PhysicsISC Class 12 · Wave Optics

Young's Double Slit Experiment: Path Difference, Fringe Positions and Fringe Width

Updated 15 October 2026 · Fact-checked

In Young's double slit experiment, light from two coherent slits a distance d apart overlaps on a screen at distance D. The path difference at height y is $\frac{yd}{D}$. Bright fringes lie at $y_n = \frac{n\lambda D}{d}$, dark fringes at $\left(n + \frac{1}{2}\right)\frac{\lambda D}{d}$, and the fringe width is $\beta = \frac{\lambda D}{d}$.

Understand Young's Double Slit Experiment and Fringe Width

Interference and coherence. When two light waves meet, their displacements add (superposition). If they arrive in phase they reinforce (constructive, bright); if half a wavelength out of step they cancel (destructive, dark). A steady pattern needs coherent sources: sources of the same frequency with a constant phase difference. Two separate lamps are not coherent, because each atom emits short random bursts. Young obtained coherent sources by letting light from one narrow slit S fall on two slits $S_{1}$ and $S_{2}$ : both are copies of the same wavefront.

Conditions for sustained interference.
1. The sources must be coherent (constant phase difference).
2. They should emit the same wavelength (monochromatic light).
3. Amplitudes should be equal or nearly equal (for good contrast: dark fringes really dark).
4. The slits should be narrow and close together, and the screen far away (so fringes are wide enough to see).

Path difference and phase difference. A path difference Δx corresponds to a phase difference $ϕ = \frac{2 π}{λ} Δ x$ .

Bright: $Δ x = nλ$ (φ = 2nπ), n = 0, 1, 2, ...
Dark: $Δ x = (n + \frac{1}{2}) λ$ (φ = (2n + 1)π).

Derivation of fringe width (asked in ISC 2024 Q13(ii)). Let O be the point on the screen opposite the midpoint of the slits, and P a point at height y. Drop a perpendicular from $S_{1}$ to $S_{2} P$ . For D ≫ d the rays $S_{1} P$ and $S_{2} P$ are nearly parallel, making angle θ with the axis, so the path difference is
$Δ x = S_{2} P - S_{1} P = d sin θ .$
As θ is small, $sin θ \approx tan θ = \frac{y}{D}$ , so $Δ x = \frac{y d}{D}$ .

Bright fringe: $\frac{y _{n} d}{D} = nλ \Rightarrow y_{n} = \frac{nλ D}{d}$ .
Dark fringe: $y_{n}^{'} = (n + \frac{1}{2}) \frac{λ D}{d}$ .

The separation of consecutive bright (or dark) fringes is
$β = y_{n + 1} - y_{n} = \frac{λ D}{d} .$
The angular fringe width is $\frac{β}{D} = \frac{λ}{d}$ .

Features of the pattern.

Fringes are equally spaced, straight and parallel to the slits; the central fringe (n = 0) is bright.
All bright fringes have (nearly) the same intensity: the intensity graph against angle is a row of equal peaks (a $cos^{2}$ shape), unlike single-slit diffraction.
β increases with λ and D, and decreases with d.
In a medium of refractive index μ, λ becomes $\frac{λ}{μ}$ , so $β^{'} = \frac{β}{μ}$ .
With white light the central fringe is white (zero path difference for every colour); the others are coloured, with violet nearer the centre (smaller β), and fade after a few orders.
Energy is not destroyed: it is redistributed from dark to bright regions.

Young's Double Slit Experiment and Fringe Width: formulas and key facts

Path difference: $Δ x = d sin θ \approx \frac{y d}{D}$; Small angles.
Phase difference: $ϕ = \frac{2 π}{λ} Δ x$; Radians.
nth bright fringe: $y_{n} = \frac{nλ D}{d}$; n = 0, 1, 2, ...
nth dark fringe: $y_{n}^{'} = (n + \frac{1}{2}) \frac{λ D}{d}$; n = 0 gives the first dark fringe.
Fringe width: $β = \frac{λ D}{d}$; Same for bright and dark.
Angular fringe width: $θ = \frac{λ}{d}$; Independent of D.
In a medium: $β^{'} = \frac{β}{μ}$; Wavelength shrinks.

How to solve Young's Double Slit Experiment and Fringe Width questions

Convert everything to metres, find β first, then count fringes in multiples of β.

1Convert: nm → $1 0^{- 9}$ m, mm → $1 0^{- 3}$ m.
2Compute $β = \frac{λ D}{d}$ .
3nth bright: $n β$ ; nth dark (counting the first as n = 1): $(n - \frac{1}{2}) β$ .
4Same side of the centre: subtract positions; opposite sides: add them.
5Path-difference questions: divide Δx by λ; a whole number means bright, a half-integer means dark.
6Coincidence of two wavelengths: find the smallest n₁, n₂ with $n_{1} λ_{1} = n_{2} λ_{2}$ .

Young's Double Slit Experiment and Fringe Width in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Fringes from a laser pointer

Suppose you scratch two fine lines 0.25 mm apart on a blackened glass slide and shine a red laser pointer (650 nm) through them onto a wall 3 m away.

How far apart are the bright fringes on the wall?

Try it first, then show the solutionSolution

In metres: λ = $650 \times 1 0^{- 9}$ m, D = 3 m, d = $0.25 \times 1 0^{- 3}$ m.
$β = \frac{650 \times 1 0 ^{- 9} \times 3}{0.25 \times 1 0 ^{- 3}} = 7.8 \times 1 0^{- 3}$ m.

Answer: About 7.8 mm, easy to see on the wall.

Memory hook: Close slits, far wall, wide fringes.

Measuring the wavelength of a laser

Imagine a student with an unlabelled green laser, a double slit of spacing 0.3 mm and a screen 1.2 m away, who measures the fringe width as 2.12 mm.

Find the wavelength.

Try it first, then show the solutionSolution

$λ = \frac{β d}{D} = \frac{2.12 \times 1 0 ^{- 3} \times 0.3 \times 1 0 ^{- 3}}{1.2}$ .
$λ \approx 5.3 \times 1 0^{- 7}$ m = 530 nm.

Answer: ≈ 530 nm, green light.

Memory hook: Rearrange β = λD/d to measure λ.

Tricks to remember Young's Double Slit Experiment and Fringe Width

Short memory aids. Say each one aloud once, then use it on a question the same day.

Lambda D over d
Big D on top, small d below: "LD over d".
Use it when: Fringe width.
Whole λ bright, half λ dark
Δx = nλ bright; Δx = (n + ½)λ dark.
Use it when: Bright or dark?
Opposite sides add
Fringes on either side of the centre: add distances.
Use it when: Avoiding the 2025 slip.
Water shrinks fringes
In a medium of index μ, β becomes β/μ.
Use it when: Medium questions.

Young's Double Slit Experiment and Fringe Width: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Changing D and d · 1 mark · MCQBoard level

In Young's experiment, the screen distance is doubled and the slit separation is halved. The fringe width becomes:
(a) the same
(b) twice
(c) four times
(d) half

Try it first, then show the solutionSolution

$β \propto \frac{D}{d}$ : $\frac{2}{0.5} = 4$ .

Answer: (c).

Example 2: Fringe width and a dark fringe · 2 marks · Short answerBoard level

In a double slit experiment, d = 0.8 mm, D = 1.6 m and λ = 500 nm. Find the fringe width and the distance of the third dark fringe from the centre.

Try it first, then show the solutionSolution

$β = \frac{500 \times 1 0 ^{- 9} \times 1.6}{0.8 \times 1 0 ^{- 3}} = 1.0 \times 1 0^{- 3}$ m = 1.0 mm. (1)
Third dark fringe: $2.5 β = 2.5$ mm. (1)

Answer: 1.0 mm; 2.5 mm.

Example 3: Derivation of fringe width · 3 marks · Long answerHarder board level

With a labelled diagram of Young's double slit experiment, show that the fringe width is $β = \frac{λ D}{d}$ .

Try it first, then show the solutionSolution

Diagram: slits $S_{1}$ , $S_{2}$ separated by d, screen at distance D, point P at height y from O, angle θ, perpendicular from $S_{1}$ on $S_{2} P$ . (1)
Path difference $Δ x = d sin θ \approx d tan θ = \frac{y d}{D}$ (θ small). (1)
Bright: $\frac{y _{n} d}{D} = nλ$ , so $y_{n} = \frac{nλ D}{d}$ ; $β = y_{n + 1} - y_{n} = \frac{λ D}{d}$ . (1)

Answer: $β = \frac{λ D}{d}$ .

Example 4: Fringes on opposite sides · 2 marks · Short answerHarder board level

In a YDSE with d = 0.6 mm, D = 1.5 m and λ = 480 nm, find the distance between the fourth bright fringe on one side of the centre and the second dark fringe on the other side.

Try it first, then show the solutionSolution

$β = \frac{480 \times 1 0 ^{- 9} \times 1.5}{0.6 \times 1 0 ^{- 3}} = 1.2$ mm. (1)
Fourth bright at 4β = 4.8 mm; second dark at 1.5β = 1.8 mm on the other side; total 4.8 + 1.8 = 6.6 mm. (1)

Answer: 6.6 mm.

Example 5: Two wavelengths together · 3 marks · Short answerHOTS / competency

A double slit (d = 1 mm, D = 1.2 m) is lit by light containing 650 nm and 520 nm. Find the least distance from the central maximum at which bright fringes of both wavelengths coincide.

Try it first, then show the solutionSolution

Coincidence needs $n_{1} \times 650 = n_{2} \times 520$ , i.e. $\frac{n _{1}}{n _{2}} = \frac{4}{5}$ . (1)
Smallest: $n_{1} = 4$ , $n_{2} = 5$ ; common path difference 2600 nm. (1)
$y = \frac{2600 \times 1 0 ^{- 9} \times 1.2}{1 0 ^{- 3}} = 3.12 \times 1 0^{- 3}$ m = 3.12 mm. (1)

Answer: 3.12 mm.

Example 6: Reading path differences · 4 marks · Case-basedHOTS / competency

Suppose a YDSE uses light of 600 nm, slits 0.9 mm apart and a screen 1.5 m away.
(i) Is a point with path difference 1.8 µm bright or dark? (1 mark)
(ii) Is a point with path difference 2.1 µm bright or dark? Give the phase difference. (1 mark)
(iii) Find the fringe width. (1 mark)
(iv) Find the fringe width if the apparatus is immersed in a liquid of refractive index 1.25. (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{1.8 \times 1 0 ^{- 6}}{600 \times 1 0 ^{- 9}} = 3$ : whole number, bright (third bright fringe).
(ii) $\frac{2.1 \times 1 0 ^{- 6}}{600 \times 1 0 ^{- 9}} = 3.5$ : dark; $ϕ = 2 π \times 3.5 = 7 π$ rad.
(iii) $β = \frac{600 \times 1 0 ^{- 9} \times 1.5}{0.9 \times 1 0 ^{- 3}} = 1.0$ mm.
(iv) $β^{'} = \frac{1.0}{1.25} = 0.8$ mm.

Answer: (i) bright; (ii) dark, 7π rad; (iii) 1.0 mm; (iv) 0.8 mm.

Where students lose marks in Young's Double Slit Experiment and Fringe Width

The slips examiners see most often, and what to write instead.

Mistake: Subtracting instead of adding across the centre
The ISC 2025 examiners (Q16) saw candidates subtract the distances of the 5th and 3rd bright fringes on opposite sides.
Wrong working
$y_{5} - y_{3} = 2 β$ .
Right working
$y_{5} + y_{3} = \frac{8 λ D}{d} = \frac{8 \times 5 \times 1 0 ^{- 7} \times 1.5}{4 \times 1 0 ^{- 3}} = 1.5$ mm.
Do this: Draw the centre and mark both fringes before calculating.
Usually costs: 1 to 2 marks.
Mistake: Mixed units
Also 2025: nm, mm and m mixed in one formula.
Do this: Convert all lengths to metres first.
Usually costs: 1 mark.
Mistake: Skipping the path-difference step
The 2024 examiners (Q13(ii)) noted unlabelled diagrams and a wrong or missing $Δ x = \frac{y d}{D}$ step.
Do this: Show $Δ x = d sin θ \approx \frac{y d}{D}$ and the small-angle approximation.
Usually costs: 1 mark.
Mistake: Not answering the "what changes" part
In 2025 Q16(ii) many ignored the effect of changing λ.
Do this: β ∝ λ: shorter λ gives narrower fringes; the pattern shrinks.
Usually costs: 1 mark.

Exam tips

Label $S_{1}$ , $S_{2}$ , d, D, y, θ and O in the diagram.
Write the bright and dark conditions before solving for positions.
Give β in mm with three significant figures.
State that bright fringes have equal intensity when drawing the intensity graph.

Practice questions on Young's Double Slit Experiment and Fringe Width

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Sustained interference requires sources that:
(a) have equal intensity only
(b) are coherent
(c) are white light sources
(d) are far apart
Hint
Constant phase difference.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Find the fringe width for λ = 550 nm, d = 1.1 mm and D = 2 m.
Hint
$\frac{λ D}{d}$ .
Answer
1.0 mm.
Q3 · 1 mark · Short answerHarder board level
Find the angular fringe width for light of 600 nm and slits 0.3 mm apart.
Hint
$\frac{λ}{d}$ .
Answer
$2 \times 1 0^{- 3}$ rad.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): With white light in YDSE, the central fringe is white and the others are coloured.
Reason (R): At the centre the path difference is zero for every wavelength, but elsewhere the fringe positions depend on λ.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
n = 0 for all colours.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Ten fringes occupy 7.2 mm on a screen 1.2 m from a double slit lit by 600 nm light. Find the slit separation.
Hint
β = 0.72 mm.
Answer
1.0 mm.
Q6 · 2 marks · Short answerHarder board level
State four conditions for a clear, sustained interference pattern.
Hint
Phase, wavelength, amplitude, geometry.
Answer
Coherent sources; same wavelength (monochromatic); equal or nearly equal amplitudes; narrow, closely spaced slits with a distant screen.
Q7 · 3 marks · Short answerHOTS / competency
When the screen is moved 0.3 m closer to the slits (d = 1 mm), the fringe width falls from 0.90 mm to 0.75 mm. Find the original screen distance and the wavelength.
Hint
β ∝ D.
Answer
$\frac{D}{D - 0.3} = 1.2$ , D = 1.8 m; λ = 500 nm.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose sodium light (589 nm) falls on slits 0.4 mm apart, with the screen 1.0 m away.
(i) Find the fringe width. (1 mark)
(ii) Find the distance of the second bright fringe from the centre. (1 mark)
(iii) What happens to the fringe width if d is doubled? (1 mark)
(iv) What is seen if one slit is covered? (1 mark)
Hint
β = λD/d.
Answer
(i) ≈ 1.47 mm. (ii) ≈ 2.95 mm. (iii) It halves (≈ 0.74 mm). (iv) The interference fringes disappear; only the single-slit diffraction pattern of the open slit remains.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Fringe width formula?Show answer
$β = \frac{λ D}{d}$ .
2.Condition for a dark fringe?Show answer
Path difference $(n + \frac{1}{2}) λ$ .
3.λ = 600 nm, D = 1 m, d = 1 mm: β?Show answer
0.6 mm.
4.Coherent sources?Show answer
Sources of the same frequency with a constant phase difference.
5.Effect of water (μ = 4/3) on β?Show answer
β becomes $\frac{3}{4}$ of its value in air.

Young's Double Slit Experiment and Fringe Width: frequently asked questions

Why can't two bulbs produce interference fringes?

Their light is emitted in random, independent bursts, so the phase difference between them changes millions of times a second; any pattern washes out.

Is energy lost at the dark fringes?

No. Energy is redistributed: what is missing at the dark fringes appears at the bright ones.

Why must the slits be narrow and close?

Narrow slits spread the light so the two beams overlap; small d makes β large enough to see.

What does the ISC 2027 syllabus need here?

Coherent sources and conditions for sustained interference, path and phase difference, the YDSE set-up and diagram, Δx = d sin θ, y_n = \frac{nD\lambda}{d}, \beta = \frac{D\lambda}{d} and the intensity graph. Deduction from wave equations is not required.

How was YDSE asked recently?

ISC 2024 Q13(ii) (3 marks) asked the derivation of β; ISC 2025 Q16 (3 marks) asked the distance between the 5th and 3rd bright fringes on opposite sides and the effect of changing λ.