Single-Slit Diffraction

PHYDiffraction & Resolution

Single-slit diffraction is the bending and spreading of light as it passes through a narrow slit of width $a$ , producing a broad bright central band flanked by progressively dimmer secondary maxima separated by dark minima on a screen. The core idea is that secondary wavelets from different parts of the same slit interfere: a dark fringe (minimum) forms when the slit can be divided into pairs of zones whose path differences cancel, giving the condition $a\sin\theta = n\lambda$ .

This topic is examined because numericals routinely ask for the angular and linear width of the central maximum and the positions of the minima, testing whether you can apply the small-angle approximation and convert units correctly.

Condition for minima (dark fringes)

a\sin\theta_n = n\lambda, \quad n = 1, 2, 3, \dots

a

= slit width,

\theta_n

= angle of the

n_{th}

minimum from the central axis,

\lambda

= wavelength of light,

n

= order of the minimum (note

n \neq 0

, since

n=0

is the centre of the bright central maximum).

Angular position of nth minimum (small angle)

\theta_n \approx \dfrac{n\lambda}{a}

\theta_n

in radians; valid when

\theta

is small so that

\sin\theta \approx \theta

. The first minimum is at

\theta_1 = \dfrac{\lambda}{a}

Half angular width of central maximum

\theta_1 = \dfrac{\lambda}{a}

\theta_1

= angular spread from the centre to the first minimum on either side; the full angular width of the central maximum is

2\theta_1 = \dfrac{2\lambda}{a}

Linear width of central maximum

W = \dfrac{2\lambda D}{a}

W

= full width of the central bright band on the screen,

D

= slit-to-screen distance, with the small-angle approximation assumed. Each secondary maximum has width

\dfrac{\lambda D}{a}

, i.e. half the central one.

Condition for secondary maxima

a\sin\theta_n = (2n+1)\dfrac{\lambda}{2}, \quad n = 1, 2, 3, \dots

\theta_n

= angle of the

n_{th}

secondary (bright) maximum; these are much fainter than the central peak and lie approximately midway between adjacent minima (e.g.

n=1

gives

\dfrac{3\lambda}{2}

The condition $a\sin\theta = n\lambda$ with $n = 1, 2, 3, \dots$ gives the MINIMA (dark fringes), not maxima. There is no minimum at $n = 0$ ; $n = 0$ corresponds to the centre of the bright central maximum.
The central maximum is twice as wide as every other (secondary) maximum, because it spans from the first minimum on one side ( $+\theta_1$ ) to the first minimum on the other side ( $-\theta_1$ ).
The small-angle approximation $\sin\theta \approx \tan\theta \approx \theta$ (with $\theta$ in radians) lets you write the linear distance on the screen as $y_n = D\tan\theta_n \approx \dfrac{n\lambda D}{a}$ . It is valid because typically $\lambda \ll a$ , so $\theta$ is small.
The linear distance of the $n_{th}$ minimum from the centre is $y_n \approx \dfrac{n\lambda D}{a}$ ; the half-width of the central maximum equals $y_1 = \dfrac{\lambda D}{a}$ and its full width is $2y_1 = \dfrac{2\lambda D}{a}$ .
Always convert to SI units before substituting: $1\,\text{mm} = 10^{-3}\,\text{m}$ and $1\,\text{nm} = 10^{-9}\,\text{m}$ . Mixing mm and nm directly is the most common source of error.
The width of the central maximum is directly proportional to $\lambda$ and $D$ but inversely proportional to the slit width $a$ : a narrower slit spreads the pattern more, while a wider slit narrows it.
Diffraction (a single source spread across one slit) is conceptually distinct from interference (two separate slits): single-slit minima follow $a\sin\theta = n\lambda$ , whereas Young's double-slit dark fringes follow $d\sin\theta = (n - \tfrac{1}{2})\lambda$ . Do not interchange the conditions.
To find the slit width $a$ from a measured central-maximum width $W$ , rearrange $W = \dfrac{2\lambda D}{a}$ to $a = \dfrac{2\lambda D}{W}$ , keeping all quantities in metres.

Where the marks go

Using a maxima condition where the minima condition is required: for single-slit DARK fringes use $a\sin\theta = n\lambda$ . Students often wrongly write $a\sin\theta = (n - \tfrac{1}{2})\lambda$ , confusing it with the Young's double-slit minima condition.
Forgetting the factor of 2: the FULL linear width of the central maximum is $W = \dfrac{2\lambda D}{a}$ , not $\dfrac{\lambda D}{a}$ . The $\dfrac{\lambda D}{a}$ value is only the distance from the centre to the first minimum (the half-width).
Unit slips: substituting $\lambda$ in nm while $a$ is in mm without converting both to metres. Convert $600\,\text{nm} \to 6 \times 10^{-7}\,\text{m}$ and $0.2\,\text{mm} \to 2 \times 10^{-4}\,\text{m}$ first.
Treating the angle $\theta$ from $\sin\theta = \tfrac{n\lambda}{a}$ as if it were already a linear distance, or using $\sin\theta$ instead of $\tan\theta$ for the screen position when the angle is not small. For ISC numericals the small-angle approximation $y \approx D\theta$ is acceptable, but you should state that you are using it.

How the board asks it

Numericallinear width of the central maximum
A parallel beam of light of wavelength $600\,\text{nm}$ is incident normally on a slit of width $0.2\,\text{mm}$ . The diffraction pattern is observed on a screen placed $1.5\,\text{m}$ away. Calculate the width of the central maximum.
Derive / provecondition for minima and central-maximum width
Derive an expression for the linear width of the central maximum in a single-slit diffraction pattern obtained on a screen at a distance $D$ from a slit of width $a$ illuminated by light of wavelength $\lambda$ .
Define / statecondition for diffraction minima
State the condition, in terms of slit width $a$ and wavelength $\lambda$ , for which dark fringes (minima) are formed in a single-slit diffraction pattern.
Numericalfinding slit width from a measured central-maximum width
In a single-slit diffraction experiment, light of wavelength $589\,\text{nm}$ falls on a slit and a screen is kept $1.0\,\text{m}$ away. If the central maximum has a width of $5.9\,\text{mm}$ , find the width of the slit.
Give reasonswidth inversely proportional to slit width
Account for the fact that the central bright band in a single-slit diffraction pattern becomes wider when the slit is made narrower.
Distinguishsingle-slit diffraction vs young's double-slit interference
Distinguish between the condition for dark fringes in single-slit diffraction and that for dark fringes in Young's double-slit interference, stating the relevant equation $a\sin\theta = n\lambda$ and $d\sin\theta = (n - \tfrac{1}{2})\lambda$ respectively.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.