9 min readConcept sheetSection A MCQ (1 mark) or objective (2 marks, two parts); Section B very short answer (2 marks) on why sound bends round corners more than light, or on interference and diffraction

Diffraction of light at a single slit (qualitative)

Light bends round the edges of a narrow slit. The pattern has a bright central maximum with fainter maxima beside it and dark minima where a sin θ = nλ.

DiffractionBending of light round the edges of an obstacle or a narrow opening into the region of geometrical shadow

When it is seenWhen the slit or obstacle is about as small as the wavelength, and the screen is far away compared with the slit width

MinimaPath difference between the edges of the slit = nλ, that is a sin θ = nλ (a = slit width, n = 1, 2, 3 ...)

The patternA brightest central maximum, twice as wide as the others, with secondary maxima that get fainter outward

What diffraction is

Look at a street light at night with your eyes almost closed. The light seems to streak out from the lamp, because it bends round the edges of your eyelids. Look at a lamp through a fine handkerchief and you see a pattern of bright spots. Diffraction is the bending of waves round the edges of an obstacle or a narrow opening, so that light enters the region of geometrical shadow. All waves diffract. Sound bends round a door easily, because its wavelength is close to a metre. Light has a tiny wavelength, about 5 × 10⁻⁷ m, so its diffraction shows only with very narrow slits or very small obstacles. That is why light usually seems to travel in straight lines. Diffraction comes from the superposition of secondary wavelets from different parts of the same wavefront.

Explaining the single slit pattern

A plane wavefront of light of one colour falls on a slit of width a. A lens focuses the light from the slit onto a screen. Think of the slit as 100 equal strips, each a source of wavelets.

1
Centre O: wavelets from strips 1 and 100 travel equal paths and arrive in phase, and so do strips 2 and 99, and so on. All add up, so the centre is the brightest: the principal (central) maximum.
2
At an angle θ, the path difference between wavelets from the two edges of the slit is a sin θ ≈ aθ.
3★
First minimum: at a point P where the path difference between the two edges is λ, strips 1 and 51 differ by λ/2 and cancel. So do 2 and 52, and every such pair. P is dark: a sin θ = λ.
4
Other minima: when the edge path difference is 2λ, divide the slit into four parts that cancel in pairs. So minima lie where a sin θ = nλ.
5
Secondary maximum: when the edge path difference is 3λ/2, divide the slit into three parts. Two parts cancel, and only one third of the slit adds up. So the point is bright, but much fainter than the centre.
6
So, further maxima lie between the minima, each fainter than the one before, because a smaller part of the slit adds up each time.

★ key step

Interference vs diffraction

Interference (double slit)Superposition of waves from two coherent sources. The fringes are equally wide (β = λD/d), and the bright fringes are about equally bright.

Diffraction (single slit)Superposition of wavelets from different parts of one wavefront. The central maximum is twice as wide as the others, and the brightness falls off quickly away from the centre.

In Young's experiment each slit also diffracts, so the interference fringes are seen inside the broad single slit pattern.

How to draw the intensity graph for a single slit

Draw a horizontal axis for θ (or distance on the screen) and a vertical axis for intensity I.
Draw a tall peak at θ = 0; label it central maximum.
Mark zero intensity on both sides at θ = ±λ/a, ±2λ/a, ±3λ/a.
Between the zeros, draw small humps; label them secondary maxima. Make each smaller than the one before.
Show the central peak twice as wide as each secondary maximum.

Worked example: slit width from the first minimum

Problem

(a) Light of wavelength 500 nm falls normally on a slit. The first minimum of the diffraction pattern is at 30° from the centre. Find the width of the slit. (b) Why does a doorway 1 m wide spread out sound of wavelength 0.7 m but not light?

Working

(a) For the first minimum: a sin θ = λ.
a = λ/sin θ = 500 nm ÷ sin 30° = 500 nm ÷ 0.5 = 1000 nm = 1 × 10⁻⁶ m.
(b) For sound, a/λ = 1 m ÷ 0.7 m ≈ 1.4. The opening is about the size of the wavelength, so the sound spreads widely.
For light, a/λ = 1 m ÷ (5 × 10⁻⁷ m) = 2 × 10⁶. The spreading angle, about λ/a, is far too small to notice.

Answer

(a) a = 1 μm; (b) the doorway is about as wide as the sound wavelength but two million times the light wavelength

In real life

You can hear a friend talking round the corner of a building but cannot see them. Sound, with a wavelength near a metre, diffracts easily; light does not. Look at a tube light through a very narrow gap between two fingers and you will see faint dark lines.

Exam angle

2 marks: “You can hear a person standing round the corner of a building but cannot see them. Explain why.” Diffraction is large only when the opening or obstacle is about the size of the wavelength. Sound has a wavelength near a metre, so it bends round the corner; light has a wavelength of about 5 × 10⁻⁷ m, so it hardly bends.

2 marks: “Give two differences between interference and diffraction of light.” Interference: waves from two coherent sources; fringes equally wide and about equally bright. Diffraction: wavelets from different parts of one wavefront; central maximum twice as wide; brightness falls off quickly.

3 marks: “Using Huygens' wavelets, explain why the centre of a single slit pattern is bright and why a minimum forms where the path difference between the edges is λ.” At the centre, wavelets from matching points of the slit travel equal paths and arrive in phase, so it is brightest. Where the edge path difference is λ, split the slit into two halves; matching points differ by λ/2 and cancel in pairs, so it is dark.

Remember it

Narrow slit, wide spread. Dark where a sin θ = nλ; the central maximum is twice as wide as the rest.