the slits, with $D$ typically about a metre or two while $d$ is a fraction of a millimetre, so $D \gg d$ throughout this chapter. At a point $P$ on the screen, light from $S_1$ and light from $S_2$ have travelled slightly different distances. The difference between these two path lengths is the path difference:

\Delta = S_2P - S_1P.

where: $\Delta$ is measured in the same length units as $\lambda$ (commonly metres or angstroms), and $S_2P$ , $S_1P$ are the straight-line distances from each slit to the point $P$ . The physical reading is simple: $\Delta$ tells you how many extra wave crests one beam has to "catch up" on before reaching $P$ , compared with the other beam. Whether the two waves reinforce or cancel at $P$ depends only on how $\Delta$ compares with the wavelength $\lambda$ :

\Delta = n\lambda \quad (\text{bright fringe}, \; n = 0, \pm1, \pm2, \dots), \qquad \Delta = \left(n + \tfrac12\right)\lambda \quad (\text{dark fringe}).

where: $n$ is the order of the fringe, an integer counting how many whole wavelengths separate the two path lengths, $\lambda$ is the wavelength of the light used, and the two conditions alternate as you move along the screen — bright, dark, bright, dark — because $\Delta$ changes smoothly and continuously with position. The syllabus's headline quantitative result is the fringe width, the distance between two consecutive bright fringes (equal to the distance between two consecutive dark fringes):

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

where: $\lambda$ is the wavelength of light, $D$ is the slit-to-screen distance, and $d$ is the slit separation. This is the one formula the board exam expects you to state and apply directly, without deriving it from the path-difference geometry above.

Figure

Two slits S one and S two lit by a single wavefront send overlapping circular wavelets toward a distant screen, forming alternating bright and dark fringe bands with the central bright fringe exactly on the axis

All fringes are equally spaced by $\beta$ , apart from the fact that the very central fringe (where $\Delta = 0$ exactly, midway between the two slits) is always bright, since zero path difference trivially satisfies $\Delta = n\lambda$ with $n = 0$ . From the coherent-addition treatment of the previous topic, the intensity does not jump abruptly between bright and dark — it varies smoothly between $I_{\max} = 4I_0$ at the bright fringes and $I_{\min} = 0$ at the dark fringes, with $\beta$ describing the spacing of the maxima, not a hard-edged stripe width.

Derivation AWhere the fringe-width formula comes from

The board exam asks only that you state and apply $\beta = \lambda D/d$ — the geometric derivation below is explicitly outside the prescribed CBSE scope and is included here purely so the formula does not feel like it fell from the sky; you are not expected to reproduce these steps in an exam answer, only the boxed result at the end and its direct use.

Step 1. Set up coordinates. Place the midpoint of the two slits at the origin, with the screen a distance $D$ away and parallel to the line joining the slits. Let $P$ be a point on the screen at a small height $y$ above the central axis, so $y \ll D$ in every realistic YDSE geometry (millimetres against metres).

Step 2. Write the two path lengths using the right-angled triangles from each slit to $P$ . With slit separation $d$ , $S_1P^2 = D^2 + \left(y - \tfrac{d}{2}\right)^2$ and $S_2P^2 = D^2 + \left(y + \tfrac{d}{2}\right)^2$ . Subtracting the two equations, the $D^2$ and $y^2$ terms cancel, leaving $S_2P^2 - S_1P^2 = 2yd$ .

Step 3. Factor the left side as $(S_2P - S_1P)(S_2P + S_1P) = 2yd$ . Because $D \gg d$ and $D \gg y$ in this geometry, both $S_1P$ and $S_2P$ are extremely close to $D$ , so their sum $S_2P + S_1P \approx 2D$ . Substituting this approximation gives the path difference directly: $\Delta = S_2P - S_1P \approx \dfrac{2yd}{2D} = \dfrac{yd}{D}$ .

Step 4. Apply the bright-fringe condition $\Delta = n\lambda$ to locate the $n$ th bright fringe: $\dfrac{y_n d}{D} = n\lambda$ , so $y_n = \dfrac{n\lambda D}{d}$ . The next bright fringe, of order $n+1$ , sits at $y_{n+1} = \dfrac{(n+1)\lambda D}{d}$ .

Step 5. Subtract consecutive fringe positions to get the spacing between them. $\beta = y_{n+1} - y_n = \dfrac{(n+1)\lambda D}{d} - \dfrac{n\lambda D}{d} = \dfrac{\lambda D}{d}$ . Because this result does not depend on $n$ , every pair of consecutive bright fringes — and, by the identical argument applied to the dark-fringe condition, every pair of consecutive dark fringes — is separated by exactly this same distance $\beta = \lambda D/d$ , which is why the pattern looks so regular.

Worked examples

Board level - fringe width and fringe position

In a Young's double-slit experiment, two slits separated by $d = 0.5\ \text{mm}$ are illuminated by monochromatic light of wavelength $\lambda = 6000\ \text{Å} = 6\times10^{-7}\ \text{m}$ . The screen is placed $D = 1.0\ \text{m}$ from the slits. Find the fringe width, and the distance of the 4th bright fringe from the central maximum.

Step 1. Write down what is given and convert everything to SI units before substituting: $d = 0.5\times10^{-3}\ \text{m}$ , $\lambda = 6\times10^{-7}\ \text{m}$ , $D = 1.0\ \text{m}$ .

Step 2. Apply the fringe-width formula directly. $\beta = \dfrac{\lambda D}{d} = \dfrac{(6\times10^{-7})(1.0)}{0.5\times10^{-3}} = \dfrac{6\times10^{-7}}{5\times10^{-4}} = 1.2\times10^{-3}\ \text{m} = 1.2\ \text{mm}$ .

Step 3. Use the fact that the $n$ th bright fringe lies a distance $n\beta$ from the central maximum (since $y_n = n\lambda D/d = n\beta$ ), so for $n = 4$ : $y_4 = 4\beta = 4 \times 1.2\ \text{mm} = 4.8\ \text{mm}$ .

The fringe width is $\beta = 1.2\ \text{mm}$ , and the 4th bright fringe sits $4.8\ \text{mm}$ from the central maximum.

Competitive level - working backward from fringe width, then shifting one slit's light source

In a YDSE setup, the slit separation is $d = 1\ \text{mm}$ and the screen is at $D = 2\ \text{m}$ . The fringe width observed is $\beta = 1.0\ \text{mm}$ . First find the wavelength of light used and confirm it is visible. Then, if the whole apparatus (same $\lambda$ , $d$ , $D$ ) is dipped in a medium and the fringe width shrinks to exactly $\beta' = 0.75\ \text{mm}$ with no change in $d$ or $D$ , find the effective wavelength inside that medium and hence the medium's refractive index for this light.

Step 1. Rearrange the fringe-width formula for wavelength: from $\beta = \lambda D/d$ , $\lambda = \dfrac{\beta d}{D}$ .

Step 2. Substitute the given board-level numbers: $\lambda = \dfrac{(1.0\times10^{-3})(1\times10^{-3})}{2} = \dfrac{1.0\times10^{-6}}{2} = 5.0\times10^{-7}\ \text{m} = 500\ \text{nm} = 5000\ \text{Å}$ . Visible light spans roughly $4000$ – $7000\ \text{Å}$ , so $5000\ \text{Å}$ falls comfortably inside it — this is green light.

Step 3. For the medium case, the geometry ( $d$ , $D$ ) is unchanged, so the shrunken fringe width must come entirely from a shorter wavelength inside the medium: $\lambda_{\text{medium}} = \dfrac{\beta' d}{D} = \dfrac{(0.75\times10^{-3})(1\times10^{-3})}{2} = 3.75\times10^{-7}\ \text{m} = 3750\ \text{Å}$ .

Step 4. The refractive index of a medium is defined as the ratio of the wavelength in vacuum (or air) to the wavelength in the medium, $\mu = \lambda_{\text{air}}/\lambda_{\text{medium}}$ , since frequency stays fixed while wavelength shortens on entering a denser medium. $\mu = \dfrac{5000}{3750} = 1.33$ .

In air, $\lambda = 500\ \text{nm}$ (visible, green light); inside the medium, $\lambda_{\text{medium}} = 3750\ \text{Å}$ and the refractive index works out to $\mu = 1.33$ , consistent with a water-like medium. Timing note: the board-level half of this problem should take under a minute — one substitution into $\lambda = \beta d/D$ . The refractive-index extension is a favourite way competitive papers dress up the same core formula; recognising that only $\lambda$ changes inside a denser medium (while $d$ and $D$ stay fixed) is the entire trick, so budget no more than two minutes total once the pattern is familiar.

Common traps

Treating $\beta = \lambda D/d$ as something to re-derive from path-difference geometry in a board answer — the syllabus specifies final expression only; state the formula, substitute, and move on. Save the geometry for curiosity, not for the answer sheet.
Mixing up $d$ (slit separation, typically sub-millimetre) and $D$ (slit-to-screen distance, typically a metre or more) — swapping them inverts the formula and produces answers that are wrong by several orders of magnitude, often an instant giveaway that something was flipped.
Forgetting that the central fringe, at $n = 0$ , is bright, not dark, since the path difference there is exactly zero — a very common slip when students assume the pattern must start with a dark band.
Assuming interference fringes are uniformly bright — only the fringe spacing is uniform; the intensity itself varies smoothly from $4I_0$ at a bright fringe down to zero at a dark fringe, it does not switch on and off like a light switch.
Forgetting to convert units before substituting — $d$ is usually given in millimetres and $\lambda$ in angstroms, and mixing unit systems inside one formula is the single most common numerical-error source in this topic.

You should be able to…

Describe the YDSE setup and explain, using Huygens' principle, why the two slits act as coherent sources.
State the path-difference conditions for bright and dark fringes, $\Delta = n\lambda$ and $\Delta = (n+\tfrac12)\lambda$ .
Recall and correctly apply $\beta = \lambda D/d$ to find fringe width, wavelength, slit separation, or screen distance given the other three quantities.
Predict how fringe width changes when $\lambda$ , $D$ , or $d$ is varied, including qualitative reasoning about which change widens or narrows the fringes.
Locate the position of the $n$ th bright or dark fringe from the central maximum using $y_n = n\beta$ .
Explain, with an example, why two independent light sources cannot produce a stable interference pattern.

Ask the tutor: Want to try a few more fringe-width problems, or see how changing the light's colour or the slit spacing reshapes the pattern step by step?

Why do we need two slits instead of just shining two separate torches at a wall?

Because two separate torches are not coherent — their light waves are emitted by huge numbers of independent atoms with randomly shifting phases, so the pattern of reinforcement and cancellation at any point on the wall changes many billions of times per second. Your eye (and even fast electronics) only ever sees the time-average, which washes out to uniform brightness. Young's trick is to derive both beams from the same original wavefront, by passing light through one narrow slit and then two more slits close together. That shared origin locks the phase relationship between the two beams, so the interference pattern stays fixed in place and becomes visible.

Does the fringe pattern go on forever, getting fainter and fainter to either side?

In the strict two-source idealisation used in this chapter, the maxima keep appearing at every integer $n$ forever, all with the same intensity $4I_0$ , so mathematically the pattern does not fade. In a real experiment it does fade toward the edges, but that fading comes from effects outside this chapter's scope — the finite width of each slit spreads out the light unevenly (a diffraction effect, covered in the next topic), and the limited coherence length of real light sources eventually blurs the fringes at large path differences. For CBSE purposes, treat the fringes from the interference formula as equally bright and equally spaced.

Is Young's double-slit experiment the same thing as diffraction?

No, though the two are closely related and often confused. Interference in YDSE is about two (or a few) discrete coherent sources overlapping to give sharp bright and dark fringes described by $\beta = \lambda D/d$ . Diffraction, the next topic in this chapter, is about a single slit of finite width, where every point across that one opening acts as a source, producing a broad central bright band flanked by much fainter secondary bands. In a real YDSE setup both effects are present together — the two-slit interference fringes sit inside a diffraction envelope from each individual slit — but CBSE treats them as two separate, independently examinable ideas.

Do I need to memorise the derivation for the board exam?

No. The 2025-26 CBSE syllabus explicitly lists the fringe-width result $\beta = \lambda D/d$ as a "final expression only" item — you are expected to know what each symbol means, state the formula correctly, and apply it in numerical problems, but you will not be asked to derive it from the path-difference geometry. Spend your revision time on getting comfortable rearranging the formula for any one of $\beta$ , $\lambda$ , $D$ , or $d$ given the other three, since that is exactly the skill both worked examples above test.