Diffraction at a single slit

Hearing round a corner

You can hear a friend who stands round the corner of a wall, but you cannot see her. The sound has spread round the corner. The light has not, or not enough to notice. Both are waves, so what is the difference? The wavelength. A sound of frequency $500$ Hz in air has a wavelength of $\dfrac{340\ \text{m/s}}{500\ \text{Hz}} = 0.68$ m, about as wide as a door. Visible light has a wavelength of about $5 \times 10^{-7}$ m, a hundred thousand times smaller than a door.

The bending and spreading of a wave when it passes an opening or the edge of an obstacle is called diffraction. It is noticeable when the opening is about as wide as the wavelength, or narrower.

The pattern of a single slit

Let a plane wave of monochromatic light fall on a narrow slit of width $a$. On a screen far behind, we see a wide bright band in the middle, the central maximum. On both sides there are dark places and then weaker bright bands.

By Huygens' principle, every point of the slit is a source of wavelets. Waves from different points of the slit reach a point of the screen at the angle $\theta$ with different path lengths, and they add up with different phases. This is interference of a great number of sources, not of two.

Why are there dark places?

Split the slit into two halves of width $a/2$. Pair a point in the top half with the point that is $a/2$ below it. The path difference between the two rays at the angle $\theta$ is $\dfrac{a}{2}\sin\theta$.

When this path difference is $\lambda/2$, the two waves of every pair cancel. All the pairs cancel, so the intensity is zero. This happens when

For the second dark place, split the slit into four parts, and the same argument gives $a\sin\theta = 2\lambda$. In general

Notice that $n = 0$ is not a dark place. At $\theta = 0$ all the waves are in step, and this is the centre of the central maximum.

Between two dark places there is a weak maximum, close to the middle: at $a\sin\theta \approx \left(n + \tfrac12\right)\lambda$. The exact intensity, as a fraction of the centre, is

The first weak maximum has $I = 0.047\,I_0$. The next ones have $0.016\,I_0$ and $0.008\,I_0$. Almost all the light is in the central maximum.

The width of the central maximum

The first minima are at $a\sin\theta = \pm\lambda$. For small angles $\sin\theta \approx \theta \approx y/D$, so they are at $y = \pm\dfrac{\lambda D}{a}$. The central maximum lies between them:

The angle to the first minimum is $\theta = \lambda/a$. The weak maxima beside the centre are only $\lambda D/a$ wide, half as wide as the central one.

Make the slit twice as narrow. What happens to the pattern? It becomes twice as wide. This is the opposite of what a ray picture says, where a smaller opening gives a smaller spot. A narrow slit spreads the light more.

A worked example. A narrow slit of width $0.14$ mm is lit by light of wavelength $560$ nm. The screen is $1.5$ m behind it. Find the angle to the first minimum, the position of the first and second minima and the width of the central maximum.

1. The angle. $\sin\theta \approx \theta = \dfrac{\lambda}{a} = \dfrac{560 \times 10^{-9}\ \text{m}}{0.14 \times 10^{-3}\ \text{m}} = 4.0\ \text{mrad}$.
2. The first minimum. $y_1 = \dfrac{\lambda D}{a} = 4.0 \times 10^{-3} \times 1.5\ \text{m} = 6.0\ \text{mm}$.
3. The second minimum. $y_2 = 2y_1 = 12.0\ \text{mm}$.
4. The central maximum. $w = 2y_1 = 12.0\ \text{mm}$.
5. Check. The angle $4.0$ mrad is far below $1$ rad, so $\sin\theta \approx \theta$ is fine. The pattern is about $86$ times as wide as the slit.

Diffraction and interference together

In a double slit, all the bright fringes are equally wide and equally bright, and their width is $\lambda D/d$. In a single slit, the central maximum is far brighter and twice as wide as the others. Both are wave effects. In the double slit, two sources interfere, and in the single slit a great number of sources do.

The Fresnel distance, in brief

A beam of light that passes through an opening of width $a$ spreads by an angle of about $\lambda/a$. After a distance $z$ the spread is about $\dfrac{\lambda z}{a}$. This equals the width $a$ of the opening itself at

This is the Fresnel distance. Up to about $z_F$ the beam keeps its width, and ray optics is a good picture. Far beyond $z_F$ the spreading by diffraction is larger than the opening, and the pattern above is what we see. For a slit $1.0$ mm wide and light of $610$ nm, $z_F = \dfrac{(1.0 \times 10^{-3}\ \text{m})^2}{610 \times 10^{-9}\ \text{m}} = 1.64$ m.

Other books. Some books call the pattern at a large distance (beyond $z_F$) the Fraunhofer pattern, and the pattern near the opening the Fresnel pattern. The slit formulas above are for the Fraunhofer case.

Practise with new numbers

Try it in Python

The program finds the weak maxima of the single-slit pattern from the exact formula, and compares them with the simple guess $\alpha = (n + \tfrac12)\pi$.

Common mistakes

• Using $n = 0$ for a dark place. $n = 0$ is the centre, which is the brightest place.
• Taking the distance to the first minimum as the width. The central maximum reaches from the first minimum on one side to the first minimum on the other side, so $w = 2\lambda D/a$.
• Thinking that a wider slit gives a wider pattern. The opposite is true. The width of the pattern is in inverse proportion to the width of the slit.
• Using the double-slit fringe width $\lambda D/d$ for a single slit. In a single slit, neighbouring dark places are $\lambda D/a$ apart, where $a$ is the width of the one slit.

Remember these

• Diffraction is the spreading of a wave at an opening about as wide as $\lambda$ or narrower.
• Dark places of a single slit: $a\sin\theta = n\lambda$, with $n = \pm1, \pm2, \dots$
• The width of the central maximum is $2\lambda D/a$, twice that of the others.
• The side maxima have $0.047\,I_0$, $0.016\,I_0$, $0.008\,I_0$.
• The Fresnel distance is $a^2/\lambda$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The slit is made half as wide. Nothing else changes. What happens to the width of the central maximum?

2. In a single-slit pattern, how wide is the central maximum compared with each of the weak maxima beside it?

3. Why can we not see round a corner, though we can hear round a corner?

4. A beam of red laser light, of wavelength $600$ nm, passes through a round hole $3.0$ mm wide. Up to about what distance does the beam keep its width?

References

• OpenStax, University Physics Volume 3, Chapter 4 (Diffraction): single-slit diffraction and its intensity.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): diffraction due to a single slit and the width of the central maximum; the Fresnel distance.