PhysicsCBSE Class 12 · Wave Optics

Young's Double Slit Experiment Class 12: Fringes, Fringe Width and Intensity

Updated 15 October 2026 · Fact-checked

In Young's experiment, light from one source passes through two narrow slits, which act as coherent sources. Their waves overlap and interfere: bright fringes form where the path difference is nλ and dark fringes where it is (n − ½)λ. The fringes are equally spaced, with fringe width β = λD/d (quoted without derivation for 2026-27).

Understand Young's Double Slit Experiment

Thomas Young showed that light passing through two close slits produces a pattern of bright and dark bands on a screen. Light added to light can give darkness: the clearest proof that light is a wave.

Superposition and coherence. When two waves meet, the resultant displacement is the sum of the individual displacements. If they arrive in phase, they add (constructive interference, brightness); if they arrive in antiphase, they cancel (destructive interference, darkness). For a steady pattern the two sources must be coherent: their phase difference must stay constant in time. Two independent lamps are not coherent, because each emits in short, randomly phased bursts, so their pattern changes far too fast to see. Young obtained coherent sources by letting light from a single source fall on two slits $S_{1}$ and $S_{2}$ ; whatever phase changes happen in the source happen equally at both slits.

Conditions for sustained interference. The sources should be coherent; they should emit the same wavelength; their amplitudes should be (nearly) equal for good contrast; the slits should be narrow and close together; and the screen should be far away compared with the slit separation.

Bright and dark fringes. Let the slits be d apart and the screen a distance D away ( $D ≫ d$ ). For a point P at height y from the centre, the path difference is $Δ = S_{2} P - S_{1} P \approx d sin θ \approx \frac{y d}{D}$ .

Bright fringe (maximum): $Δ = nλ$ (n = 0, 1, 2, …), so $y_{n} = \frac{nλ D}{d}$ .
Dark fringe (minimum): $Δ = (2 n - 1) \frac{λ}{2}$ (n = 1, 2, …), so $y_{n} = \frac{( 2 n - 1 ) λ D}{2 d}$ .

The centre (Δ = 0) is always bright.

Fringe width. The distance between two consecutive bright (or dark) fringes is $β = \frac{λ D}{d}$ . For 2026-27 this result is quoted without derivation. Consequences:

β ∝ λ: red light gives wider fringes than blue.
β ∝ D and β ∝ 1/d.
In a liquid of refractive index n, λ becomes λ/n, so β becomes β/n.
With white light, the central fringe is white; on each side a few coloured fringes appear (violet edge nearer the centre), then the pattern washes out.
The angular width $\frac{β}{D} = \frac{λ}{d}$ does not depend on D.

Intensity. For two equal waves of intensity $I_{0}$ with phase difference φ, the resultant intensity is $I = 4 I_{0} cos^{2} \frac{ϕ}{2}$ , with $ϕ = \frac{2 π}{λ} Δ$ : from $4 I_{0}$ at bright fringes to 0 at dark fringes. Energy is not lost, only redistributed: the average is $2 I_{0}$ . For unequal intensities $I_{1}$ and $I_{2}$ , $I_{ma x} = (I_{1} + I_{2})^{2}$ and $I_{min} = (I_{1} - I_{2})^{2}$ , so the dark fringes are not completely dark. Slit widths affect intensities ( $I \propto$ width), and amplitude $\propto I$ .

Young's Double Slit Experiment: formulas and key facts

Path difference: $Δ = d sin θ \approx \frac{y d}{D}$; For D ≫ d.
Bright fringes: $Δ = nλ, y_{n} = \frac{nλ D}{d}$; n = 0, 1, 2, …
Dark fringes: $Δ = (2 n - 1) \frac{λ}{2}, y_{n} = \frac{( 2 n - 1 ) λ D}{2 d}$; n = 1, 2, …
Fringe width (no derivation needed): $β = \frac{λ D}{d}$; In a medium of index n: β/n.
Phase and path difference: $ϕ = \frac{2 π}{λ} Δ$; Δ = λ corresponds to φ = 2π.
Resultant intensity: $I = 4 I_{0} cos^{2} \frac{ϕ}{2}, I_{ma x, min} = (I_{1} \pm I_{2})^{2}$; Equal slits: I_min = 0.

How to solve Young's Double Slit Experiment questions

Use this method for YDSE numericals and reasoning questions.

1List λ (in m), d and D (in m). Convert nm and mm carefully.
2Fringe width: $β = \frac{λ D}{d}$ ; nth bright fringe at $n β$ , nth dark fringe at $(n - \frac{1}{2}) β$ .
3For a given point, find the path difference and divide by λ: a whole number means bright, a half-odd number means dark.
4For two wavelengths, find the smallest whole numbers with $n_{1} λ_{1} = n_{2} λ_{2}$ ; that gives the first coincidence.
5For intensity, find φ from Δ and use $4 I_{0} cos^{2} \frac{ϕ}{2}$ ; for unequal slits use $(I_{1} \pm I_{2})^{2}$ .
6For changes (liquid, D, d, λ), write β ∝ λD/d and scale.
7Give answers in mm or μm as sensible.

Young's Double Slit Experiment 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 shine a red laser pointer (λ = 650 nm) through a pair of slits 0.25 mm apart, made by scratching a blackened glass slide, onto a wall 3 m away.

(a) Find the fringe width. (b) Over what length of wall would 10 fringes be spread? (c) How would the pattern change with a green laser of shorter wavelength?

Try it first, then show the solutionSolution

$β = \frac{λ D}{d} = \frac{650 \times 1 0 ^{- 9} \times 3}{0.25 \times 1 0 ^{- 3}} = 7.8 \times 1 0^{- 3}$ m = 7.8 mm.
10 fringes: 7.8 cm.
β ∝ λ, so the green fringes would be closer together.

Answer: (a) 7.8 mm; (b) about 7.8 cm; (c) narrower fringes.

Memory hook: Long λ, long screen distance, close slits: wide fringes.

Coherence

Imagine switching on two identical torch bulbs side by side and pointing them at the same patch of wall, hoping to see interference fringes.

(a) Will you see a steady fringe pattern? (b) Why not? (c) How did Young get round this problem?

Try it first, then show the solutionSolution

No steady pattern: the wall is just evenly lit (intensities add).
Each bulb emits light in short bursts with random phases, so the phase difference between the two bulbs changes randomly and very rapidly; any pattern shifts too fast to be seen.
Young used a single source to illuminate two slits; the slits then act as coherent sources with a fixed phase difference.

Answer: (a) No; (b) the bulbs are not coherent; (c) by deriving both sources from one source.

Memory hook: Two slits, one source: coherence for free.

Tricks to remember Young's Double Slit Experiment

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

Beta = lambda D over d
'Big D on top, little d below': β = λD/d.
Use it when: Fringe width.
Whole for white, half for dark
Path difference a whole number of λ: bright; a half-odd number of λ: dark.
Use it when: Identifying fringes.
Four times, cos squared
Equal slits: I = 4I₀cos²(φ/2), peak four times one slit.
Use it when: Intensity questions.
Liquid squeezes fringes
In a medium of index n, λ and β both shrink by n.
Use it when: Immersion questions.

Young's Double Slit Experiment: solved examples

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

Example 1: Fringe width and fringe positions · 2 marks · Very short answerBoard level

In a Young's double slit experiment, λ = 600 nm, d = 0.5 mm and D = 1 m. Find (a) the fringe width and (b) the distance of the third bright fringe from the centre.

Try it first, then show the solutionSolution

(a) $β = \frac{600 \times 1 0 ^{- 9} \times 1}{0.5 \times 1 0 ^{- 3}} = 1.2 \times 1 0^{- 3}$ m = 1.2 mm. (1)
(b) $y_{3} = 3 β = 3.6$ mm. (1)

Answer: (a) 1.2 mm; (b) 3.6 mm.

Example 2: Coherent sources and conditions · 3 marks · Short answerBoard level

(a) What are coherent sources? Why can two independent sources not produce a sustained interference pattern? (b) How did Young obtain coherent sources? (c) Write the conditions for bright and dark fringes in terms of path difference.

Try it first, then show the solutionSolution

(a) Sources that emit waves of the same frequency with a constant phase difference. Independent sources emit in random bursts, so their phase difference changes rapidly and the pattern does not stay still. (1)
(b) A single source illuminates two narrow slits; any phase change in the source reaches both slits equally, so they stay coherent. (1)
(c) Bright: Δ = nλ; dark: Δ = (2n − 1)λ/2. (1)

Answer: Coherent sources keep a constant phase difference; Young split one source with two slits; bright at Δ = nλ, dark at Δ = (2n − 1)λ/2.

Example 3: Two wavelengths together · 3 marks · Short answerHarder board level

Light containing wavelengths 650 nm and 520 nm is used in a double slit experiment with d = 1 mm and D = 1 m. Find the least distance from the central maximum where bright fringes of the two wavelengths coincide.

Try it first, then show the solutionSolution

Coincidence needs $n_{1} λ_{1} = n_{2} λ_{2}$ : $650 n_{1} = 520 n_{2}$ , so $\frac{n _{1}}{n _{2}} = \frac{4}{5}$ . (1)
The smallest whole numbers: n₁ = 4, n₂ = 5. (1)
$y = \frac{4 \times 650 \times 1 0 ^{- 9} \times 1}{1 0 ^{- 3}} = 2.6 \times 1 0^{- 3}$ m = 2.6 mm. (1)

Answer: 2.6 mm.

Example 4: Intensity in the pattern · 3 marks · Short answerHarder board level

(a) The widths of two slits are in the ratio 1 : 4. Find the ratio of maximum to minimum intensity in the pattern. (b) For two equal slits, each giving intensity I₀ on the screen, find the intensity at a point where the path difference is λ/3.

Try it first, then show the solutionSolution

(a) Intensity ∝ width, so I₁ : I₂ = 1 : 4 and amplitudes are 1 : 2. (1)
$\frac{I _{ma x}}{I _{min}} = \frac{( 1 + 2 ) ^{2}}{( 2 - 1 ) ^{2}} = 9$ . (1)
(b) $ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$ ; $I = 4 I_{0} cos^{2} \frac{π}{3} = 4 I_{0} \times \frac{1}{4} = I_{0}$ . (1)

Answer: (a) 9 : 1; (b) I₀.

Example 5: Changing the set-up · 5 marks · Long answerHOTS / competency

In a YDSE the fringe width is 1.2 mm in air.
(a) Find the fringe width when the whole apparatus is immersed in water (n = 4/3). (2)
(b) In water, the slit separation is also halved. Find the new fringe width. (1)
(c) What would you see if one slit were covered? (1)
(d) What would you see with white light in air? (1)

Try it first, then show the solutionSolution

(a) In water λ' = λ/n, so β' = β/n. (1)
β' = 1.2 × 3/4 = 0.9 mm. (1)
(b) β ∝ 1/d: 0.9 × 2 = 1.8 mm. (1)
(c) No interference; the single slit gives a diffraction pattern: a broad central bright band with weaker bands on either side. (1)
(d) A white central fringe (all colours bright there), then a few coloured fringes with violet nearer the centre and red farther out, after which the colours overlap and the screen looks uniformly lit. (1)

Answer: (a) 0.9 mm; (b) 1.8 mm; (c) a single-slit diffraction pattern; (d) white centre with a few coloured fringes.

Example 6: Bright or dark? · 4 marks · Case-basedHarder board level

Read the passage and answer.

In a YDSE with light of wavelength 600 nm, two points P and Q on the screen have path differences 1.8 μm and 1.5 μm from the two slits.
(a) Is P on a bright or dark fringe? Which order? (1)
(b) Is Q bright or dark? Which order? (1)
(c) What is the phase difference at Q? (1)
(d) If the intensity at the central maximum is I_m, what is the intensity at P and at Q (equal slits)? (1)

Try it first, then show the solutionSolution

(a) $\frac{1.8 \times 1 0 ^{- 6}}{6 \times 1 0 ^{- 7}} = 3$ : bright, third order. (1)
(b) $\frac{1.5 \times 1 0 ^{- 6}}{6 \times 1 0 ^{- 7}} = 2.5$ : dark, third dark fringe (n = 3 in (2n − 1)λ/2). (1)
(c) $ϕ = 2 π \times 2.5 = 5 π$ (equivalent to π). (1)
(d) P: I_m (a maximum); Q: zero. (1)

Answer: (a) third bright; (b) third dark; (c) 5π; (d) I_m at P, zero at Q.

Where students lose marks in Young's Double Slit Experiment

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

Mistake: Using mm for λ or d without converting.
Mixed units give answers off by powers of ten.
Do this: Convert λ to m (600 nm = 6 × 10⁻⁷ m) and d to m before substituting.
Usually costs: The final mark.
Mistake: Writing the dark condition as nλ/2.
Missing 'odd'.
Do this: Dark: (2n − 1)λ/2, an odd multiple of λ/2.
Usually costs: 1 mark.
Mistake: Deriving the fringe width at length.
Older syllabi required it.
Do this: For 2026-27, quote β = λD/d; use the time for the numerical.
Usually costs: Time.
Mistake: Adding intensities instead of amplitudes.
Thinking I_max = I₁ + I₂.
Do this: I_max = (√I₁ + √I₂)², I_min = (√I₁ − √I₂)².
Usually costs: 1 to 2 marks.
Mistake: Saying energy is destroyed at dark fringes.
Misreading destructive interference.
Do this: Energy is redistributed: dark fringes have less, bright fringes more; the average is unchanged.
Usually costs: 1 mark in reasoning.

Exam tips

Young's experiment is the most heavily tested topic in Wave Optics in our 2023-26 bank: 17 board-paper questions (for example 2023 55/1/1 Q33(a), 2024 55/5/1 Q33(a) and 2025 55/1/1 Q33(a)).
State 'β = λD/d (no derivation required)' and spend your effort on applications.
In reasoning questions about changes (D, d, λ, liquid), write β ∝ λD/d first.
Draw the set-up: source S, slits S₁ and S₂, screen, d, D, point P and the path difference.
Keep the intensity formulas handy; I_max/I_min items are short and quick to score.

Practice questions on Young's Double Slit Experiment

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
In a YDSE 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
Hint
β ∝ D/d.
Answer
(c) four times.
Q2 · 1 mark · Very short answerBoard level
Write the condition (path difference) for the nth dark fringe and its distance from the centre.
Hint
Half-odd multiple of λ.
Answer
Δ = (2n − 1)λ/2; y = (2n − 1)λD/(2d).
Q3 · 1 mark · Very short answerHarder board level
A fringe width of 0.9 mm is observed with D = 1.5 m and d = 1 mm. Find the wavelength.
Hint
λ = βd/D.
Answer
λ = 0.9 × 10⁻³ × 10⁻³/1.5 = 6 × 10⁻⁷ m = 600 nm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Two independent sodium lamps cannot produce a sustained interference pattern.
Reason (R): Light from independent sources is not coherent.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Constant phase difference?
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
In a fixed region of the screen, 10 bright fringes are seen with light of wavelength λ. How many are seen in the same region with light of wavelength 1.2λ?
Hint
Number ∝ 1/β ∝ 1/λ.
Answer
10/1.2 ≈ 8.3, so about 8 complete fringes.
Q6 · 3 marks · Short answerHOTS / competency
The ratio of maximum to minimum intensity in an interference pattern is 25 : 9. Find the ratio of the intensities of the two interfering waves.
Hint
(a₁ + a₂)/(a₁ − a₂) = 5/3.
Answer
(a₁ + a₂)/(a₁ − a₂) = 5/3 gives a₁ = 4a₂. Intensity ratio = (a₁/a₂)² = 16 : 1.
Q7 · 2 marks · Very short answerHOTS / competency
With white light, why is the central fringe white, and which colour forms the edge of the first coloured fringe nearest to the centre?
Hint
Path difference zero; β ∝ λ.
Answer
At the centre the path difference is zero for every wavelength, so all colours are bright together: white. The fringe width is least for violet, so the first fringe's inner edge is violet (red appears farther out).

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 in YDSE?Show answer
β = λD/d.
2.Condition for a bright fringe?Show answer
Path difference = nλ.
3.What are coherent sources?Show answer
Sources with the same frequency and a constant phase difference.
4.Fringe width in a liquid of refractive index n?Show answer
β/n.
5.Intensity at a point with phase difference φ (equal slits)?Show answer
4I₀cos²(φ/2).
6.Colour of the central fringe with white light?Show answer
White.

Young's Double Slit Experiment: frequently asked questions

Is the derivation of fringe width in the 2026-27 syllabus?

No. CBSE asks for the expression for fringe width without derivation (final expression only). You should know how to use β = λD/d and the conditions for bright and dark fringes.

Why are the fringes equally spaced?

For small angles the path difference grows in proportion to the distance y from the centre (Δ ≈ yd/D), so successive bright fringes occur at equal steps of λD/d.

What happens if the slits are made wider?

More light passes, so the pattern is brighter, but the fringes become less sharp, and very wide slits wash out the pattern because each slit no longer acts as a narrow source.

Does interference violate conservation of energy?

No. Energy missing from the dark fringes appears in the bright fringes; the average intensity over the pattern equals the sum of the two individual intensities.

Why must the screen be far from the slits?

For D much larger than d, the path difference is approximately yd/D and the fringes are equally spaced and wide enough to see.