A full lesson: a cartoon movie at the school science fair, then the path difference x d/D, bright and dark fringe positions, fringe width, measuring the wavelength of a laser yourself, red, blue and white-light fringes, what changes when you dip the set-up in water or cover a slit with glass, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Two scratches on a sooty slide turn one laser beam into a row of equally spaced bright stripes. Move the wall back and the stripes spread out; use green light and they squeeze together. All of it is one formula: β = λD/d.
At the school science fair, Arjun has the simplest stall: a red laser pointer, a glass slide blackened with candle soot, and two thin scratches on it made with a blade held together. He shines the laser through the scratches in a dark corner. On the wall Meera sees not one red spot but a neat row of red stripes, all the same distance apart.
This is Thomas Young's 1801 experiment, the one that proved light is a wave. The two scratches are the two coherent sources from Lesson 3. In this lesson you will find exactly where each stripe sits, why they are equally spaced, and how one formula, β = λD/d, predicts what happens when you move the wall, change the colour or move the scratches closer.
Smoke a glass slide over a candle, let it cool, and draw two lines very close together with two blades taped together. Shine a laser pointer (never into anyone's eyes) through them onto a wall 2–3 m away. Then try one scratch only.
Two scratches give a row of equally spaced fringes inside a broad glow. One scratch gives only the broad glow with a few wide bands: that is single-slit diffraction (Lesson 5). The fine, equal stripes need two slits.
"One formula runs this whole lesson: β = λD/d. Longer wavelength, bigger screen distance, closer slits: wider fringes."
Two slits S₁ and S₂, a distance d apart, light a screen a distance D away. P is a point at height x above the centre O. Drag P up and down the screen (or use the slider). The drawing exaggerates d; the numbers use d = 0.5 mm, D = 1 m and λ = 600 nm.
O is the point on the screen opposite the midpoint of S₁S₂.
The D² and the squares cancel. Factorise the left side as (S₂P − S₁P)(S₂P + S₁P).
In real set-ups d ≈ 0.1–1 mm and D ≈ 1 m, so the error is tiny (see the readout).
"At the centre x = 0, so the path difference is zero and the centre is always bright. Every other fringe is counted from there."
Put the bright and dark conditions from Lesson 3 into Δ = x d / D. Change λ, d and D and watch the fringes on the screen and the intensity graph.
Δ = nλ. The central fringe (n = 0) is bright.
Δ = (n + ½)λ. The first dark fringe is at β/2, the second at 3β/2.
Distance between two neighbouring bright (or dark) fringes. Bright and dark fringes are equally wide and equally spaced.
"Both formulas jump by the same step, λD/d, from one fringe to the next. That is why the stripes are equally spaced."
"Which is wider in Young's experiment, a bright fringe or a dark fringe?"
Equal. Each is β/2 wide (from the middle of one dark fringe to the next is β). That is different from single-slit diffraction, where the central maximum is twice as wide.
This is how a lab measures λ. The slits are d = 0.30 mm apart and the screen is D = 1.20 m away. Drag the two cross-wires (A and B) onto the middles of two bright fringes, count the fringe widths between them, and compute λ = β d / D.
Measuring across several fringes and dividing reduces the error.
N = number of fringe widths between A and B (count the gaps, not the fringes).
NCERT's example: d = 0.28 mm, D = 1.4 m, centre to 4th bright fringe = 1.2 cm gives λ = 600 nm.
"Convert everything to metres before you divide. Most wrong answers in this question are unit mistakes, not physics mistakes."
Pick the light. Then move the screen with the slider: the fringes on the screen grow, but the angle between neighbouring fringes, seen from the slits, does not change.
β_red / β_blue = 650/450 ≈ 1.44 in the same set-up.
Independent of D. Moving the screen only scales the picture.
Δ = 0 for every colour, so the centre is white. Blue goes dark first (smallest β), so the fringe next to the centre looks red; farther out it looks bluish (NCERT). After a few fringes the colours overlap and the screen looks white again.
"Energy is not lost at dark fringes, only moved. Averaged over many fringes the intensity is 2I₀, the same as two incoherent sources."
NEET loves "what if" questions. The faint pattern is the original; the bright one is after the change. Pick a change and use its slider.
| Change | Fringe width β | Pattern |
|---|---|---|
| λ increased (blue → red) | increases (∝ λ) | wider |
| D increased | increases (∝ D) | wider, angular width same |
| d increased | decreases (∝ 1/d) | narrower |
| Set-up in liquid μ | β/μ | narrower (λ becomes λ/μ) |
| Glass sheet over S₁ | unchanged | whole pattern shifts towards S₁ by (μ−1)tD/d |
| One slit wider | unchanged | dark fringes not fully dark (poorer contrast) |
| One slit closed | no Young fringes | single-slit diffraction pattern |
| White light | differs for each colour | white centre, coloured fringes, then uniform |
"A glass sheet slows the light from S₁, so S₁ needs a shorter path to stay in step. The zero-path-difference point, the central fringe, moves towards S₁. The fringe width does not change."
Change factors multiply: d halved and D doubled → β × 4.
3rd bright to 5th dark, same side: 4.5β − 3β = 1.5β. Opposite sides: add.
600 nm, d = 0.2 mm: θ = 3 × 10⁻³ rad ≈ 0.17°. No D in it.
β = 1.2 mm in air → 0.9 mm in water (μ = 4/3).
Shift by m fringes when (μ − 1)t = mλ. Towards the covered slit.
650 nm and 520 nm: 4 × 650 = 5 × 520 = 2600 nm. First coincidence at x = 4λ₁D/d.
"Write every position as a multiple of β first, then multiply by λD/d at the very end. It saves time and avoids most mistakes."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.
In Young's experiment λ = 600 nm, d = 0.5 mm and D = 1 m. Find the fringe width, the distance of the 3rd bright fringe and of the 2nd dark fringe from the centre.
β = λD/d; n-th bright at nβ; n-th dark at (n − ½)β.
In a double-slit experiment the slits are 0.28 mm apart and the screen is 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Find the wavelength.
4β = 1.2 cm, then λ = βd/D.
The fringe width in air is 2.0 mm. The whole apparatus is immersed in water (μ = 4/3). What is the new fringe width?
In water λ becomes λ/μ, so β becomes β/μ.
Light of λ = 600 nm is used. A thin glass sheet (μ = 1.5) over one slit shifts the central fringe to where the 5th bright fringe was. Find the thickness of the sheet.
Extra optical path (μ − 1)t = number of fringes shifted × λ.
Light of 650 nm and 520 nm is used together, with d = 2 mm and D = 1.2 m. (a) How far from the centre is the 3rd bright fringe of 650 nm? (b) What is the least distance from the centre where bright fringes of both wavelengths coincide?
(a) x = 3λD/d. (b) n₁ × 650 = n₂ × 520 with the smallest whole numbers.
Write every position as a multiple of β first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "fringe width" | β = λD/d | Writing λd/D |
| "n-th dark fringe" | (n − ½)β from the centre | Using (n + ½)β with n starting at 1 |
| "distance between two fringes" | positions in β, subtract (same side) or add (opposite sides) | Forgetting which side each fringe is on |
| "angular fringe width" | λ/d (no D) | Changing it when the screen moves |
| "apparatus immersed in liquid" | β' = β/μ | Multiplying by μ |
| "thin sheet placed over one slit" | shift = (μ − 1)tD/d, β unchanged | Saying the fringe width changes, or the shift is away from the sheet |
| "white light is used" | central fringe white; nearest fringe reddish, then bluish | Saying the central fringe is coloured or dark |
| "bright fringes of two wavelengths coincide" | n₁λ₁ = n₂λ₂ (smallest whole numbers) | Equating fringe widths instead of positions |
Close one slit and the fine fringes vanish, leaving a broad single-slit pattern. Lesson 5 explains that pattern: diffraction at a single slit.
Values used: red 650 nm, blue 450 nm, sodium/laser 589–600 nm, n(water) = 4/3. All positions are measured from the central bright fringe.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.