Diffraction and two slits.

Step 1 of 4

A narrow gap

In the water tank a straight bar sends out straight waves, at a frequency of 10 Hz and a speed of 25 cm/s. Their wavelength is therefore $\lambda = \dfrac{v}{f} = 2.5\ \mathrm{cm}$$\lambda =$$\dfrac{v}{f} =$$2.5\ \mathrm{cm}$⁠. The waves meet a wall with one opening, 10 cm wide, and pass through it as a straight beam, with almost calm water in the wall’s shadow on either side.

When a wave passes through an opening or past an edge, it spreads a little into the space behind it. That spreading is diffraction. Predict what will happen when the opening is narrower than one wavelength, then watch the Scene narrow it.

Step 2 of 4

Sound and light

Every point across the opening sends out ripples of its own. In a wide opening, ripples from different parts of the opening cancel each other out to the sides, leaving a straight beam. In an opening no wider than the wavelength, no two parts are far enough apart to cancel, and the wave spreads in every direction. So what matters is the ratio of the opening’s width to the wavelength.

Try it: slide Opening width to 1.5 cm, 2.5 cm (one wavelength) and 10 cm, and choose On under Show the calm water to see where the wave does not reach.

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Step 3 of 4

Two slits

Now the wall has two slits, 10 cm apart. Each slit spreads half-circles of its own, and they interfere: in a direction $\theta$⁠ the path from one slit is longer than the path from the other by $d \sin\theta$⁠, where $d$⁠ is the distance between the slits. When that difference is a whole wavelength, crest meets crest and the wave is strong; when it is half a wavelength, crest meets trough and the water is calm. Show the calm water colours the calm lines.

Step 4 of 4

Young’s experiment

In 1801 Thomas Young did the same experiment with light, and the bright and dark fringes on the wall persuaded physicists that light is a wave. The bright fringes lie at the angles where $d \sin\theta = m\lambda$$d \sin\theta =$$m\lambda$⁠. At small angles $\sin\theta \approx \tan\theta = \dfrac{y}{L}$$\sin\theta \approx$$\tan\theta =$$\dfrac{y}{L}$⁠, where $y$⁠ is the distance from the middle on a wall a distance $L$⁠ away, so:

$y_m = \frac{m \lambda L}{d}$$y_m =$$\frac{m \lambda L}{d}$⁠

Visible light has a wavelength of a few hundred nanometres, so to see the fringes the slits have to be very close together, tenths of a millimetre apart.