β = λD/d, where d is slit separation, D is screen distance, λ is wavelength. Bright fringes: x = nλD/d (n integer). Dark fringes: x = (n+½)λD/d.
-- NCERT Class 12 Physics, Ch. 10, p. 266Interference YDSE
8 MCQs4 revision cards9-step worked example
Source: NCERT OpticsPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 24 Sep 2026
Interference YDSE, explained for NEET
The costly habit in YDSE is treating path difference and phase difference as interchangeable labels for the same quantity. They are not. Path difference is a length; phase difference is an angle. A point on the screen with path difference Δx has phase difference φ = (2π/λ)Δx. Bright fringes need Δx = nλ, equivalently φ = 2nπ. Dark fringes need Δx = (n + ½)λ, equivalently φ = (2n + 1)π. Writing "path difference = π" or "phase difference = λ/2" is the error that turns a solvable question into a wrong option.
The physics (NCERT Class 12 Physics Part 2, Chapter 10, page 262): two slits illuminated by one monochromatic source emit waves that arrive at a screen point P having travelled unequal distances. Geometry with D ≫ d gives Δx = yd/D, where y is measured from the central point. Setting Δx = nλ gives the bright-fringe positions y_n = nλD/d. Consecutive bright fringes are therefore separated by β = λD/d — the fringe-width result developed in its own lesson.
Intensity is the part that repeats in NEET. Superposing two waves of amplitude a each gives resultant intensity I = 4I₀cos²(φ/2), where I₀ is the intensity from one slit alone. At a maximum this is 4I₀, not 2I₀: amplitudes add first, then square. At a minimum it is exactly zero for equal amplitudes. Average intensity across the pattern is 2I₀, so energy is redistributed, not created or destroyed.
Watch out for two more things. The central fringe at y = 0 has Δx = 0 for every wavelength, so it is white in white light while the outer fringes spread into colours. And an unequal-amplitude pair gives minima that are dim rather than dark — I_min = (a₁ − a₂)², which is zero only when a₁ = a₂.
Can you answer these Interference YDSE MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Young's double-slit experiment with monochromatic light, a point on the screen shows a bright fringe when the path difference between the two arriving waves equals
Show answer and why every option is right or wrong
Answer: D. Constructive superposition requires the two waves to arrive in phase, which happens when the extra distance travelled is a whole number of wavelengths, Δx = nλ. NCERT Class 12 Physics Part 2, Chapter 10, page 262 states this bright-fringe condition.
Why A is wrong: A is wrong because an odd multiple of λ/2 is the condition for a dark fringe, not a bright one — the two waves then arrive exactly out of phase.
Why B is wrong: B is wrong because λ/4 path difference corresponds to a quarter-cycle phase lag; it is neither a maximum nor a minimum condition.
Why C is wrong: C is wrong because integral multiples of λ/2 include the odd multiples, which are minima. Only the even multiples — that is, whole multiples of λ — give maxima.
MCQ 2Direct ApplicationPractice
At a point on the screen in a double-slit experiment, the light from the two slits arrives with a path difference of 3λ/2. The phase difference at that point is
Show answer and why every option is right or wrong
Answer: C. Phase difference and path difference are related by φ = (2π/λ)Δx. Substituting Δx = 3λ/2 gives φ = (2π/λ)(3λ/2) = 3π rad, an odd multiple of π — a dark fringe. The relation is set out in NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why A is wrong: A is wrong because it inverts the ratio, computing 2π ÷ 3 instead of 2π × (3/2). Inverting the conversion factor is the arithmetic form of confusing the two quantities.
Why B is wrong: B is wrong because it drops the factor 2 in the conversion, effectively using φ = (π/λ)Δx. That halves every phase difference and would place minima at even multiples of π.
Why D is wrong: D is wrong because it reports the path difference itself and attaches radians to it. A path difference is a length in metres; a phase difference is an angle. They are different physical quantities and cannot share a value.
MCQ 3Concept TrapPractice
In a double-slit experiment each slit alone produces intensity I₀ at the centre of the screen. With both slits open, the intensity at the central maximum is
Show answer and why every option is right or wrong
Answer: B. Amplitudes, not intensities, add for coherent waves in phase: the resultant amplitude is 2a, so intensity is proportional to (2a)² = 4a², giving 4I₀. NCERT Class 12 Physics Part 2, Chapter 10, page 262 derives I = 4I₀cos²(φ/2), which equals 4I₀ when φ = 0.
Why A is wrong: A is wrong because it ignores the second slit's contribution entirely; with both slits open and waves in phase the resultant amplitude is doubled.
Why C is wrong: C is wrong because √2 I₀ would arise from adding amplitudes in quadrature, which describes waves a quarter-cycle apart, not waves in phase.
Why D is wrong: D is wrong because it adds the two intensities directly. Intensity addition applies to incoherent sources; for coherent waves in phase the amplitudes add first and the square is taken afterwards, giving 4I₀. 2I₀ is the average intensity across the whole pattern, not the peak.
MCQ 4Easy RecallPractice
In a double-slit experiment illuminated with white light, the fringe formed at the point equidistant from the two slits is
Show answer and why every option is right or wrong
Answer: B. At the equidistant point the path difference is zero for every wavelength simultaneously, so all colours interfere constructively there and the central fringe is white. NCERT Class 12 Physics Part 2, Chapter 10, page 262 notes this wavelength-independent central maximum.
Why A is wrong: A is wrong because violet would dominate only if the position depended on wavelength. The zero-path-difference condition is satisfied by every wavelength at the same point, so no single colour is selected.
Why C is wrong: C is wrong for the same reason as B — no colour is preferentially reinforced at zero path difference. Red appears at the outer edges of the coloured fringes, where the larger wavelength pushes maxima further out.
Why D is wrong: D is wrong because zero path difference means the waves arrive exactly in phase, which is maximum brightness, not darkness.
MCQ 5CalculationPractice
In a Young's double-slit setup the central maximum has intensity I_max. At a point where the path difference is λ/3, the intensity is
Show answer and why every option is right or wrong
Answer: C. First convert: φ = (2π/λ)(λ/3) = 2π/3 rad. Then I = I_max cos²(φ/2) = I_max cos²(π/3) = I_max × (1/2)² = I_max/4. NCERT Class 12 Physics Part 2, Chapter 10, page 262 gives the intensity distribution I = 4I₀cos²(φ/2).
Why A is wrong: A is wrong because it uses cos(φ/2) without squaring: cos(π/3) = 1/2 gives I_max/2. The intensity depends on the square of the cosine, since intensity goes as amplitude squared.
Why B is wrong: B is wrong because it assumes intensity falls in direct proportion to the fraction of a wavelength in the path difference. The dependence is cos²(φ/2), not linear.
Why D is wrong: D is wrong because it treats the path difference λ/3 as leaving two-thirds of the intensity. No such proportionality exists; the cos² law must be applied through the phase difference.
MCQ 6Concept TrapPractice
A double-slit interference pattern is produced. Compared with the sum of the energies the two slits would deliver acting separately, the total light energy arriving on the whole screen
Show answer and why every option is right or wrong
Answer: A. The average of 4I₀cos²(φ/2) over the pattern is 2I₀, exactly the sum of the two slits acting alone. Interference redistributes energy across the screen without creating or destroying it — the conservation statement in NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why B is wrong: B is wrong for the mirror reason — it counts the missing energy at the minima without counting the surplus at the maxima.
Why C is wrong: C is wrong because it generalises the peak value to the whole screen. The maxima do reach 4I₀, but that excess is exactly balanced by the zeros at the minima; the spatial average is 2I₀.
Why D is wrong: D is wrong because light carries energy whether it is described as a wave or otherwise, and conservation of energy is not suspended by interference.
MCQ 7CalculationPractice
In a Young's double-slit experiment the two slits are given unequal widths, so the amplitudes reaching the screen are in the ratio 3 : 1. The ratio of the maximum intensity to the minimum intensity in the pattern is
Show answer and why every option is right or wrong
Answer: D. With amplitudes a₁ = 3a and a₂ = a, I_max ∝ (a₁ + a₂)² = (4a)² = 16a² and I_min ∝ (a₁ − a₂)² = (2a)² = 4a², so the ratio is 16 : 4 = 4 : 1. The amplitude-superposition basis for this appears in NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why A is wrong: A is wrong because it quotes the amplitude ratio itself. The intensity ratio requires forming the sum and difference of amplitudes and squaring each.
Why B is wrong: B is wrong because it squares the amplitude ratio directly, giving the ratio of the two slits' individual intensities — not the maximum-to-minimum ratio of the pattern.
Why C is wrong: C is wrong because it does not follow from any step of the superposition; no combination of 3a and a yields 2 : 1 for max to min.
MCQ 8Direct ApplicationPractice
A thin transparent sheet is placed over one slit of a double-slit apparatus, introducing an additional path difference of exactly one wavelength for light through that slit. The pattern on the screen
Show answer and why every option is right or wrong
Answer: A. An added path difference of λ corresponds to a phase difference of 2π, which leaves the interference condition at every point identical to before — a bright fringe remains where a bright fringe was. The φ = (2π/λ)Δx relation in NCERT Class 12 Physics Part 2, Chapter 10, page 262 makes a full-wavelength shift equivalent to no shift.
Why B is wrong: B is wrong because a fixed, constant extra path difference preserves the constant phase relationship. Coherence is lost only if the added phase varies randomly with time.
Why C is wrong: C is wrong because a half-wavelength addition, not a full wavelength, would put a minimum at the centre. One full wavelength returns the phase to where it started.
Why D is wrong: D is wrong because fringe width is set by λ, D and d; adding a fixed path difference in one arm shifts the pattern at most, and here not even that.
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Interference YDSE: quick recall before you leave
How do you solve a Interference YDSE question? A worked example
- 1
Given
Two slits, separation d = 5.00 × 10⁻⁴ m, screen distance D = 1.50 m, monochromatic light of wavelength λ = 6.00 × 10⁻⁷ m. Intensity from either slit alone at the screen, I₀ = 2.00 W m⁻². Point P lies 4.50 × 10⁻⁴ m from the centre of the pattern.
- 2
Required
The phase difference between the two waves arriving at P, and the resultant intensity there.
- 3
Concept
For D ≫ d the path difference at a screen point a distance y from the centre is Δx = yd/D. The phase difference follows from φ = (2π/λ)Δx, and the resultant of two equal-amplitude coherent waves is I = 4I₀cos²(φ/2). The two quantities Δx and φ are distinct: one is computed from geometry, the other from Δx.
- 4
Formula
Δx = yd/D; φ = (2π/λ)Δx; I = 4I₀cos²(φ/2). The fringe-width relation β = λD/d comes from the same geometry (
formula: ydse fringe, NCERT Class 12 Physics Part 2, Chapter 10, page 262). - 5
Substitution
Δx = (4.50 × 10⁻⁴ m)(5.00 × 10⁻⁴ m) / (1.50 m)
φ = (2π / 6.00 × 10⁻⁷ m) × Δx
I = 4 × (2.00 W m⁻²) × cos²(φ/2) - 6
Calculation
Δx = (4.50 × 10⁻⁴ × 5.00 × 10⁻⁴) / 1.50 = (2.25 × 10⁻⁷) / 1.50 = 1.50 × 10⁻⁷ m.
As a fraction of a wavelength, Δx / λ = (1.50 × 10⁻⁷) / (6.00 × 10⁻⁷) = 0.250, i.e. λ/4.
φ = 2π × 0.250 = π/2 rad = 1.57 rad (3 s.f.).
cos²(φ/2) = cos²(π/4) = (1/√2)² = 0.500.
I = 4 × 2.00 × 0.500 = 4.00 W m⁻².
The 4 in I = 4I₀cos²(φ/2) is a counting factor from squaring the doubled amplitude, and the 2 in 2π/λ is part of a mathematical constant; π and √2 are mathematical constants. None of these contributes to the significant-figure count. The three significant figures come from the measured quantities d, D, λ and I₀. - 7
Final answer
φ = 1.57 rad (= π/2 rad); I = 4.00 W m⁻², which is half the central-maximum intensity of 8.00 W m⁻².
- 8
Common trap
Reporting the answer as "phase difference = λ/4" or as "path difference = π/2". The quarter-wavelength figure 0.250 is a path result; π/2 is the phase result; the conversion φ = (2π/λ)Δx sits between them. A second version of the same slip is applying cos² to the path difference directly, writing cos²(Δx/2) — the cosine takes an angle, never a length.
- 9
Similar NEET-style question
In a double-slit arrangement the two slits are 4.00 × 10⁻⁴ m apart and the screen is 1.20 m away, illuminated by light of wavelength 5.00 × 10⁻⁷ m. Each slit alone would produce intensity 3.00 W m⁻² on the screen. At a point 5.00 × 10⁻⁴ m from the centre of the pattern, find the phase difference between the arriving waves and the resultant intensity. *(Answer: Δx = 1.667 × 10⁻⁷ m = λ/3, φ = 2π/3 rad, I = 12.0 × cos²(π/3) = 3.00 W m⁻².)*
What to remember before solving Interference YDSE questions
Which Interference YDSE formulas do you need for NEET?
1 formula — click to collapse
YDSE fringe width
Fringe width in Young's double-slit experiment. Wavelength lambda, slit separation d, screen distance D.
| Symbol | Quantity | SI Unit |
|---|---|---|
| beta | fringe width | m |
| lambda | wavelength | m |
| D | screen distance | m |
| d | slit separation | m |
Valid when
- D >> d
- Coherent monochromatic source
- Small angles
More in Optics: 9 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Interference YDSE questions from past NEET papers
2 questions from NEET 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
If the monochromatic source in Young’s double slit experiment is replaced by white light, then
1Interference pattern will disappear.
2There will be a central dark fringe surrounded by a few coloured fringes.
3There will be a central bright white fringe surrounded by a few coloured fringes.
4All bright fringes will be of equal width.
NTA Answer: Option 3(revised_final)
NEET 2023
1Both Statement I and Statement II are false.
2Statement I is true but Statement II is false.
3Statement I is false but Statement II is true.
4Both Statement I and Statement II are true.
NTA Answer: Option 2(final)
How does NEET ask about Interference YDSE?
1 recurring pattern from past papers — click to collapse
YDSE fringe width β = λD/d. Compute new fringe given changes in λ, D, d.
Multi StepMedium
Common distractors
inverts D d relation
Uses d/D instead of D/d
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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