Young's experiment and diffraction

End of September 2026: in Year 4 of the liceo scientifico (science-oriented high school) the block on mechanical waves and sound is coming to a close, and wave optics — interference, diffraction, gratings — starts between late October and November. The test typically falls in December: about nine to ten weeks from now. The same topic, with the same weight, is also covered in Year 4 of the applied sciences option.

The chapter takes up a single concept from the previous block — the superposition of waves, and the path difference that decides constructive or destructive interference — and applies it to light instead of sound. If you have arrived here without being clear on that concept, it is worth reviewing it first in mechanical waves and sound in Year 4: here I do not go over it from scratch, I only pick up the idea of path difference and build the double slit on top of it.

Why light interferes (and why it took time to understand)

In Young’s experiment, a monochromatic light source illuminates two thin, very close slits made in an opaque screen. Behind the two slits, on an observation screen, there is not a simple overlap of two bright spots: there appears a series of alternating bright and dark fringes. It is the experimental proof that light behaves as a wave, because only a wave produces interference.

Why it happens: each slit behaves as a secondary source (Huygens’ principle), and the two waves leaving the two slits, in phase with each other because they are generated by the same incident wave, travel paths of different lengths to reach a given point on the screen. It is again the path difference $Δ r$ seen with sound, applied here to the distance between the two slits and the point on the screen.

The conditions of coherence

For a stable interference pattern to form (fringes fixed in time, not a random flicker) the two sources must be coherent: the same frequency and a phase difference that is constant in time. That is why the experiment uses a single source split in two, instead of two independent sources — two different light bulbs never produce an observable interference pattern, because their phase difference changes randomly and too quickly.

The path difference in the double slit

If the two slits are a distance $d$ apart (the distance is traditionally called this; do not confuse it with the width of a single slit, which we shall shortly see called $a$ ) and the screen is at a distance $L$ much greater than $d$ , for a point on the screen that sees the two slits at an angle $θ$ from the central direction, the path difference between the two rays is:

Δ r = d sin θ

There is constructive interference (bright fringe) when the path difference is a whole number of wavelengths, destructive (dark fringe) when it is a whole number plus half a wavelength:

d sin θ = mλ (maxima, bright fringes), d sin θ = (m + \frac{1}{2}) λ (minima, dark fringes), m = 0, \pm 1, \pm 2, \dots

The $m = 0$ maximum is the central one, on the axis of symmetry between the two slits: it is always a bright fringe, never a dark one — a point worth fixing well before moving on to the single slit, where the exact opposite happens.

The position of the fringes on the screen

When the angle $θ$ is small (an approximation valid in almost all school exercises, because $d ≪ L$ ), you can write $sin θ \approx tan θ \approx y / L$ , where $y$ is the distance of the fringe from the centre of the screen. Substituting into the condition for maxima gives the position of the bright fringes:

y_{m} = \frac{mλ L}{d}

The distance between two consecutive bright fringes is called the fringe spacing $Δ y$ , and it does not depend on $m$ : it is the same across the whole screen (as long as the small-angle approximation holds), because it is the difference $y_{m + 1} - y_{m}$ :

Δ y = \frac{λ L}{d}

This is the most used formula in exercises: given three of the four quantities ( $λ$ , $L$ , $d$ , $Δ y$ ), you find the fourth.

Single-slit diffraction: watch the meaning of $m$

If instead of the double slit the light passes through a single slit of width $a$ (here the symbol changes on purpose, so as not to confuse it with the distance $d$ of the double slit), you do not observe interference between two sources but diffraction: the light spreads out beyond the slit, and on the screen appears a pattern with a wide, intense central maximum, flanked by progressively weaker secondary maxima, separated by minima.

The condition for the minima of single-slit diffraction is:

a sin θ = mλ, m = \pm 1, \pm 2, \pm 3, \dots (m \neq = 0)

Here is the point that causes the most confusion in the whole chapter: this formula has the same form as the one for Young’s maxima ( $d sin θ = mλ$ ), but describes the exact opposite — minima, not maxima — and excludes $m = 0$ , because the centre of the diffraction pattern is always a maximum, the most intense of all, never a minimum. Confusing the two formulas (using the one for diffraction minima as if it were the one for Young’s maxima, or vice versa) is the most frequent mistake in tests, precisely because they are algebraically identical.

The diffraction grating

A diffraction grating is a series of very many parallel slits, equally spaced by an amount $d$ (the grating spacing). The effect of having many slits instead of two is that the interference maxima, which in the double slit are wide and blurred, become narrow and much more intense: the more slits take part, the sharper the pattern. The angular position of the maxima is still described by the same Young formula:

d sin θ = mλ

but with $d$ now meaning the grating spacing, often given indirectly as a number of lines per millimetre (for example “500 lines/mm” means $d = 1/500 mm$ ).

Typical mistakes

Mistake	How you notice it
Confusing $d$ (distance between slits/grating spacing) and $a$ (width of a single slit)	Always ask yourself: am I working with two (or more) separate slits, or with one wide slit? They are two different experiments with different letters
Using the diffraction minima ( $a sin θ = mλ$ , $m \neq = 0$ ) as if they were Young’s maxima	Re-check what the problem asks — maxima (bright fringes, Young) or minima (dark gaps in the diffraction pattern) — before choosing the formula
Getting the units wrong (nm, mm, m)	$λ$ is almost always given in nanometres (nm), $d$ in millimetres or micrometres, $L$ in metres: convert everything to a single unit (usually metres) before substituting into the formula
Angles in degrees instead of radians on the calculator	If you use $sin θ \approx y / L$ you do not need to convert anything, because you never go through the value of the angle in degrees; if instead you have to find $θ$ explicitly from an arcsine, check the DEG/RAD mode of the calculator before reading the result
Forgetting the small-angle approximation	The formula $y_{m} = mλ L / d$ holds only when $θ$ is small, that is when $d ≪ L$ : it is almost always the case in school exercises, but if the text gives an explicit angle that is already large, you work with $sin θ$ directly, not with $y / L$

Three worked exercises

1. Fringe spacing in the double slit. Two slits $d = 0.20 mm$ apart are illuminated by monochromatic light of wavelength $λ = 600 nm$ . The screen is $L = 2.0 m$ from the slits. Find the fringe spacing.

First I convert everything to metres: $d = 0.20 mm = 2.0 \times 1 0^{- 4} m$ , $λ = 600 nm = 6.0 \times 1 0^{- 7} m$ .

Δ y = \frac{λ L}{d} = \frac{6.0 \times 1 0 ^{- 7} \times 2.0}{2.0 \times 1 0 ^{- 4}}

Δ y = \frac{1.2 \times 1 0 ^{- 6}}{2.0 \times 1 0 ^{- 4}} = 6.0 \times 1 0^{- 3} m = 6.0 mm

The fringes are 6 mm apart: enough to be seen with the naked eye on a screen two metres away, which is exactly why the experiment can be reproduced in a lab without any special instruments.

2. Finding the wavelength from an observed fringe. In the same set-up as before ( $d = 0.20 mm$ , $L = 2.0 m$ ), the third bright fringe (beyond the central one) is observed at $y_{3} = 1.8 cm$ from the centre. Find the wavelength of the light used.

I use the formula for the position of the bright fringes with $m = 3$ , isolating $λ$ :

y_{m} = \frac{mλ L}{d} ⟹ λ = \frac{y _{m} d}{m L}

I convert $y_{3} = 1.8 cm = 1.8 \times 1 0^{- 2} m$ :

λ = \frac{1.8 \times 1 0 ^{- 2} \times 2.0 \times 1 0 ^{- 4}}{3 \times 2.0} = \frac{3.6 \times 1 0 ^{- 6}}{6.0}

λ = 6.0 \times 1 0^{- 7} m = 600 nm

Six hundred nanometres is in the yellow-orange part of the visible spectrum: a good plausibility check, besides the calculation, is to verify that the value found falls between about 400 nm (violet) and 700 nm (red) — if the result comes out, for example, as 6000 nm or 60 nm, you have almost certainly got a power of ten wrong in a conversion.

You can practise more exercises like these in the Year 4 wave optics test: the correct answer is free, and the guided step-by-step solution is unlocked with the first lesson.

3. The first minimum of single-slit diffraction. A slit of width $a = 0.040 mm$ is illuminated by the same light of $λ = 600 nm$ . Find the angle at which the first diffraction minimum ( $m = 1$ ) is observed.

I convert $a = 0.040 mm = 4.0 \times 1 0^{- 5} m$ and use $a sin θ = mλ$ with $m = 1$ :

sin θ = \frac{mλ}{a} = \frac{1 \times 6.0 \times 1 0 ^{- 7}}{4.0 \times 1 0 ^{- 5}}

sin θ = 0.015

θ = arcsin (0.015) \approx 0.8 6^{\circ}

A very small angle, and the central diffraction maximum extends between $- 0.8 6^{\circ}$ and $+ 0.8 6^{\circ}$ : twice the angular width of each secondary maximum. Watch the value of $m$ : here $m = 1$ is a minimum, whereas in the double slit $m = 1$ was a maximum.

The rest of the physics syllabus

If you want to review the rest of the Year 4 physics syllabus too, you will find everything on the physics page. Light as an electromagnetic wave, which lies behind these fringes, comes back in Year 5 with Maxwell: the map is in electromagnetism for the maturità. The mistakes that lose the most marks in physics problems, in general, are in typical physics mistakes in high school.

If you would like a hand setting up this block before the December test, the first introductory call is free and lasts half an hour.

Interference and diffraction: Young's experiment

Why light interferes (and why it took time to understand)

The conditions of coherence

The path difference in the double slit

The position of the fringes on the screen

Single-slit diffraction: watch the meaning of $m$

The diffraction grating

Typical mistakes

Three worked exercises

The rest of the physics syllabus

Geometric optics: mirrors, refraction and lenses

Year 4 physics: mechanical waves and sound

First law of thermodynamics with worked exercises

Want to put what you read into practice?

Why light interferes (and why it took time to understand)

The conditions of coherence

The path difference in the double slit

The position of the fringes on the screen

Single-slit diffraction: watch the meaning of m

The diffraction grating

Typical mistakes

Three worked exercises

The rest of the physics syllabus

Geometric optics: mirrors, refraction and lenses

Year 4 physics: mechanical waves and sound

First law of thermodynamics with worked exercises

Want to put what you read into practice?

Single-slit diffraction: watch the meaning of $m$