Chapter 10 · Wave Optics

10.3 Diffraction at a Single Slit and Polarisation

Light passing through a narrow slit spreads out into a bright central band flanked by fainter ones. The wavelets from different parts of the slit interfere: the first dark band is where $a\sin\theta=\lambda$, and the central maximum has width $2\lambda D/a$ — wider the narrower the slit.

You can hear a friend talking in the next room through an open door, even though you cannot see them. Sound bends round the edges of the doorway; light, it seems, does not — a doorway casts a sharp shadow. Yet look closely at the shadow of a razor blade lit by a distant point of light, or squint at a street lamp through the narrow gap between two fingers, and you will see faint bright and dark bands at the edges: light does bend a little. Both are examples of diffraction, and the difference between them is simply a matter of wavelength: sound waves are about a metre long, comparable with a doorway, while light waves are a million times shorter.

Diffraction finally settled the debate about the nature of light. In 1818 Augustin Fresnel entered a prize competition in Paris with a wave theory of diffraction. One of the judges, Siméon Poisson, pointed out what seemed an absurd consequence: the shadow of a small round disc should have a bright spot at its very centre. François Arago did the experiment — and found the spot. In this section we use Huygens' wavelets together with interference to explain the diffraction pattern of a single slit: why its central bright band is twice as wide as the others, where its dark bands lie, and how it sets the overall shape of the fringes in Young's experiment. The same physics limits how fine a detail a microscope or a telescope can show, and lets a laser measure the width of a hair.

Diffraction

In Interactive 1 of Section 10.1, a plane wave passing through a gap curled round its edges into the geometrical shadow. Huygens' construction tells us where the wavefront goes, but not how bright the light is in each direction. For that we must add up the secondary wavelets from every point of the opening, each with its own phase — the idea Fresnel added to Huygens' principle.

📘 Definition 10.5 — Diffraction of light

Diffraction is the spreading of a wave into the region of geometrical shadow when it passes the edge of an obstacle or goes through an opening, together with the pattern of bright and dark regions this produces. It results from the interference of the secondary wavelets from the unobstructed parts of the wavefront. Its effects are large when the size of the opening or obstacle is comparable with the wavelength, and small when it is many wavelengths across.

For an opening of width $a$, the spreading angle turns out to be about $\lambda/a$. A doorway $0.9$ m wide spreads sound of wavelength $0.34$ m through angles of tens of degrees, but spreads light of wavelength $500$ nm by less than a millionth of a radian — far too little to notice. That is why ray optics works so well for ordinary objects: it is the limit of wave optics when every opening is huge compared with $\lambda$. To see light diffract, we need slits a fraction of a millimetre wide.

Diffraction at a Single Slit

Let a plane wave of wavelength $\lambda$ fall normally on a long, narrow slit of width $a$. We look at the light leaving the slit at an angle $\theta$ to the straight-through direction, on a screen a distance $D$ away that is far compared with $a$ (or in the focal plane of a lens placed just after the slit). Every point across the width of the slit is a source of secondary wavelets, all in phase, since the incident wavefront lies across the slit.

Straight through ($\theta=0$). All the wavelets travel the same distance to the central point $O$ of the screen and arrive in phase. They all reinforce: this is the bright central maximum.

At an angle $\theta$. The wavelet from the lower edge of the slit travels further than the one from the upper edge by $a\sin\theta$, and wavelets from points in between have intermediate extra paths, in proportion to their distance across the slit.

The first minimum. Divide the slit into an upper half and a lower half. Pair each point of the upper half with the point a distance $a/2$ below it, in the lower half. If $\dfrac a2\sin\theta=\dfrac\lambda2$, every such pair has a path difference of half a wavelength, and the two wavelets of each pair cancel. Every wavelet has a partner, so the whole slit sends no light in this direction: a dark band, at $$a\sin\theta=\lambda.$$ Notice the surprise: the edges of the slit are a whole wavelength out of step there, which would mean constructive interference for two slits — but here every point in between contributes too.

Further minima. If $a\sin\theta=2\lambda$, divide the slit into four quarters; the first and second quarters cancel in pairs as above, and so do the third and fourth. In general, dividing the slit into $2n$ equal strips shows that the intensity is zero whenever $a\sin\theta=n\lambda$.

Secondary maxima. Between the minima the light does not cancel completely. If $a\sin\theta=\tfrac32\lambda$, divide the slit into three thirds: the first two thirds cancel each other, but the last third is left over and sends some light — a weak secondary maximum. Its amplitude is at most about a third of the central amplitude (less, because the wavelets within that third are themselves spread in phase), so its intensity is only a few per cent of the central one. Similarly there are weaker maxima near $a\sin\theta=\tfrac52\lambda,\ \tfrac72\lambda,\dots$

📐 Result 10.9 — Minima and maxima of single-slit diffraction

For a plane wave of wavelength $\lambda$ incident normally on a slit of width $a$, with the pattern observed far away:

$$\text{minima (dark): }\ a\sin\theta=n\lambda,\qquad n=\pm1,\,\pm2,\,\pm3,\dots$$

(not $n=0$, which is the central maximum). Secondary maxima lie approximately midway between the minima, at $a\sin\theta\approx\left(n+\tfrac12\right)\lambda$, $n=1,\,2,\dots$, with intensities falling rapidly — about $4.7\%$, $1.6\%$, $0.8\%$ of the central maximum. For small angles on a screen at distance $D$, the minima are at

$$y_n=\frac{n\lambda D}{a}.$$

📐 Result 10.10 — Width of the central maximum

The central maximum extends between the first minima on either side, $\sin\theta=\pm\dfrac{\lambda}{a}$. Its angular width (small angles) and its linear width on a screen at distance $D$ (or in the focal plane of a lens of focal length $f$, with $D$ replaced by $f$) are

$$2\theta_1=\frac{2\lambda}{a},\qquad w=\frac{2\lambda D}{a}.$$

The central maximum is twice as wide as each secondary maximum (which spans $\lambda D/a$), and the narrower the slit, the wider it is.

(A careful numerical sum puts the secondary maxima slightly inside the midpoints — at $a\sin\theta=1.43\lambda$ and $2.46\lambda$ — but the midpoint rule is accurate enough for every school calculation.)

The width of the central maximum, $2\lambda D/a$, is the quantitative form of "diffraction is important when $a$ is comparable with $\lambda$". If $a\le\lambda$ there is no minimum at all ($\sin\theta$ would have to exceed $1$): the slit acts like a single line source and the light spreads into the whole half-space beyond it — exactly the secondary wavelet of Huygens' principle.

✏️ Example 1 — The pattern of a narrow slit

Light of wavelength $600$ nm falls normally on a slit $0.20$ mm wide. The pattern is seen on a screen $2.0$ m away. Find (a) the angle of the first minimum, (b) the width of the central maximum, (c) the distance of the second minimum from the centre of the pattern, and (d) the approximate position of the first secondary maximum.

💡 Show solution▶

(a) First minimum $\sin\theta_1=\dfrac{\lambda}{a}=\dfrac{600\times10^{-9}}{0.20\times10^{-3}}=3.0\times10^{-3}$, so $\theta_1\approx3.0\times10^{-3}$ rad $\approx0.17^\circ$ — a small angle.

(b) Central maximum $$w=\frac{2\lambda D}{a}=\frac{2(600\times10^{-9})(2.0)}{0.20\times10^{-3}}=1.2\times10^{-2}\ \text{m}=12\ \text{mm}.$$

(c) Second minimum $y_2=\dfrac{2\lambda D}{a}=12$ mm from the centre (the same number as the full width of the central maximum).

(d) Secondary maximum At $a\sin\theta\approx1.5\lambda$: $y\approx\dfrac{1.5\lambda D}{a}=9.0$ mm (the exact peak, at $1.43\lambda$, is at $8.6$ mm).

Check Interactive 1 starts with $a=0.20$ mm, $\lambda=600$ nm and $D=2.0$ m and finds the first minima at $\pm6.0$ mm by adding $200$ wavelets.

✏️ Example 2 — Finding the slit width

Light of wavelength $500$ nm passes through a slit and produces a central maximum $5.0$ mm wide on a screen $1.0$ m away. Find the slit width. What would the width of the central maximum be if the slit were twice as wide?

💡 Show solution▶

Slit width $a=\dfrac{2\lambda D}{w}=\dfrac{2(500\times10^{-9})(1.0)}{5.0\times10^{-3}}=2.0\times10^{-4}$ m $=0.20$ mm.

Wider slit $w\propto\dfrac1a$: $2.5$ mm. A wider slit gives a narrower diffraction pattern.

✏️ Example 3 — When the spreading is large

(a) How wide must a slit be for the first minimum of $600$ nm light to lie at $30^\circ$? (b) Why can't the small-angle formula $w=2\lambda D/a$ be used here? (c) What happens if the slit is narrower than $600$ nm?

💡 Show solution▶

(a) $a=\dfrac{\lambda}{\sin30^\circ}=2\lambda=1.2\ \mu\text{m}$.

(b) The central maximum spans $60^\circ$; $\sin\theta\approx\theta$ fails badly at such angles ($\sin30^\circ=0.50$ but $30^\circ=0.52$ rad), and $\tan\theta=y/D$ differs from both. Use $a\sin\theta=n\lambda$ directly.

(c) $\sin\theta_1=\lambda/a\gt1$: there is no minimum. The light spreads into all directions in front of the slit, with no dark band anywhere.

The slit is cut into $200$ strips, each a source of a secondary wavelet. Left: for the point $P$ on the screen, the $200$ small phasors laid head to tail (only every fifth one is drawn). Straight ahead they form a straight line; further out they curl round — when the edges of the slit are one wavelength out of step the chain closes into a circle and the resultant is zero. Right: the intensity on a screen $D=2.0$ m away, from the same sum. Change the slit width and the wavelength.

$a$ (mm):0.20$\lambda$ (nm):600

central maximum $2\lambda D/a$—

probe $y$ and $a\sin\theta/\lambda$—

$I/I_0$ (sum of 200 wavelets)—

✏️ Example 4 — How bright are the secondary maxima?

Use the chain of phasors of Interactive 1. Straight ahead, the many tiny phasors from the strips of the slit lie in a straight line of total length $A_0$, the amplitude of the central maximum. In a direction where the edges of the slit differ in path by $a\sin\theta$, the phase changes steadily across the slit by $\dfrac{2\pi}{\lambda}a\sin\theta$ in total, so the chain (of the same total length $A_0$) bends into an arc of a circle that turns through this angle. (a) Show that the chain closes into a full circle at $a\sin\theta=\lambda$. (b) Estimate the intensity at $a\sin\theta=\tfrac32\lambda$ as a fraction of the central intensity $I_0$. (c) Repeat for $a\sin\theta=\tfrac52\lambda$.

💡 Show solution▶

(a) Full circle At $a\sin\theta=\lambda$ the total phase change is $2\pi$: the chain turns exactly once and its end meets its start. Resultant zero — the first minimum ✓.

(b) 1.5 turns The phase change is $3\pi$, one and a half turns, so the chain wraps round a circle of circumference $\dfrac{A_0}{1.5}$ and ends diametrically opposite its start. The resultant is the diameter: $$A=\frac{A_0}{1.5\pi},\qquad\frac{I}{I_0}=\frac{1}{(1.5\pi)^2}\approx0.045.$$

(c) 2.5 turns Circumference $\dfrac{A_0}{2.5}$, resultant $\dfrac{A_0}{2.5\pi}$: $\dfrac{I}{I_0}=\dfrac{1}{(2.5\pi)^2}\approx0.016$.

Lesson The secondary maxima carry only about $4.5\%$ and $1.6\%$ of the central intensity; almost all the light goes into the central maximum. (Adding $200$ wavelets numerically gives the slightly higher peak $4.7\%$ at $1.43\lambda$, where the chain is a little less than one and a half turns.)

Diffraction and Interference Together

Interference and diffraction are not different kinds of physics: both are superposition of coherent waves. We usually say interference when a few separate sources are added (two slits), and diffraction when a continuous wavefront is added up across an opening. The patterns differ in important ways.

Interference (Young's double slit)	Diffraction (single slit)
Superposition of waves from two coherent sources	Superposition of wavelets from every point of one wavefront
All fringes have the same width, $\beta=\lambda D/d$	Central maximum $2\lambda D/a$, twice as wide as the others
Bright fringes (ideally) of equal intensity	Intensity falls rapidly: $100\%$, $4.7\%$, $1.6\%$…
Dark fringes completely dark (equal slits)	Minima are dark; maxima are of unequal brightness
Path difference $n\lambda$ between the two slits gives a maximum	Path difference $n\lambda$ between the edges gives a minimum

A real double slit. In Young's experiment each slit has a small width $a$, so each one produces its own single-slit diffraction pattern, and the two patterns overlap on the screen. The light from the two slits then interferes within that overlap. The result is the double-slit fringes, spaced $\beta=\lambda D/d$, with their brightness controlled by the single-slit pattern: the fringes are bright inside the central diffraction maximum and fade under the weak secondary maxima — the diffraction pattern is the envelope of the interference fringes (Interactive 2).

Inside the central diffraction maximum ($|a\sin\theta|\lt\lambda$) the interference maxima occur at $d\sin\theta=n\lambda$, so they satisfy $|n|\lt\dfrac{d}{a}$. If $d/a$ happens to be a whole number $m$, the $m$th-order interference maximum falls exactly on the first diffraction minimum and is missing. The central diffraction maximum then contains $2m-1$ bright fringes.

✏️ Example 5 — Counting fringes inside the envelope

A double slit has slits $0.15$ mm wide whose centres are $0.60$ mm apart. It is lit by light of wavelength $600$ nm and the screen is $1.0$ m away. Find (a) the interference fringe width, (b) the width of the central diffraction maximum, (c) the number of bright fringes inside it, and (d) the relative brightness of the first three fringes on each side.

💡 Show solution▶

(a) β $\beta=\dfrac{\lambda D}{d}=\dfrac{(600\times10^{-9})(1.0)}{0.60\times10^{-3}}=1.0$ mm.

(b) Envelope $w=\dfrac{2\lambda D}{a}=\dfrac{2(600\times10^{-9})(1.0)}{0.15\times10^{-3}}=8.0$ mm, from $-4.0$ to $+4.0$ mm.

(c) Count $\dfrac da=4$: the 4th-order fringes, at $\pm4.0$ mm, fall on the diffraction minima and are missing. Orders $0,\ \pm1,\ \pm2,\ \pm3$ remain: $7$ bright fringes.

(d) Brightness Adding the wavelets of both slits numerically, the fringes of orders $1,\ 2,\ 3$ have $0.81,\ 0.41$ and $0.09$ of the central brightness: the single-slit envelope dims them.

✏️ Example 6 — Why we hear round corners but do not see round them

A doorway is $0.90$ m wide. Find the angle of the first diffraction minimum (a) for sound of frequency $1000$ Hz (speed $340$ m/s), and (b) for light of wavelength $500$ nm. (c) Comment.

💡 Show solution▶

(a) Sound $\lambda=\dfrac{340}{1000}=0.34$ m; $\sin\theta=\dfrac{0.34}{0.90}\approx0.38$, so $\theta\approx22^\circ$. Lower-pitched sounds, with longer wavelengths, spread even more.

(b) Light $\sin\theta=\dfrac{500\times10^{-9}}{0.90}\approx5.6\times10^{-7}$ — a few ten-millionths of a radian.

(c) Comment Sound fills the room beyond the door; light goes straight through and casts a sharp shadow. Diffraction is the same for both — only the ratio $\lambda/a$ differs.

Two slits, each of width $a$, with centres $d$ apart ($\lambda=600$ nm, $D=1.0$ m). The purple curve is the full pattern, computed by adding wavelets from both slits; the dashed curve is the single-slit pattern of one slit, which forms its envelope. Change $d$ and $a$ and count the fringes inside the central diffraction maximum. Choose "single slit" to switch the second slit off.

$d$ (mm):0.60$a$ (mm):0.15

fringe width $\lambda D/d$—

central diffraction max $2\lambda D/a$—

bright fringes inside it—

⚠️ Common mistakes with single-slit diffraction

1. Swapping the conditions. For a single slit, $a\sin\theta=n\lambda$ gives minima. For a double slit, $d\sin\theta=n\lambda$ gives maxima.

2. Including $n=0$ as a minimum. $\theta=0$ is the bright centre; the minima start at $n=\pm1$.

3. Forgetting the factor 2. The width of the central maximum is $2\lambda D/a$; $\lambda D/a$ is its half-width (centre to first minimum).

4. Thinking a wider slit spreads light more. The pattern scales as $1/a$: a narrower slit gives a wider central maximum.

5. Treating secondary maxima as equally bright. They carry only a few per cent of the central intensity.

Polarisation

Interference and diffraction happen for every kind of wave. Polarisation shows something more: light is a transverse wave. In an electromagnetic wave the electric field $\mathbf E$ oscillates perpendicular to the direction of travel (Chapter 8). Light from the Sun or a bulb is unpolarised: $\mathbf E$ points in random directions in the transverse plane, changing rapidly. A longitudinal wave such as sound could never be polarised — there is no transverse direction for it to choose.

📘 Definition 10.6 — Plane-polarised light and polaroids

Light whose electric field oscillates along one fixed line perpendicular to the direction of travel is plane-polarised (linearly polarised). A polaroid is a sheet of long, aligned molecules that transmits only the component of $\mathbf E$ along its transmission axis (pass-axis) and absorbs the component perpendicular to it.

Unpolarised light of intensity $I_0$ passing through one polaroid emerges plane-polarised along the axis, with intensity $I_0/2$ (the average of $\cos^2$ over all directions), whatever way the polaroid is turned. A second polaroid, the analyser, with its axis at angle $\theta$ to the plane of vibration, passes only the component $E\cos\theta$ of the field, and the intensity goes as the square of the amplitude.

📐 Law 10.11 — Malus's law

Plane-polarised light of intensity $I_1$ falling on a polaroid whose transmission axis makes an angle $\theta$ with its plane of vibration emerges with intensity $$I=I_1\cos^2\theta.$$ Unpolarised light of intensity $I_0$ through a single polaroid gives $I_1=I_0/2$.

Turning the analyser through a full circle, the transmitted light is brightest twice (at $\theta=0^\circ$ and $180^\circ$) and dark twice — this is how one tests whether light is plane-polarised. Crossed polaroids ($\theta=90^\circ$) transmit nothing. Three polaroids. The surprise: put a third polaroid between two crossed ones, at $45^\circ$ to both, and light gets through — $\dfrac{I_0}{2}\cos^245^\circ\cos^245^\circ=\dfrac{I_0}{8}$. The middle sheet creates a new component along the last axis; Malus's law is simply applied once per sheet. Polaroids are used in sunglasses, LCD screens, 3-D cinema glasses and in the stress analysis of transparent models.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — Brewster's law

Polarisation by reflection is not in the NCERT book or the CBSE syllabus, but Brewster's law is listed for NEET, JEE Main and JEE Advanced.

Light reflected from glass or water is partly polarised, and at one angle of incidence, the Brewster angle $i_B$, the reflected light is completely plane-polarised (with $\mathbf E$ parallel to the surface). At this angle the reflected and refracted rays are perpendicular, $i_B+r=90^\circ$, which with $\sin i_B=n\sin r=n\cos i_B$ gives Brewster's law, $\tan i_B=n$. For glass ($n=1.5$), $i_B\approx56.3^\circ$; for water ($n=1.33$), $i_B\approx53.1^\circ$. Polaroid sunglasses, with vertical transmission axes, block the horizontally polarised glare reflected from roads and water.

Exercises

Minima and the central maximum.

1 Light of wavelength $500$ nm falls normally on a slit $0.10$ mm wide. Find the angle of the first minimum.
2 For part 1, find the width of the central maximum on a screen $1.5$ m away.
3 For the same set-up, how far from the centre is the third minimum?
4 The first minimum of $650$ nm light from a slit is at $0.40^\circ$. Find the slit width.
5 The slit width is halved. What happens to the width of the central maximum?
6 Red light is replaced by blue light. Does the central maximum become wider or narrower? Why?
7 The apparatus of part 2 is immersed in water ($n=1.33$). Find the new width of the central maximum.
8 A slit has width $a=2\lambda$. Find the angular width of the central maximum exactly, and compare it with the small-angle value $2\lambda/a$.

💡 Show solutions▶

1 $\sin\theta_1=\dfrac{500\times10^{-9}}{0.10\times10^{-3}}=5.0\times10^{-3}$: $\theta_1\approx5.0\times10^{-3}$ rad $\approx0.29^\circ$.

2 $w=\dfrac{2\lambda D}{a}=\dfrac{2(500\times10^{-9})(1.5)}{0.10\times10^{-3}}=15$ mm.

3 $y_3=\dfrac{3\lambda D}{a}=22.5$ mm (to two figures, $22$ mm by the round-to-even rule).

4 $a=\dfrac{\lambda}{\sin0.40^\circ}=\dfrac{650\times10^{-9}}{6.98\times10^{-3}}\approx9.3\times10^{-5}$ m $\approx93\ \mu\text{m}$.

5 $w\propto1/a$: it doubles.

6 Narrower: $w\propto\lambda$ and blue light has the shorter wavelength.

7 $\lambda\to\lambda/n$, so $w\to\dfrac{15}{1.33}\approx11$ mm.

8 $\sin\theta_1=\dfrac12$, $\theta_1=30^\circ$: angular width $60^\circ\approx1.05$ rad. The small-angle value is $1.0$ rad $\approx57^\circ$ — about $5\%$ low; the approximation is poor at such angles.

Secondary maxima and the shape of the pattern.

1 A slit $0.20$ mm wide is lit with $600$ nm light; the screen is $2.0$ m away. Find the approximate distance of the first secondary maximum from the centre.
2 How does the width of a secondary maximum compare with the width of the central maximum?
3 Explain, by dividing the slit into three parts, why the first secondary maximum is much fainter than the central maximum.
4 Use the phasor-chain argument of Example 4 to estimate the intensity, as a fraction of $I_0$, near $a\sin\theta=\tfrac72\lambda$.
5 In which direction is the intensity of a single-slit pattern greatest, and why?
6 Describe the single-slit pattern in white light. Which colour lies at the outer edge of the central maximum?

💡 Show solutions▶

1 $y\approx\dfrac{1.5\lambda D}{a}=\dfrac{1.5(600\times10^{-9})(2.0)}{0.20\times10^{-3}}=9.0$ mm.

2 A secondary maximum spans from one minimum to the next, $\lambda D/a$: half the width $2\lambda D/a$ of the central maximum.

3 At $a\sin\theta=\tfrac32\lambda$, the first two thirds of the slit are half a wavelength out of step strip by strip and cancel. Only one third of the slit contributes, and even its wavelets are spread over half a cycle in phase, so the amplitude is well under a third of the central amplitude and the intensity only a few per cent.

4 $3.5$ turns: $\dfrac{I}{I_0}\approx\dfrac{1}{(3.5\pi)^2}\approx0.0083$, under $1\%$.

5 Straight ahead ($\theta=0$): every wavelet travels the same distance, all arrive in phase and the phasors add in a straight line.

6 The centre is white (all colours have their maximum there). The edges of the central maximum are coloured: since the first minimum is at $\lambda D/a$, violet light stops first and red extends furthest, so the outer edge is reddish. The fainter secondary maxima are coloured fringes.

III

Interference and diffraction together.

1 State three differences between the fringes of Young's double-slit experiment and the single-slit diffraction pattern.
2 A double slit has $d=0.50$ mm and slit width $a=0.10$ mm. How many bright interference fringes lie inside the central diffraction maximum?
3 For part 2 with $\lambda=500$ nm and $D=1.0$ m, find the interference fringe width and the width of the central diffraction maximum.
4 A single slit of width $a$ and a double slit of separation $1.0$ mm are used with the same light and screen distance. Find $a$ if the width of the single slit's central maximum equals the double slit's fringe width.
5 In a double slit with $d=4a$, which orders of interference maxima are missing?
6 Explain why, in a real double-slit experiment, the fringes fade away from the centre.
7 Sound of frequency $500$ Hz (speed $340$ m/s) and light of wavelength $600$ nm pass through a doorway $0.85$ m wide. Find the angle of the first diffraction minimum for each.

💡 Show solutions▶

1 (i) Double slit: fringes of equal width $\lambda D/d$; single slit: central maximum twice as wide as the rest. (ii) Double slit: bright fringes equally bright; single slit: intensity falls rapidly away from the centre. (iii) Double slit: two coherent sources; single slit: wavelets from all points of one wavefront.

2 $\dfrac da=5$: orders $0,\pm1,\dots,\pm4$ lie inside and $\pm5$ fall on the diffraction minima: $9$ bright fringes.

3 $\beta=\dfrac{(500\times10^{-9})(1.0)}{0.50\times10^{-3}}=1.0$ mm; $w=\dfrac{2(500\times10^{-9})(1.0)}{0.10\times10^{-3}}=10$ mm.

4 $\dfrac{2\lambda D}{a}=\dfrac{\lambda D}{d}\Rightarrow a=2d=2.0$ mm.

5 Orders $\pm4,\ \pm8,\ \pm12,\dots$ (where $d\sin\theta=4k\lambda$ coincides with $a\sin\theta=k\lambda$).

6 Each slit has a finite width, so the light from each slit is itself spread into a single-slit pattern whose intensity falls away from the centre. The interference fringes are modulated by that envelope.

7 Sound: $\lambda=\dfrac{340}{500}=0.68$ m, $\sin\theta=\dfrac{0.68}{0.85}=0.80$, $\theta\approx53^\circ$. Light: $\sin\theta=\dfrac{600\times10^{-9}}{0.85}\approx7.1\times10^{-7}$, $\theta\approx7.1\times10^{-7}$ rad.

Applications and longer problems.

1 Modelling. A thin wire or hair produces the same diffraction pattern as a slit of the same width (a result known as Babinet's principle). A laser of wavelength $650$ nm shines on a hair and the central maximum on a wall $2.0$ m away is $26$ mm wide (illustrative data). Estimate the thickness of the hair.
2 Microwaves of wavelength $3.0$ cm pass through a gap $10$ cm wide. Find the angles of the first minima.
3 A laser beam of wavelength $633$ nm leaves a $2.0$ mm aperture. Using the single-slit result as an estimate, find the angle at which the beam spreads, and the size of the spot it would make on the Moon, $3.8\times10^8$ m away.
4 With a slit $0.25$ mm wide and a screen $1.5$ m away, the two first minima are $7.2$ mm apart. Find the wavelength.
5 Light of wavelength $550$ nm passes through a slit $0.11$ mm wide; the screen is $1.0$ m away. Find the distance between the first and the third minima on the same side of the centre.
6 How narrow must a slit be for its central maximum to spread over the whole half-space beyond it (first minimum at $90^\circ$)?

💡 Show solutions▶

1 $a=\dfrac{2\lambda D}{w}=\dfrac{2(650\times10^{-9})(2.0)}{26\times10^{-3}}=1.0\times10^{-4}$ m $=0.10$ mm $=100\ \mu\text{m}$ — a typical hair.

2 $\sin\theta=\dfrac{3.0}{10}=0.30$: $\theta\approx\pm17.5^\circ$.

3 $\theta\approx\dfrac{\lambda}{a}=\dfrac{633\times10^{-9}}{2.0\times10^{-3}}\approx3.2\times10^{-4}$ rad. Spot half-width $\approx(3.8\times10^8)(3.2\times10^{-4})\approx1.2\times10^5$ m: a spot some $240$ km across. (Real lunar-ranging lasers are first widened by a telescope to make $a$ much larger and the spot smaller.)

4 $w=\dfrac{2\lambda D}{a}\Rightarrow\lambda=\dfrac{wa}{2D}$ $=\dfrac{(7.2\times10^{-3})(0.25\times10^{-3})}{3.0}$ $=6.0\times10^{-7}$ m $=600$ nm.

5 $y_3-y_1=\dfrac{2\lambda D}{a}=\dfrac{2(550\times10^{-9})(1.0)}{0.11\times10^{-3}}=10$ mm.

6 $a\sin90^\circ=\lambda$: $a=\lambda$. For any narrower slit there is no minimum at all.

Polarisation by polaroids.

1 Unpolarised light of intensity $40\ \text{W/m}^2$ falls on two polaroids whose transmission axes are at $30^\circ$ to each other. Find the intensity after the first and after the second polaroid.
2 Plane-polarised light passes through an analyser and its intensity falls to one quarter. Find the angle between the analyser's axis and the plane of vibration.
3 Two polaroids are crossed. A third is placed between them with its axis at $30^\circ$ to the first. (a) What fraction of the incident unpolarised intensity $I_0$ emerges? (b) At what angle of the middle polaroid is the emerging intensity greatest, and what is it?

💡 Show solutions▶

1 First polaroid: $I_1=\dfrac{40}{2}=20\ \text{W/m}^2$. Second: $I=20\cos^230^\circ=20\times\dfrac34=15\ \text{W/m}^2$.

2 $\cos^2\theta=\dfrac14\Rightarrow\cos\theta=\dfrac12$, so $\theta=60^\circ$.

3 (a) The middle sheet is at $30^\circ$ to the first and $60^\circ$ to the last: $I=\dfrac{I_0}{2}\cos^230^\circ\cos^260^\circ=\dfrac{I_0}{2}\cdot\dfrac34\cdot\dfrac14=\dfrac{3I_0}{32}\approx0.094\,I_0$. (b) With the middle axis at $\phi$, $I=\dfrac{I_0}{2}\cos^2\phi\sin^2\phi=\dfrac{I_0}{8}\sin^22\phi$, greatest at $\phi=45^\circ$, where $I=\dfrac{I_0}{8}$.

🎯 Exam corner — single-slit diffraction and polarisation BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Explain, using Huygens' principle and the idea of dividing the slit into two halves, why the first minimum of the diffraction pattern of a single slit of width $a$ occurs at $a\sin\theta=\lambda$. Hence write the width of the central maximum on a screen at distance $D$, and find it for $\lambda=500$ nm, $a=0.10$ mm and $D=1.0$ m.

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Idea Every point of the slit sends out a secondary wavelet; in direction $\theta$ the two edges differ in path by $a\sin\theta$. Divide the slit into two halves and pair each point of the upper half with the point $a/2$ below it.

Condition If $\dfrac a2\sin\theta=\dfrac\lambda2$, each pair is in antiphase and cancels; every point has a partner, so the intensity is zero: $a\sin\theta=\lambda$.

Width First minima at $\theta\approx\pm\dfrac{\lambda}{a}$, so $w=\dfrac{2\lambda D}{a}=\dfrac{2(500\times10^{-9})(1.0)}{0.10\times10^{-3}}=10$ mm.

Exam tip Draw the slit split into two halves with one pair of corresponding points marked — it is what the examiner looks for.

E2. Board · 1 mark Assertion (A): When the width of a single slit is reduced, its central diffraction maximum becomes wider. Reason (R): The angular width of the central maximum is $2\lambda/a$. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

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Concept $2\lambda/a$ grows as $a$ shrinks — R directly explains A.

Answer (A). Exam tip: narrower slit, wider pattern; the same is true for any aperture.

E3. JEE Main Light of wavelength $600$ nm falls normally on a single slit; the first minimum is observed at $30^\circ$. The slit width is (A) $0.30\ \mu\text{m}$ (B) $0.60\ \mu\text{m}$ (C) $1.2\ \mu\text{m}$ (D) $2.4\ \mu\text{m}$.

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Equation $a\sin30^\circ=\lambda\Rightarrow a=\dfrac{600\ \text{nm}}{0.5}=1.2\ \mu\text{m}$.

Answer (C). Exam tip: at $30^\circ$ do not use the small-angle form; (B) uses $a\sin\theta=\lambda/2$, the double-slit dark-fringe condition.

E4. NEET JEE Main JEE Advanced Unpolarised light of intensity $I_0$ passes through two polaroids whose transmission axes are at $60^\circ$ to each other. The emerging intensity is (A) $I_0/2$ (B) $I_0/4$ (C) $I_0/8$ (D) $I_0/16$.

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Concept The first polaroid passes $I_0/2$; Malus's law then gives $\dfrac{I_0}{2}\cos^260^\circ=\dfrac{I_0}{2}\cdot\dfrac14=\dfrac{I_0}{8}$.

Answer (C). Exam tip: (B) forgets the factor $\tfrac12$ at the first polaroid; Malus's law applies only to light that is already plane-polarised.

E5. JEE Advanced (One or more correct.) In a double-slit experiment each slit has width $a$ and the slit centres are $d=3a$ apart. Then (A) the interference maxima are at $d\sin\theta=n\lambda$; (B) the third-order interference maximum is missing; (C) the central diffraction maximum contains $5$ bright fringes; (D) the interference fringe width depends on $a$.

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Options (A) true. (B) $d\sin\theta=3\lambda$ gives $a\sin\theta=\lambda$ — the first diffraction minimum, so that order is missing: true. (C) Inside $|a\sin\theta|\lt\lambda$: $|n|\lt3$, orders $0,\pm1,\pm2$ — $5$ fringes: true. (D) $\beta=\lambda D/d$ does not involve $a$ (only the brightness does): false.

Answer (A), (B), (C). Exam tip: with $d/a=m$ a whole number, the central envelope holds $2m-1$ fringes.

E6. Board · Case study In a physics laboratory a laser of wavelength $630$ nm shines on an adjustable slit set to $0.15$ mm, and the pattern is observed on a wall $1.5$ m away. (i) (2 marks) Find the distance of the first minimum from the centre and the width of the central maximum. (ii) (1 mark) The slit is narrowed to $0.10$ mm. Find the new width of the central maximum. (iii) (1 mark) With the $0.15$ mm slit, the laser is replaced by a blue one of wavelength $450$ nm. Find the width of the central maximum.

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(i) $y_1=\dfrac{\lambda D}{a}=\dfrac{(630\times10^{-9})(1.5)}{0.15\times10^{-3}}=6.3$ mm; $w=2y_1=12.6$ mm.

(ii) $w\propto\dfrac1a$: $12.6\times\dfrac{0.15}{0.10}=18.9$ mm $\approx19$ mm.

(iii) $w\propto\lambda$: $12.6\times\dfrac{450}{630}=9.0$ mm. Exam tip: use ratios in case studies — each part then takes one line.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Can laser diffraction measure the thickness of a human hair as reliably as a micrometer screw gauge — and how does hair thickness vary between people?

Why it matters. Laser diffraction gauges measure the diameter of wires, optical fibres and synthetic fibres on production lines without touching them, in factories from Germany to China; forensic laboratories compare hair and fibre samples; and the same principle is used in particle-size analysers for pharmaceuticals and cement. A thin hair diffracts light just like a slit of the same width (Babinet's principle), so the single-slit formula applies.

What to do.

Tape a single hair vertically across the opening of a cardboard frame and shine a laser pointer on it, with the pattern falling on a wall $2$–$3$ m away. (Never look into the beam.)
Mark the centres of the dark bands on paper. Measure the distance between the $n$th minima on either side for $n=1,\,2,\,3$, and $D$.
Use $y_n=\dfrac{n\lambda D}{a}$ to find $a$ from each $n$ and average. Why does using a larger $n$ give a smaller percentage error?
Measure the same hair with a micrometer screw gauge and compare. Collect results for hairs from several volunteers (with consent) and present the spread of values.

Tools: Laser pointer (class 2), cardboard, tape, metre rule, micrometer screw gauge; spreadsheet.

Hand in: Your table of minima positions, the value of $a$ with an uncertainty, the micrometer comparison and a short discussion.

Going further. Replace the hair by a thin copper wire of known gauge and use it to calibrate your set-up. How well do the two methods agree?

Instructor note (illustrative sample data, not real measurements): $\lambda=650$ nm, $D=2.50$ m; distance between the second minima on either side $64$ mm, so $y_2=32$ mm and $a=\dfrac{2\lambda D}{y_2}=\dfrac{2(650\times10^{-9})(2.50)}{0.032}\approx1.0\times10^{-4}$ m $=0.10$ mm. Using higher $n$ makes $y_n$ larger for the same reading error, so the percentage error falls as $1/n$. Typical hairs are $50$–$120\ \mu\text{m}$; micrometer readings usually agree within $10\%$ but can be low if the hair is squashed.

🧪 Self-check quiz

0/5 correct

Q1. For a single slit of width $a$, the first dark band of the diffraction pattern is at an angle $\theta$ given by

Q2. The width of a single slit is halved. The width of the central diffraction maximum

Q3. Compared with each secondary maximum, the central maximum of a single-slit pattern is

Q4. Diffraction effects are most noticeable when the size of the opening is

Q5. In a double-slit experiment the slit separation is twice the slit width ($d=2a$). The number of bright interference fringes within the central diffraction maximum is

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