PhysicsISC Class 12 · Wave Optics

Single Slit Fraunhofer Diffraction: Minima, Secondary Maxima and Width of the Central Maximum

Updated 15 October 2026 · Fact-checked

When light of wavelength λ passes through a slit of width a, minima occur where $a\sin\theta = n\lambda$ (n = 1, 2, 3, ...) and weak secondary maxima near $a\sin\theta = \left(n + \frac{1}{2}\right)\lambda$. The bright central maximum has angular width $\frac{2\lambda}{a}$ and linear width $\frac{2\lambda D}{a}$ on a screen at distance D.

Understand Single Slit Diffraction and the Central Maximum

What diffraction is. Diffraction is the bending and spreading of waves round the edges of an obstacle or aperture into the region of geometrical shadow. It is noticeable only when the size of the opening is comparable to the wavelength. Light's wavelength is under a micrometre, so everyday openings show little diffraction, while sound (wavelength around a metre) bends easily round doors.

Fraunhofer set-up. Monochromatic light from a source at the focus of a convex lens forms a plane wavefront falling normally on a narrow slit of width a. A second convex lens (or a distant screen) collects light leaving the slit at each angle θ to a point on the screen. Source and screen are effectively at infinity.

First minimum: the half-slit argument (qualitative). Divide the slit into two equal halves. At angle θ the path difference between waves from the two edges is $a sin θ$ . If $a sin θ = λ$ , each wavelet from the upper half has a partner in the lower half exactly $\frac{a}{2}$ further down, with path difference $\frac{λ}{2}$ . Every pair cancels, so the intensity is zero: the first minimum, $sin θ_{1} = \frac{λ}{a}$ .

Dividing into 4, 6, ... parts gives further minima: $a sin θ_{n} = nλ$ .
At $a sin θ = \frac{3 λ}{2}$ , divide into three parts: two cancel, one third survives: a secondary maximum. Generally $a sin θ \approx (n + \frac{1}{2}) λ$ .
At θ = 0 all wavelets arrive in phase: the central maximum.

Intensity pattern (asked in ISC 2024 Q6). A broad, very bright central maximum, flanked on each side by much weaker secondary maxima (each under 5% of the central peak, falling off further out), with dark minima between. Plot intensity (y-axis) against angle θ or $sin θ$ (x-axis), and mark minima at $\pm \frac{λ}{a}$ , $\pm \frac{2 λ}{a}$ .

Width of the central maximum. The central band extends from the first minimum on one side to that on the other:
$angular width = 2 θ_{1} \approx \frac{2 λ}{a}, linear width = \frac{2 λ D}{a} .$
It is twice the width of each secondary maximum. A narrower slit or a longer wavelength spreads the pattern; a very wide slit shrinks it to a sharp geometrical image (ray optics).

Interference versus diffraction.

Interference (YDSE)	Diffraction (single slit)
Superposition of waves from two coherent sources	Superposition of wavelets from different parts of one wavefront
Fringes of equal width	Central maximum twice as wide as the others
Bright fringes of equal intensity	Intensity falls rapidly away from the centre
Dark fringes perfectly dark	Minima dark, but pattern dominated by one peak

In a medium of index μ, λ becomes $\frac{λ}{μ}$ and the pattern narrows by the same factor.

Single Slit Diffraction and the Central Maximum: formulas and key facts

Minima: $a sin θ_{n} = nλ$; n = 1, 2, 3, ...
Secondary maxima: $a sin θ_{n} \approx (n + \frac{1}{2}) λ$; n = 1, 2, ...
First minimum: $θ_{1} \approx \frac{λ}{a}$; Small angles.
Angular width of central maximum: $2 θ_{1} = \frac{2 λ}{a}$; Radians.
Linear width of central maximum: $w = \frac{2 λ D}{a}$; Screen at D.
Position of nth minimum: $y_{n} = \frac{nλ D}{a}$; From the centre.

How to solve Single Slit Diffraction and the Central Maximum questions

Convert to metres, find $\frac{λ}{a}$ first; everything else is a multiple of it.

1Compute $\frac{λ}{a}$ : this is the angle (rad) of the first minimum.
2Minima at $n \frac{λ}{a}$ ; secondary maxima near $(n + \frac{1}{2}) \frac{λ}{a}$ .
3Multiply angles by D for positions on the screen.
4Central maximum: angular width $\frac{2 λ}{a}$ , linear width $\frac{2 λ D}{a}$ .
5To find λ or a from a measured width, rearrange: $λ = \frac{w a}{2 D}$ .
6For coincidence of features of two wavelengths, equate path differences, e.g. $2 λ_{1} = 1.5 λ_{2}$ .

Single Slit Diffraction and the Central Maximum 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.

A street lamp through a finger gap

Suppose you look at a distant street lamp through a narrow gap of about 0.1 mm between two fingers, with light of about 550 nm.

Find the angle from the centre to the first dark band and explain the streaks you see.

Try it first, then show the solutionSolution

$θ_{1} \approx \frac{λ}{a} = \frac{550 \times 1 0 ^{- 9}}{0.1 \times 1 0 ^{- 3}} = 5.5 \times 1 0^{- 3}$ rad ≈ 0.32°.
The gap is a single slit; the lamp is spread into a central band with fainter bands on each side, spread out at right angles to the gap (the bands themselves run parallel to it).

Answer: About 0.3°; the faint dark and bright lines are a single-slit diffraction pattern.

Memory hook: Narrow gap, wide spread.

AM radio behind a hill

Imagine a valley behind a hill a few hundred metres across. An AM station broadcasts at 1 MHz and an FM station at 100 MHz.

Which signal is more likely to reach the valley? Why?

Try it first, then show the solutionSolution

AM: $λ = \frac{3 \times 1 0 ^{8}}{1 0 ^{6}} = 300$ m, comparable to the hill, so it diffracts strongly round it.
FM: $λ = 3$ m, much smaller than the hill, so it diffracts little and the valley lies in its shadow.

Answer: The AM signal: diffraction is strong when λ is comparable to the obstacle.

Memory hook: Long waves bend round big things.

Tricks to remember Single Slit Diffraction and the Central Maximum

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

Minima use n, maxima use n + ½
Single slit is the reverse of YDSE: $a sin θ = nλ$ is dark.
Use it when: Avoiding the classic swap.
Central is double
Central maximum = $\frac{2 λ D}{a}$ , twice any other band.
Use it when: Width questions.
Halve the slit, double the spread
Width ∝ $\frac{1}{a}$ .
Use it when: Change questions.
One big tower, small hills
Intensity graph: tall central peak, small side peaks falling away.
Use it when: Drawing the graph.

Single Slit Diffraction and the Central Maximum: solved examples

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

Example 1: Narrowing the slit · 1 mark · MCQBoard level

In single-slit diffraction, if the slit width is halved, the width of the central maximum:
(a) halves
(b) doubles
(c) stays the same
(d) becomes four times

Try it first, then show the solutionSolution

$w = \frac{2 λ D}{a}$ , so w ∝ $\frac{1}{a}$ .

Answer: (b).

Example 2: First minimum and central width · 2 marks · Short answerBoard level

Light of wavelength 500 nm falls normally on a slit 0.25 mm wide; the screen is 1.5 m away. Find the distance of the first minimum from the centre and the width of the central maximum.

Try it first, then show the solutionSolution

$y_{1} = \frac{λ D}{a} = \frac{500 \times 1 0 ^{- 9} \times 1.5}{0.25 \times 1 0 ^{- 3}} = 3 \times 1 0^{- 3}$ m = 3 mm. (1)
Central width $= 2 y_{1} = 6$ mm. (1)

Answer: 3 mm; 6 mm.

Example 3: Explaining the first minimum · 3 marks · Long answerHarder board level

With a labelled diagram of Fraunhofer diffraction at a single slit, explain why the first minimum occurs at $a sin θ = λ$ , and why a secondary maximum occurs near $a sin θ = \frac{3 λ}{2}$ .

Try it first, then show the solutionSolution

Diagram: plane wavefront on slit of width a, rays leaving at angle θ, focused by a lens on the screen; path difference between edge rays $a sin θ$ . (1)
If $a sin θ = λ$ , split the slit into two halves: each point in the upper half has a partner $\frac{a}{2}$ below with path difference $\frac{λ}{2}$ ; all pairs cancel, giving darkness. (1)
If $a sin θ = \frac{3 λ}{2}$ , split into three parts; two cancel in pairs and the third survives, giving a weak maximum. (1)

Answer: Pairwise cancellation of half-slits gives the minimum; one uncancelled third gives a secondary maximum.

Example 4: Second minimum and first secondary maximum · 3 marks · Short answerHarder board level

Light of 450 nm passes through a slit 0.09 mm wide onto a screen 2 m away. Find the angular positions of the second minimum and the first secondary maximum, and their distances from the centre.

Try it first, then show the solutionSolution

$\frac{λ}{a} = \frac{450 \times 1 0 ^{- 9}}{0.09 \times 1 0 ^{- 3}} = 5 \times 1 0^{- 3}$ rad. (1)
Second minimum: $2 \times 5 \times 1 0^{- 3} = 0.01$ rad, at 0.01 × 2 = 20 mm. (1)
First secondary maximum: about $1.5 \times 5 \times 1 0^{- 3} = 7.5 \times 1 0^{- 3}$ rad, at about 15 mm. (1)

Answer: 0.01 rad, 20 mm; ≈ $7.5 \times 1 0^{- 3}$ rad, ≈ 15 mm.

Example 5: Finding the wavelength · 3 marks · Short answerHOTS / competency

A slit of width 0.12 mm gives a pattern on a screen 1.8 m away in which the first minima on either side of the centre are 15 mm apart. Find the wavelength. What would this distance become if the experiment were done in water (μ = 4/3)?

Try it first, then show the solutionSolution

Distance between first minima = central width $= \frac{2 λ D}{a} = 15$ mm, so $y_{1} = 7.5$ mm. (1)
$λ = \frac{a y _{1}}{D} = \frac{0.12 \times 1 0 ^{- 3} \times 7.5 \times 1 0 ^{- 3}}{1.8} = 5.0 \times 1 0^{- 7}$ m = 500 nm. (1)
In water λ becomes $\frac{3}{4}$ as large, so the width is $\frac{3}{4} \times 15 = 11.25$ mm. (1)

Answer: 500 nm; 11.25 mm.

Example 6: Green laser through a slit · 4 marks · Case-basedHOTS / competency

Suppose a 532 nm green laser shines normally on a slit.
(i) For a = 0.133 mm, find $sin θ$ for the first minimum. (1 mark)
(ii) Find the width of the central maximum on a screen 2.5 m away. (1 mark)
(iii) What slit width would make the central maximum span 60° in total? (1 mark)
(iv) What happens to the pattern as a is made very large? (1 mark)

Try it first, then show the solutionSolution

(i) $sin θ_{1} = \frac{532 \times 1 0 ^{- 9}}{0.133 \times 1 0 ^{- 3}} = 4 \times 1 0^{- 3}$ .
(ii) $w = 2 \times 4 \times 1 0^{- 3} \times 2.5 = 0.02$ m = 20 mm.
(iii) First minimum at 30°: $a = \frac{λ}{s i n 30°} = 2 λ \approx 1.06$ µm.
(iv) $\frac{λ}{a} \to 0$ : the pattern shrinks to a sharp image of the slit; ray optics applies.

Answer: (i) $4 \times 1 0^{- 3}$ ; (ii) 20 mm; (iii) ≈ 1.06 µm; (iv) shrinks to the geometrical image.

Where students lose marks in Single Slit Diffraction and the Central Maximum

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

Mistake: Wrong x-axis on the intensity graph
The ISC 2024 examiners (Q6) saw the x-axis marked incorrectly.
Do this: x-axis: angle θ (or sin θ) with minima at $\pm \frac{λ}{a}$ , $\pm \frac{2 λ}{a}$ ; y-axis: intensity.
Usually costs: 1 mark.
Mistake: Secondary maxima drawn as tall as the central one
Also 2024 Q6: no clear fall in intensity away from the centre.
Do this: Draw the central peak many times taller, with side peaks shrinking outwards.
Usually costs: 1 mark.
Mistake: Using the YDSE condition for minima
Students write $a sin θ = (n + \frac{1}{2}) λ$ for dark bands.
Do this: Single slit: $a sin θ = nλ$ gives minima.
Usually costs: 1 mark.
Mistake: Not linking diffraction to aperture size
In 2025 Q1(A)(vii) candidates could not connect the assertion with the reason that the obstacle must be comparable to λ.
Do this: Diffraction is noticeable when a ≈ λ; light's λ is tiny, so it needs very narrow slits.
Usually costs: 1 mark.

Exam tips

Mark at least two minima and two maxima on each side in the graph.
State "central maximum is twice as wide as the secondary maxima".
Keep angles in radians for small-angle work.
Remember the syllabus treats single-slit diffraction qualitatively: explain, do not integrate.

Practice questions on Single Slit Diffraction and the Central Maximum

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
For a single slit of width a, the first minimum occurs when:
(a) $a sin θ = \frac{λ}{2}$
(b) $a sin θ = λ$
(c) $a sin θ = \frac{3 λ}{2}$
(d) $a sin θ = 0$
Hint
Half-slit pairing.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Find the angular position of the first minimum for light of 700 nm and a slit 0.35 mm wide.
Hint
$\frac{λ}{a}$ .
Answer
$2 \times 1 0^{- 3}$ rad.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The secondary maxima in single-slit diffraction are much fainter than the central maximum.
Reason (R): At a secondary maximum, only a fraction of the slit's wavelets remain uncancelled.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Divide into three parts.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Give two differences between the interference pattern of YDSE and the single-slit diffraction pattern.
Hint
Widths, intensities.
Answer
YDSE fringes have equal width; in diffraction the central maximum is twice as wide. YDSE bright fringes have equal intensity; diffraction maxima fall rapidly in intensity away from the centre.
Q5 · 2 marks · Short answerHarder board level
The central maximum of a single-slit pattern is 4.8 mm wide on a screen 1.2 m away with 480 nm light. Find the slit width.
Hint
$a = \frac{2 λ D}{w}$ .
Answer
0.24 mm.
Q6 · 2 marks · Short answerHOTS / competency
For the same slit, the second minimum of 400 nm light falls at the same angle as the first secondary maximum of another wavelength. Find that wavelength.
Hint
$2 λ_{1} = 1.5 λ_{2}$ .
Answer
≈ 533 nm.
Q7 · 2 marks · Short answerHOTS / competency
Explain why AM radio (λ ≈ 300 m) can be heard behind a hill while FM (λ ≈ 3 m) often cannot.
Hint
Compare λ with obstacle size.
Answer
Diffraction is appreciable only when the wavelength is comparable to the obstacle. AM waves are as long as the hill and bend round it; FM waves are much shorter and leave a shadow behind it.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student uses red light of 640 nm on a slit 0.16 mm wide and must draw the intensity pattern.
(i) What goes on each axis? (1 mark)
(ii) At what angles are the first two minima on each side? (1 mark)
(iii) How do the heights of the maxima compare? (1 mark)
(iv) Where is the first minimum if blue light of 480 nm is used? (1 mark)
Hint
$\frac{λ}{a} = 4 \times 1 0^{- 3}$ rad.
Answer
(i) Intensity on the y-axis, angle θ (or sin θ) on the x-axis. (ii) $\pm 4 \times 1 0^{- 3}$ rad and $\pm 8 \times 1 0^{- 3}$ rad. (iii) The central peak is far taller; secondary peaks are small and fall off further out. (iv) $\pm 3 \times 1 0^{- 3}$ rad: the pattern narrows.

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?Show answer
$a sin θ = nλ$ .
2.Linear width of the central maximum?Show answer
$\frac{2 λ D}{a}$ .
3.λ = 500 nm, a = 0.5 mm: θ₁?Show answer
$1 0^{- 3}$ rad.
4.Effect of a longer wavelength?Show answer
Wider pattern.
5.How wide is the central maximum compared with others?Show answer
Twice as wide.

Single Slit Diffraction and the Central Maximum: frequently asked questions

Why is the central maximum the brightest?

At θ = 0 all wavelets from the slit arrive in phase and reinforce; at other maxima most wavelets cancel and only a fraction survive.

What makes it "Fraunhofer" diffraction?

Source and screen are effectively at infinity (plane wavefronts, using lenses), so the wavefront at the slit is plane.

Is diffraction just interference?

Both are superposition. Interference usually means waves from separate coherent sources; diffraction means wavelets from different parts of the same wavefront.

What does the ISC 2027 syllabus need here?

An elementary, qualitative treatment of Fraunhofer diffraction at a single slit: set-up and diagram, minima a\sin\theta_n = n\lambda, secondary maxima (n + \frac{1}{2})\lambda, the intensity distribution and the angular width of the central maximum.

How was this asked recently?

ISC 2024 Q6 (2 marks) asked a labelled intensity-angle graph; ISC 2025 Q1(A)(vii) (1 mark) was an assertion-reason item on the condition for diffraction.