PhysicsCBSE Class 12 · Wave Optics

Single Slit Diffraction Class 12: Pattern and Width of Central Maximum

Updated 15 October 2026 · Fact-checked

When light passes through a single narrow slit of width a, it spreads out and forms a broad bright central maximum with weaker bright and dark bands on either side. Minima occur where a sin θ = nλ, so the central maximum has angular width 2λ/a. A narrower slit or longer wavelength gives a wider pattern. CBSE treats this qualitatively.

Understand Single Slit Diffraction

Sound bends round a doorway so you hear someone in the next room, but light through the same doorway casts a fairly sharp shadow. Light does bend round edges, but only noticeably when the opening is tiny, comparable to its wavelength. This bending and spreading is diffraction.

The pattern. Send monochromatic light through a single narrow slit of width a onto a distant screen. You see a broad, very bright central maximum straight ahead, flanked by dark bands (minima) and much fainter secondary maxima, getting dimmer as you go outwards. The 2026-27 syllabus treats this pattern and the width of the central maximum qualitatively.

Why there is a central maximum. Straight ahead (θ = 0), light from every part of the slit travels the same distance to the screen, so all the secondary wavelets (Huygens) arrive in phase and add up to the brightest point.

Why minima appear. At an angle θ, the path difference between waves from the two edges of the slit is $a sin θ$ . When $a sin θ = λ$ , divide the slit into two halves: each point in the upper half has a partner in the lower half exactly $\frac{λ}{2}$ farther away, so the pairs cancel and the screen is dark. Similarly, dark bands occur whenever $a sin θ = nλ$ (n = ±1, ±2, …), i.e. $θ \approx \frac{nλ}{a}$ for small angles.

Secondary maxima. Roughly midway between minima (near $a sin θ = \frac{3 λ}{2}, \frac{5 λ}{2}, \dots$ ) there are weaker bright bands. Divide the slit into three parts: two cancel and only one-third contributes, so the first secondary maximum is much fainter than the central one, and later ones fainter still.

Width of the central maximum. It extends between the first minima on either side, from $θ = - \frac{λ}{a}$ to $+ \frac{λ}{a}$ : angular width $\frac{2 λ}{a}$ , and linear width $\frac{2 λ D}{a}$ on a screen at distance D. It is twice as wide as each secondary maximum. Consequences:

Narrower slit (smaller a) → wider central maximum, more spreading.
Longer wavelength (red) → wider pattern than shorter (blue).
In a liquid of index n, λ becomes λ/n and the pattern shrinks.
With a wide slit ( $a ≫ λ$ ) the spreading is negligible: light travels in straight lines.

Interference versus diffraction. Both come from superposition of waves. Interference (YDSE) is the superposition of waves from two coherent sources; its bright fringes are (ideally) of equal width and equal brightness. Diffraction is the superposition of wavelets from different points of a single wavefront (one slit); its central maximum is twice as wide as the others and the intensity falls off rapidly. In a real double slit experiment both happen together: the interference fringes are seen inside the single-slit diffraction envelope.

Single Slit Diffraction: formulas and key facts

Minima: $a sin θ = nλ (n = \pm 1, \pm 2, \dots)$; a = slit width.
Secondary maxima (approximately): $a sin θ \approx (n + \frac{1}{2}) λ$; Much weaker than the central maximum.
Angular width of central maximum: $2 θ \approx \frac{2 λ}{a}$; Twice the width of a secondary maximum.
Linear width of central maximum: $w = \frac{2 λ D}{a}$; D = distance to the screen.
Interference fringes inside the central maximum: $N \approx \frac{2 λ D / a}{λ D / d} = \frac{2 d}{a}$; For a double slit with separation d and slit width a.

How to solve Single Slit Diffraction questions

Use this method for single-slit questions, most of which are qualitative.

1Identify a (slit width), λ and D, all in metres.
2For the angle of the nth minimum, use $sin θ = \frac{nλ}{a}$ (small angles: θ ≈ nλ/a).
3For the central maximum, angular width = 2λ/a and linear width = 2λD/a.
4For changes, write width ∝ λD/a and scale (halve a, double the width; liquid, divide λ by n).
5For comparisons with YDSE, contrast number of sources, fringe widths and intensities.
6In explanations, use the 'divide the slit into halves (or thirds)' argument.

Single Slit Diffraction 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.

Width of the central maximum

Suppose you shine a red laser (λ = 650 nm) through a narrow 0.1 mm gap between two razor blades onto a wall 2 m away.

(a) Find the width of the central bright band on the wall. (b) What happens to it if the gap is narrowed to 0.05 mm? (c) Why does a wide window not produce a visible pattern?

Try it first, then show the solutionSolution

$w = \frac{2 λ D}{a} = \frac{2 \times 650 \times 1 0 ^{- 9} \times 2}{1 0 ^{- 4}} = 0.026$ m = 2.6 cm.
Halving a doubles the width: about 5.2 cm (and the band is dimmer).
For a window, a is enormous compared with λ, so 2λ/a is far too small to notice; light travels essentially straight.

Answer: (a) about 2.6 cm; (b) about 5.2 cm; (c) the spreading 2λ/a is negligible for a wide opening.

Memory hook: Squeeze the slit, spread the light.

Diffraction depends on wavelength

Imagine standing outside an open doorway 0.85 m wide while someone inside talks at about 500 Hz. Take the speed of sound as 340 m/s.

(a) Find the wavelength of the sound. (b) Estimate the angle of the first diffraction minimum for the sound and for light of wavelength 500 nm. (c) Explain why you can hear but not see round the edge of the door.

Try it first, then show the solutionSolution

$λ = \frac{340}{500} = 0.68$ m.
Sound: $sin θ = \frac{0.68}{0.85} = 0.8$ , θ ≈ 53°. Light: $sin θ = \frac{5 \times 1 0 ^{- 7}}{0.85} \approx 5.9 \times 1 0^{- 7}$ , a tiny angle.
Sound's wavelength is comparable to the doorway, so it spreads widely; light's wavelength is a million times smaller, so it hardly spreads.

Answer: (a) 0.68 m; (b) about 53° for sound, about 6 × 10⁻⁷ rad for light; (c) sound diffracts strongly through the doorway, light does not.

Memory hook: Diffraction is big only when the gap is about the size of the wave.

Tricks to remember Single Slit Diffraction

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

a sin θ = nλ is DARK
In single slit, nλ gives minima (the opposite of YDSE, where nλ gives maxima).
Use it when: Avoiding the classic swap.
Central is double
Central maximum: 2λ/a wide; every secondary maximum: λ/a.
Use it when: Width comparisons.
Halves cancel
At a sin θ = λ, split the slit into two halves whose wavelets cancel in pairs.
Use it when: Explaining the first minimum.
Small gap, big spread
Width ∝ λ/a.
Use it when: Effect of slit width and wavelength.

Single Slit Diffraction: solved examples

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

Example 1: Position of the first minimum and central width · 2 marks · Very short answerBoard level

Light of wavelength 600 nm falls on a slit 0.2 mm wide, and the pattern is seen on a screen 2 m away. Find (a) the angle of the first minimum, (b) its distance from the centre, and (c) the width of the central maximum.

Try it first, then show the solutionSolution

(a) $θ \approx \frac{λ}{a} = \frac{6 \times 1 0 ^{- 7}}{2 \times 1 0 ^{- 4}} = 3 \times 1 0^{- 3}$ rad; (b) y = Dθ = 6 mm. (1)
(c) Width = 2 × 6 mm = 12 mm. (1)

Answer: (a) 3 × 10⁻³ rad; (b) 6 mm; (c) 12 mm.

Example 2: Explaining the single-slit pattern · 3 marks · Short answerBoard level

Explain, using Huygens' wavelets, why a single slit gives (a) a bright central maximum, (b) a minimum at a sin θ = λ, and (c) a much weaker first secondary maximum. Sketch the intensity pattern.

Try it first, then show the solutionSolution

(a) At θ = 0 all wavelets from the slit travel equal paths and arrive in phase: maximum brightness. (1)
(b) At a sin θ = λ, divide the slit into two halves; each wavelet from the upper half pairs with one from the lower half with path difference λ/2, so all pairs cancel: dark. (1)
(c) Near a sin θ = 3λ/2, divide the slit into three parts; two cancel and only one-third contributes, so the maximum is much fainter. Sketch: tall central peak of width 2λ/a, smaller peaks on each side falling off quickly. (1)

Answer: All wavelets in phase at the centre; pairwise cancellation at a sin θ = λ; only a third of the slit contributes to the first secondary maximum.

Example 3: Changing the slit width · 3 marks · Short answerHarder board level

In a single-slit experiment, the slit width is halved. What happens to (a) the angular width of the central maximum, (b) its brightness, and (c) the positions of the minima?

Try it first, then show the solutionSolution

(a) Angular width 2λ/a doubles. (1)
(b) Less light passes through the slit and it is spread over a wider region: the central maximum becomes dimmer. (1)
(c) The minima move outwards: θ_n ≈ nλ/a doubles for each n. (1)

Answer: (a) doubles; (b) dimmer; (c) all minima move twice as far out.

Example 4: Interference versus diffraction · 3 marks · Short answerHarder board level

Give three differences between the interference pattern in Young's double slit experiment and the diffraction pattern of a single slit.

Try it first, then show the solutionSolution

Interference: superposition of waves from two coherent sources. Diffraction: superposition of wavelets from different parts of the same wavefront. (1)
Interference fringes are (ideally) of equal width; in diffraction the central maximum is twice as wide as the others. (1)
Interference bright fringes are of (nearly) equal intensity; in diffraction the intensity of the maxima falls off rapidly away from the centre. (1)

Answer: Two sources versus one wavefront; equal versus unequal widths; equal versus falling intensities.

Example 5: Double slit inside a single-slit envelope · 5 marks · Long answerHOTS / competency

In a double slit experiment, each slit is 0.2 mm wide and the slits are 1 mm apart (centre to centre).
(a) Write the width of the central diffraction maximum and the interference fringe width on a screen at distance D. (2)
(b) How many interference fringes lie within the central diffraction maximum? (2)
(c) How would you change the slit width to fit 20 fringes inside it? (1)

Try it first, then show the solutionSolution

(a) Central diffraction width = 2λD/a; interference fringe width β = λD/d. (2)
(b) $N = \frac{2 λ D / a}{λ D / d} = \frac{2 d}{a}$ . (1)
$N = \frac{2 \times 1}{0.2} = 10$ fringes. (1)
(c) N = 20 needs a = 2d/20 = 0.1 mm: halve the slit width. (1)

Answer: (a) 2λD/a and λD/d; (b) about 10; (c) make each slit 0.1 mm wide.

Example 6: Hearing round corners · 4 marks · Case-basedHarder board level

Read the passage and answer.

You can hear a radio playing in the next room through an open door, but you cannot see the radio. Take the sound frequency as 1000 Hz, the speed of sound as 340 m/s and the door width as 0.85 m.
(a) Find the wavelength of the sound. (1)
(b) Find the angle of the first diffraction minimum for this sound. (1)
(c) Why is there no comparable spreading of light from the radio? (1)
(d) What size of opening would make visible light (500 nm) spread through about 30° to its first minimum? (1)

Try it first, then show the solutionSolution

(a) λ = 340/1000 = 0.34 m. (1)
(b) sin θ = 0.34/0.85 = 0.4: θ ≈ 23.6°. (1)
(c) Light's wavelength (about 500 nm) is about a million times smaller than the door, so λ/a is tiny and the light hardly spreads. (1)
(d) a = λ/sin 30° = 1 μm. (1)

Answer: (a) 0.34 m; (b) about 23.6°; (c) λ ≪ a for light; (d) about 1 μm.

Where students lose marks in Single Slit Diffraction

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

Mistake: Writing a sin θ = nλ as the condition for maxima.
Carrying over the YDSE condition.
Do this: In single-slit diffraction, a sin θ = nλ gives minima.
Usually costs: 1 mark.
Mistake: Giving the central maximum width as λD/a.
Forgetting it spans both sides.
Do this: Central width = 2λD/a (between the first minima on either side).
Usually costs: 1 mark.
Mistake: Saying diffraction and interference are unrelated.
Missing that both are superposition.
Do this: Both are superposition; they differ in the number and nature of the sources.
Usually costs: ½ to 1 mark.
Mistake: Long mathematical derivations of the intensity pattern.
Over-preparing.
Do this: The syllabus treats single-slit diffraction qualitatively; give the halves/thirds argument and the width 2λ/a.
Usually costs: Time.

Exam tips

Single-slit diffraction was set in 2024 (55/5/1 Q33(b)), 2025 (55/1/1 Q18(a)) and 2026 (55/1/1 Q18(a) and 55/5/1 Q33(a)).
Draw the intensity curve with the central peak twice as wide as the side peaks and much taller.
For 'explain' questions, use the halves argument for minima and the thirds argument for secondary maxima.
A three-row table is the clearest way to answer 'differentiate interference and diffraction'.
Resolving power and the validity of ray optics are not in the 2026-27 syllabus.

Practice questions on Single Slit Diffraction

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
If the slit width in a single-slit experiment is doubled, the angular width of the central maximum
(a) doubles (b) halves (c) stays the same (d) becomes four times
Hint
Width ∝ 1/a.
Answer
(b) halves.
Q2 · 1 mark · Very short answerBoard level
Why is diffraction of light not commonly noticed in everyday life?
Hint
Compare λ with everyday openings.
Answer
Light's wavelength (about 400 to 700 nm) is very small compared with everyday openings and obstacles, so the spreading λ/a is too small to notice.
Q3 · 1 mark · Very short answerHarder board level
The first minimum of light of wavelength 600 nm occurs at 30° for a single slit. Find the slit width.
Hint
a sin θ = λ.
Answer
a = 600 nm/0.5 = 1.2 μm.
Q4 · 1 mark · Very short answerHarder board level
Find the angular width of the central maximum for light of wavelength 450 nm passing through a slit 0.15 mm wide.
Hint
2λ/a.
Answer
2 × 4.5 × 10⁻⁷/(1.5 × 10⁻⁴) = 6 × 10⁻³ rad.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The central maximum in single-slit diffraction is twice as wide as the secondary maxima.
Reason (R): The central maximum lies between the first minima at θ = ±λ/a, while successive secondary minima are λ/a apart.
(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
Count the λ/a steps.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
A single-slit experiment is repeated with the whole set-up under water (n = 4/3). How does the width of the central maximum change, and why?
Hint
λ in water.
Answer
The wavelength in water is λ/n, so the width 2λD/a falls to 3/4 of its value in air.
Q7 · 2 marks · Very short answerHOTS / competency
Why are the secondary maxima of a single slit much weaker than the central maximum?
Hint
Divide the slit into parts.
Answer
At the first secondary maximum (a sin θ ≈ 3λ/2) the slit can be divided into three equal parts; the wavelets from two parts cancel and only one-third of the slit contributes. At higher orders an even smaller fraction contributes, so the maxima get weaker.

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.Condition for minima in single-slit diffraction?Show answer
a sin θ = nλ.
2.Angular width of the central maximum?Show answer
2λ/a.
3.Linear width of the central maximum?Show answer
2λD/a.
4.Effect of reducing slit width?Show answer
The central maximum widens.
5.How wide is the central maximum compared with secondary maxima?Show answer
Twice as wide.
6.Why is the centre bright?Show answer
All wavelets reach it in phase.

Single Slit Diffraction: frequently asked questions

What is the difference between interference and diffraction?

Interference is the superposition of waves from two coherent sources; diffraction is the superposition of wavelets from different parts of the same wavefront. Interference fringes are of equal width; in diffraction the central maximum is twice as wide and intensities fall off quickly.

Why is a sin θ = nλ the condition for darkness and not brightness?

At that angle the slit can be divided into 2n equal parts whose wavelets cancel in pairs, so the screen is dark. In YDSE the same kind of condition gives brightness because there are only two sources.

How much of single-slit diffraction is in the 2026-27 syllabus?

Diffraction due to a single slit and the width of the central maximum are to be treated qualitatively. Know the pattern, the minima condition, the width 2λ/a and the effect of changing a and λ.

Why does sound diffract more than light?

Diffraction is significant when the opening is comparable to the wavelength. Sound wavelengths are around a metre, similar to doors; light wavelengths are less than a micrometre.

What happens if white light is used?

The central maximum is white, and the side bands are coloured, with red spread out more than violet because the pattern width is proportional to wavelength.