Key Fact
Single-slit diffraction
Central maximum width 2λD/a (a = slit width, D = distance). Subsequent minima at x = nλD/a (n ≠ 0). Pattern is significantly different from interference.
-- NCERT Class 12 Physics, Ch. 10, p. 267Diffraction Single Slit
8 MCQs9-step worked example
Source: NCERT OpticsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Diffraction Single Slit, explained for NEET
The trap here is importing the double-slit habit. In Young's experiment the condition d sin θ = nλ gives a maximum. At a single slit the same-looking condition a sin θ = nλ gives a minimum — a dark band. Students who memorised one equation and its meaning together arrive at the single slit, recognise the algebra, and mark the wrong kind of band.
The reason the sign flips is the mechanism. A single slit of width a is not two sources; it is a continuous strip of secondary sources across the aperture. Divide the slit into two halves. When the extra path from the top edge to the bottom edge equals exactly one wavelength, every point in the upper half has a partner in the lower half exactly half a wavelength behind it. The strip cancels itself pairwise, and the screen is dark. Generalising, path difference a sin θ = nλ for n = ±1, ±2, ±3 … marks the minima. n = 0 is excluded — that is the central maximum, where every secondary source arrives in phase.
That geometry gives the pattern its distinctive shape: one broad, bright central band flanked by markedly fainter secondary maxima. The central maximum spans from the first minimum on one side to the first on the other, so it is twice the width of any subsequent band. NCERT Class 12 Physics Part 2, Chapter 10, page 268, records that the intensity of the first secondary maximum falls to only a small fraction of the central peak.
For NEET this topic pays about 0.4 questions a year, usually recall or one-step application. The whole weight sits on the two facts above. Watch out for the width relationships: widening the slit a narrows the pattern, and reaching a ≈ λ spreads the light so far that the geometric-shadow picture collapses entirely.
Can you answer these Diffraction Single Slit 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 the diffraction pattern produced by a single slit, the condition a sin θ = nλ with n = 1, 2, 3, … locates the positions of the
Show answer and why every option is right or wrong
Answer: C. C is correct. For a single slit this condition marks the dark bands, because the aperture divides into pairs of secondary sources half a wavelength out of step, which cancel. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because the secondary maxima fall roughly midway between successive minima, not at the minima themselves.
Why B is wrong: B is wrong because the principal (central) maximum sits at θ = 0, which this condition explicitly excludes by requiring n ≥ 1.
Why D is wrong: D is wrong because the pattern's intensity falls steeply away from the centre; points satisfying this condition are dark, not merely equal.
MCQ 2Concept TrapPractice
A student states that a single-slit minimum at a sin θ = λ arises because the two edges of the slit are exactly one wavelength apart in path, so they interfere destructively. The statement's conclusion is right but its reasoning is flawed. The correct reasoning is that
Show answer and why every option is right or wrong
Answer: A. A is correct. A one-wavelength edge-to-edge difference puts the edges in phase; the darkness comes from pairing each point in the upper half of the aperture with the point exactly a/2 below it, which trails by λ/2. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why B is wrong: B is wrong because a full-wavelength path difference produces constructive, not destructive, interference — and in any case the interior of the aperture contributes.
Why C is wrong: C is wrong because a single slit is a continuous strip of secondary sources, not two discrete ones; treating it as a double slit imports the wrong condition entirely.
Why D is wrong: D is wrong because the minima appear at specific angles determined by λ and a, which absorption by the slit edges could not explain.
MCQ 3Direct ApplicationPractice
Monochromatic light of wavelength 6.00 × 10⁻⁷ m falls normally on a slit of width 2.40 × 10⁻⁴ m. The angular position of the first minimum is closest to
Show answer and why every option is right or wrong
Answer: B. B is correct. For the first minimum, sin θ = λ/a = (6.00 × 10⁻⁷)/(2.40 × 10⁻⁴) = 2.50 × 10⁻³, and at this size sin θ ≈ θ. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because it doubles the result — the value 2λ/a locates the second minimum, not the first.
Why C is wrong: C is wrong because it halves the ratio; λ/2a corresponds to no minimum of this pattern.
Why D is wrong: D is wrong because it inverts the ratio, a/λ = 400, which cannot be the angle of a minimum: sin θ can never exceed 1.
MCQ 4Easy RecallPractice
Compared with the width of any one of the secondary bright bands in a single-slit pattern, the width of the central maximum is
Show answer and why every option is right or wrong
Answer: D. D is correct. The central maximum runs from the first minimum on one side to the first on the other, spanning two minimum-spacings, while each secondary band spans one. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because it reverses the relationship; the central band is the widest feature of the pattern, not the narrowest.
Why B is wrong: B is wrong because equal widths would describe an interference fringe pattern, where fringes are evenly spaced — a single slit is not evenly spaced about the centre.
Why C is wrong: C is wrong because the factor comes from counting minimum-spacings on each side of the centre, which gives two, not three.
MCQ 5Direct ApplicationPractice
In a single-slit diffraction experiment the slit width is increased while the wavelength and the screen distance are held fixed. The central maximum
Show answer and why every option is right or wrong
Answer: A. A is correct. The half-angular width of the central maximum is set by sin θ = λ/a, so increasing a decreases θ and the pattern contracts. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why B is wrong: B is wrong because it inverts the dependence — widening the slit spreads the light less, approaching the geometric-shadow limit.
Why C is wrong: C is wrong because the angular width depends explicitly on the ratio λ/a; changing a must change it.
Why D is wrong: D is wrong because no change in slit width splits the central maximum; the pattern retains one central band throughout.
MCQ 6Concept TrapPractice
A slit is illuminated first by red light and then by blue light, with everything else unchanged. Relative to the red pattern, the blue pattern has
Show answer and why every option is right or wrong
Answer: C. C is correct. sin θ = λ/a increases with λ, and blue light has the shorter wavelength, so its pattern is more tightly compressed about the centre. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because although the stated dependence on wavelength is correct, blue light has the shorter wavelength of the two, so its angle is smaller, not larger.
Why B is wrong: B is wrong because the angle is set by the ratio λ/a — wavelength matters as much as slit width.
Why D is wrong: D is wrong because diffraction occurs for every wavelength; photon energy plays no part in this geometry.
MCQ 7CalculationPractice
Light of wavelength 5.0 × 10⁻⁷ m passes through a slit of width 1.0 × 10⁻⁴ m onto a screen 2.0 m away. The distance on the screen from the centre of the pattern to the second minimum is
Show answer and why every option is right or wrong
Answer: B. B is correct. For n = 2, sin θ = 2λ/a = 1.0 × 10⁻², and the screen distance is y = D sin θ = 2.0 × 1.0 × 10⁻² = 2.0 × 10⁻² m. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because it uses λ/a and omits the screen distance, conflating an angle with a length.
Why C is wrong: C is wrong because it uses n = 1, locating the first minimum rather than the second.
Why D is wrong: D is wrong because 2.5 × 10⁻³ m is one eighth of the correct distance and does not follow from x = nλD/a with any whole n. For the second minimum, x = 2 × 5.0 × 10⁻⁷ × 2.0 / 1.0 × 10⁻⁴ = 2.0 × 10⁻² m.
MCQ 8CalculationPractice
In a single-slit experiment the first minimum for light of wavelength 6.0 × 10⁻⁷ m is observed at a certain angle. To place the second minimum of a different monochromatic source at that same angle, that source must have wavelength
Show answer and why every option is right or wrong
Answer: D. D is correct. Equal angle with the same slit means equal n λ, so 1 × (6.0 × 10⁻⁷) = 2 × λ′, giving λ′ = 3.0 × 10⁻⁷ m. NCERT Class 12 Physics Part 2, Chapter 10, page 268.
Why A is wrong: A is wrong because it multiplies by the order instead of dividing — a longer wavelength would push the second minimum beyond the stated angle.
Why B is wrong: B is wrong because the same wavelength would put its second minimum at roughly twice the angle, not the same one.
Why C is wrong: C is wrong because it divides by 1.5 rather than by the order 2; no order-matching gives this value.
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How do you solve a Diffraction Single Slit question? A worked example
- 1
Given
Wavelength of light, λ = 6.00 × 10⁻⁷ m
Slit width, a = 1.20 × 10⁻⁴ m
Screen distance, D = 1.50 m - 2
Required
The total width of the central maximum on the screen.
- 3
Concept
A single slit acts as a continuous strip of secondary sources. Dark bands form where the strip cancels pairwise. The central maximum is bounded by the first minimum on each side, so its total width is twice the distance from the centre to the first minimum.
- 4
Formula
First minimum: a sin θ = λ, hence sin θ = λ/a.
On-screen distance for small angles: y = D sin θ.
Total central width: W = 2y. - 5
Substitution
sin θ = (6.00 × 10⁻⁷ m)/(1.20 × 10⁻⁴ m)
y = (1.50 m) × sin θ
W = 2 × y - 6
Calculation
sin θ = 5.00 × 10⁻³
y = 1.50 × 5.00 × 10⁻³ = 7.50 × 10⁻³ m
W = 2 × 7.50 × 10⁻³ = 1.50 × 10⁻² m
The factor 2 here is a counting number — it comes from there being one first minimum on each side of the centre — and the order n = 1 is likewise exact. Neither limits the significant figures; the three given measurements each carry three, so the answer carries three. - 7
Final answer
W = 1.50 × 10⁻² m (15.0 mm).
- 8
Common trap
Reporting 7.50 × 10⁻³ m. That is the half-width — the distance from the centre to one edge. The question asked for the full central maximum, which spans both sides. The half-width is always offered as a distractor, and it is the answer a student gets by computing correctly and then stopping one step early.
- 9
Similar NEET-style question
Monochromatic light of wavelength 5.00 × 10⁻⁷ m passes through a slit of width 2.50 × 10⁻⁴ m onto a screen 2.00 m away. Find the total width of the central maximum. *(Answer: 8.00 × 10⁻³ m.)*
What to remember before solving Diffraction Single Slit questions
More in Optics: 9 exam traps and mistakes · 11 formulas · 7 question patterns from its other lessons.
Diffraction Single Slit questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
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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