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. 266Fringe Width YDSE
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsPYQ coverage: NEET 2020, 2022Official key: NTA-verifiedLast updated: 25 Sep 2026
Fringe Width YDSE, explained for NEET
The trap in this topic is one letter deep: writing β = λd/D instead of β = λD/d. It survives because both symbols are distances and both appear in the same fraction, so the wrong version looks dimensionally plausible. It is not. Check it against the apparatus: move the screen further back and the pattern visibly spreads out. If your formula makes β shrink when D grows, your formula is upside down.
The fringe width is the distance between two adjacent bright fringes — equivalently, between two adjacent dark fringes. NCERT Class 12 Physics Part 2, Chapter 10, page 262 gives it as β = λD/d, where λ is the wavelength, D the slit-to-screen distance, and d the separation between the two slits. Read the ratio physically: D is large (order of a metre), d is small (order of a fraction of a millimetre), and their ratio is what magnifies a wavelength of 10⁻⁷ m into a fringe you can see.
Three consequences follow directly, and NEET tests them as ratio questions rather than as full substitutions. Doubling D doubles β. Doubling d halves β. Switching from red to blue light shortens λ, so the fringes crowd together. Nothing else in the setup matters to the spacing — slit width and source brightness change how the pattern looks, not how far apart the fringes sit.
The unit trap sits alongside the inversion trap. Stems give d in millimetres and λ in nanometres or ångströms while asking for β in millimetres. Convert everything to metres before dividing, then convert the answer once at the end. Mixing a millimetre d with a metre D silently scales the result by a thousand — and the factor-of-1000 value is almost always one of the four options.
Watch-out: when a question changes two quantities at once, work in ratios. β₂/β₁ = (λ₂/λ₁)(D₂/D₁)(d₁/d₂) — note d is inverted in that ratio, which is exactly where the inversion trap resurfaces.
Can you answer these Fringe Width 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, the fringe width β is given by
Show answer and why every option is right or wrong
Answer: D. D is correct. NCERT Class 12 Physics Part 2, Chapter 10, page 262 states the fringe spacing as β = λD/d, with D the slit-to-screen distance in the numerator and d the slit separation in the denominator.
Why A is wrong: A is wrong because it inverts D and d. This is the standard D-vs-d inversion: it predicts that moving the screen away would narrow the fringes, which contradicts the observed pattern.
Why B is wrong: B is wrong because it places both distances in the denominator. Increasing the screen distance would then reduce β, again inverting the observed dependence on D.
Why C is wrong: C is wrong because it puts the wavelength in the denominator, predicting that longer-wavelength red light gives narrower fringes than blue. The opposite is observed.
MCQ 2Concept TrapPractice
In a double-slit experiment, only the distance between the screen and the slits is doubled. The fringe width becomes
Show answer and why every option is right or wrong
Answer: C. C is correct. From β = λD/d (NCERT Class 12 Physics Part 2, Chapter 10, page 262), β is directly proportional to D with λ and d fixed, so doubling D doubles β.
Why A is wrong: A is wrong because β depends on D to the first power, not the square. There is no squared term anywhere in β = λD/d.
Why B is wrong: B is wrong because it applies β ∝ 1/D, the inverted form β = λd/D. Under the correct formula D sits in the numerator.
Why D is wrong: D is wrong because β is not independent of D; the screen distance is precisely the geometric factor that magnifies the path-difference pattern.
MCQ 3Direct ApplicationPractice
In a Young's double-slit setup, light of wavelength 6.00 × 10⁻⁷ m illuminates two slits separated by 3.00 × 10⁻⁴ m. The screen is 1.50 m from the slits. The fringe width is
Show answer and why every option is right or wrong
Answer: C. C is correct. β = λD/d = (6.00 × 10⁻⁷ × 1.50)/(3.00 × 10⁻⁴) = 9.00 × 10⁻⁷/3.00 × 10⁻⁴ = 3.00 × 10⁻³ m, using the formula on NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why A is wrong: A is wrong because it enters the slit separation as 0.300 m instead of 3.00 × 10⁻⁴ m (0.300 mm), so the result is 1000 times too small.
Why B is wrong: B is wrong because it omits the screen distance, reporting λ/d = 2.00 × 10⁻³ — a pure number, the angular fringe width — as if it were a length; β = λD/d.
Why D is wrong: D is wrong because it uses 2λD/d, twice the fringe width — the width of the central maximum in single-slit diffraction, a different formula.
MCQ 4Concept TrapPractice
The separation between the two slits in a double-slit experiment is increased while the wavelength and the screen distance are held fixed. The fringes on the screen become
Show answer and why every option is right or wrong
Answer: B. B is correct. In β = λD/d (NCERT Class 12 Physics Part 2, Chapter 10, page 262), d is in the denominator, so increasing d decreases β and the fringes crowd together.
Why A is wrong: A is wrong because it treats d as if it were in the numerator. That is the D-vs-d inversion applied to the slit separation.
Why C is wrong: C is wrong because β depends explicitly on d; changing d with λ and D fixed must change the spacing.
Why D is wrong: D is wrong because under the small-angle conditions of the standard treatment the fringes are uniformly spaced across the pattern; the spacing does not vary with position.
MCQ 5CalculationPractice
A double-slit pattern is formed with light of wavelength 4.80 × 10⁻⁷ m. The wavelength is then changed to 7.20 × 10⁻⁷ m and, at the same time, the slit separation is doubled. Everything else is unchanged. The new fringe width, as a multiple of the original, is
Show answer and why every option is right or wrong
Answer: D. D is correct. β ∝ λ/d, so β₂/β₁ = (λ₂/λ₁) × (d₁/d₂) = (7.20/4.80) × (1/2) = 1.50 × 0.500 = 0.750, following from β = λD/d on NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why A is wrong: A is wrong because it multiplies by d₂/d₁ = 2 instead of dividing, giving 1.50 × 2. That is the D-vs-d inversion appearing in the ratio form, where d must be inverted.
Why B is wrong: B is wrong because it applies only the wavelength change and ignores the doubling of the slit separation.
Why C is wrong: C is wrong because it applies only the slit-separation change and ignores the increase in wavelength.
MCQ 6Direct ApplicationPractice
In a double-slit experiment, the slits are 1.00 mm apart and the screen is 2.00 m away. The observed fringe width is 1.10 mm. The wavelength of the light used is
Show answer and why every option is right or wrong
Answer: A. A is correct. Rearranging β = λD/d gives λ = βd/D = (1.10 × 10⁻³ × 1.00 × 10⁻³)/2.00 = 5.50 × 10⁻⁷ m, using the relation on NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why B is wrong: B is wrong because it multiplies by the screen distance instead of dividing: βdD = 1.10 × 10⁻³ × 1.00 × 10⁻³ × 2.00 = 2.20 × 10⁻⁶ m.
Why C is wrong: C is wrong by a factor of 1000, the signature of leaving one of the millimetre quantities unconverted before dividing.
Why D is wrong: D is wrong because it leaves out the screen distance, reporting βd = 1.10 × 10⁻⁶ m; the rearranged relation is λ = βd/D.
MCQ 7Easy RecallPractice
In a double-slit interference pattern, the fringe width is defined as the distance between
Show answer and why every option is right or wrong
Answer: A. A is correct. NCERT Class 12 Physics Part 2, Chapter 10, page 262 defines the fringe width as the separation between two consecutive bright fringes, which equals the separation between two consecutive dark fringes.
Why B is wrong: B is wrong because the slit separation d is an apparatus dimension, not the fringe spacing; it appears in the denominator of β = λD/d.
Why C is wrong: C is wrong because that distance is half the fringe width; a dark fringe lies midway between adjacent bright fringes.
Why D is wrong: D is wrong because the first and third bright fringes are separated by two fringe widths, not one.
MCQ 8CalculationPractice
A double-slit arrangement produces fringes of width 6.0 × 10⁻⁴ m. The screen is then moved to three times its original distance and the slit separation is reduced to one half of its original value, with the same light. The new fringe width is
Show answer and why every option is right or wrong
Answer: B. B is correct. β ∝ D/d, so β₂ = β₁ × (D₂/D₁) × (d₁/d₂) = 6.0 × 10⁻⁴ × 3 × 2 = 3.6 × 10⁻³ m, from β = λD/d on NCERT Class 12 Physics Part 2, Chapter 10, page 262.
Why A is wrong: A is wrong because it divides by 3 and multiplies by ½, inverting both factors; this is the D-vs-d inversion applied to both changes at once.
Why C is wrong: C is wrong because 9.0 × 10⁻⁴ m is 6.0 × 10⁻⁴ × 3 × ½: it scales by ½ for the halved slit separation. Since d is in the denominator, halving d doubles β.
Why D is wrong: D is wrong because 3.0 × 10⁻³ m is 6.0 × 10⁻⁴ × (3 + 2): it adds the two scale factors instead of multiplying them. β ∝ D/d, so β = 6.0 × 10⁻⁴ × 3 × 2 = 3.6 × 10⁻³ m.
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Fringe Width YDSE: quick recall before you leave
How do you solve a Fringe Width YDSE question? A worked example
- 1
Given
• Wavelength of light, λ = 5.40 × 10⁻⁷ m• Slit separation, d = 4.50 × 10⁻⁴ m• Screen distance, D = 1.20 m• Number of fringes spanned in the measurement, N = 5 (exact, a count)
- 2
Required
The fringe width β, and the total width of the region spanned by 5 fringe spacings.
- 3
Concept
Two coherent slits produce a pattern of evenly spaced bright and dark fringes. The spacing between adjacent bright fringes is fixed by the wavelength and by the geometry ratio D/d. Since D ≫ d here (1.20 m against 4.50 × 10⁻⁴ m), the small-angle treatment of NCERT Class 12 Physics Part 2, Chapter 10, page 262 applies.
- 4
Formula
β = λD/d
- 5
Substitution
β = (5.40 × 10⁻⁷ m × 1.20 m) / (4.50 × 10⁻⁴ m)
All three quantities are already in metres, so no conversion is needed before dividing. This is the step to check before touching the calculator. - 6
Calculation
Numerator: 5.40 × 10⁻⁷ × 1.20 = 6.48 × 10⁻⁷
Divide: (6.48 × 10⁻⁷) / (4.50 × 10⁻⁴) = 1.44 × 10⁻³
So β = 1.44 × 10⁻³ m.
Width of 5 spacings = 5 × 1.44 × 10⁻³ = 7.20 × 10⁻³ m.
The factor 5 is a counting number and the exponents are exact, so neither contributes to the significant-figure count. The three data values each carry three significant figures, so the answers are reported to three. - 7
Final answer
β = 1.44 × 10⁻³ m (1.44 mm). Five fringe spacings span 7.20 × 10⁻³ m (7.20 mm).
- 8
Common trap
Writing β = λd/D. With these numbers that gives (5.40 × 10⁻⁷ × 4.50 × 10⁻⁴)/1.20 ≈ 2.03 × 10⁻¹⁰ m — a "fringe width" smaller than an atom. The physical check catches it instantly: a fringe pattern is something you look at, so β should come out on the millimetre scale. Any answer in micrometres or smaller means D and d have been swapped. The second trap here is unit mixing: had d been quoted as 0.450 mm and used as 0.450, the answer would have come out 1000 times too small, and that value is a standard distractor.
- 9
Similar NEET-style question
In a double-slit experiment, light of wavelength 6.30 × 10⁻⁷ m gives a fringe width of 1.40 × 10⁻³ m on a screen 1.00 m from the slits. The slits are then moved to a separation of 0.900 mm with everything else unchanged. Find the new fringe width. (Work in ratios: find d first from β = λD/d, then scale by the inverse ratio of slit separations.)
What to remember before solving Fringe Width YDSE questions
Which Fringe Width 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
Where do students lose marks on Fringe Width YDSE?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student writes β = λd/D instead of λD/d. Consequences: doubling D should INCREASE fringe width, not decrease.
When it triggers
YDSE fringe width question with changes to slit separation or screen distance.
How to avoid
β = λD/d. D is screen distance (large); d is slit separation (small). β scales with D/d. Mnemonic: 'D for distance increases width'.
Root cause: formula misuse
Correction
β = λD/d. D = screen distance (large), d = slit separation (small). β ∝ D/d. Larger D → wider fringes.
More in Optics: 7 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Fringe Width YDSE questions from past NEET papers
2 questions from NEET 2020, 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
112
26
38
49
NTA Answer: Option 1(final)
NEET 2020
1one-fourth
2double
3half
4four times
NTA Answer: Option 4(final)
How does NEET ask about Fringe Width 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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