Hold a microscope slide over a candle until it is black with soot. With a sharp blade, scratch two fine lines side by side, a quarter of a millimetre apart. Shine a green laser pointer through the two scratches onto a wall two metres away. Never look into the laser; watch only the wall. Instead of one spot you see a row of spots, bright and dark in turn, evenly spaced across the wall.
Young's arrangement
The last lesson found that two coherent sources are needed, and that they must come from one source. Thomas Young's arrangement does exactly this. Light from a lamp falls on a narrow slit S. The light from S falls on two narrow, parallel slits \(S_1\) and \(S_2\), close together and equally far from S. Their light spreads out and overlaps on a screen at a distance \(D\).
Both slits receive their light from the same wavefront of S. Any jump of phase in the lamp reaches \(S_1\) and \(S_2\) together, so the two slits keep a fixed phase difference. They are coherent sources, and in phase, because they are equally far from S. If \(S_1\) and \(S_2\) were lit by two separate lamps, the pattern would jump about and wash out, like the two torches.
On the screen appear bright and dark bands, called fringes, parallel to the slits and equally spaced. Fig 4.1 shows how the two slits paint them. Play it with each chip.
Look along a dashed line in Fig 4.1. Every point on it is the same number of wavelengths farther from one slit than from the other. The middle line is equally far from both. Between the dashed lines lie lines where crest meets trough, and there the screen is dark. Moving the slits apart turns every dashed line closer to the middle, so the fringes crowd together.
Why is a single narrow slit S placed in front of the lamp, before the two slits \(S_1\) and \(S_2\)?
Thomas Young was an English physician who read many languages as a boy and later helped to decipher the Egyptian hieroglyphs. In 1801 he stated the principle of interference in a lecture to the Royal Society in London. In 1803 he split a thin beam of sunlight with the edge of a card and saw bands of light and dark. His Lectures of 1807 describe light from one source passing through two close openings. From the spacing of the bands he worked out the wavelengths of the colours: about \(700\,\text{nm}\) for red and a little over \(400\,\text{nm}\) for violet. Many in Britain rejected his work, because the wave idea went against Newton.
The path difference at a point
Take a point P on the screen at a height \(y\) above O, the point opposite the middle of the two slits. Let the slits be a distance \(d\) apart. The line from the middle of the slits to P makes an angle \(\theta\) with the axis. Fig 4.2 shows the geometry, with \(d\) drawn far too large.
\[ S_1P \approx NP \]
Why. the screen is very far away compared with \(d\), so the two paths are nearly parallel. The short line \(S_1N\) is at right angles to both.
\[ \Delta x = S_2P - S_1P \approx S_2N = d\sin\theta \]
Why. in the small right triangle \(S_1NS_2\), the hypotenuse is \(d\). The angle at \(S_1\) is \(\theta\), because its sides are at right angles to those of \(\theta\).
\[ \sin\theta \approx \tan\theta = \dfrac{y}{D} \]
Why. \(y\) is a few millimetres and \(D\) a metre, so \(\theta\) is tiny; for small angles the sine and the tangent are equal.
\[ \Delta x \approx \dfrac{yd}{D} \]
Read it. substitute step 3 in step 2.
Check the sizes. With \(d = 1\,\text{mm}\), \(D = 1\,\text{m}\) and \(y = 1.5\,\text{mm}\), the path difference is \(1.5 \times 10^{-3} \times 10^{-3}/1\) \(= 1.5 \times 10^{-6}\,\text{m}\) \(= 1.5\,\mu\text{m}\). For \(600\,\text{nm}\) light that is \(2.5\) wavelengths. A point only a millimetre and a half from the centre is already two and a half wavelengths out of step.
The fringes are equally spaced only near the centre. The exact bright condition is \(d\sin\theta = n\lambda\), while the position on the screen is \(y = D\tan\theta\). Near the centre \(\sin\theta \approx \tan\theta\), so \(y\) grows in equal steps. Far out, \(\tan\theta\) grows faster than \(\sin\theta\), and the fringes widen. With slits \(10\,\mu\text{m}\) apart and \(500\,\text{nm}\) light, the tenth bright fringe has \(\sin\theta = 0.5\). It lies at \(D\tan 30° = 0.577D\), not at the \(0.5D\) the simple rule gives.
Where the fringes are
Put the path difference \(yd/D\) into the two conditions of the last lesson. A point is bright when \(yd/D = n\lambda\), and dark when \(yd/D = (2n - 1)\lambda/2\). Solving for \(y\) gives the heights of the fringes.
The number \(n\) of a bright fringe is its order. At the centre O the two slits are equally far away, so the path difference is zero for every wavelength. The waves always arrive in phase, and O is always bright. It is the central fringe, the fringe of order zero. The first bright fringes lie one on each side of it, then the second, and so on, symmetrically above and below O.
In a double slit experiment \(d = 1\,\text{mm}\), \(D = 1\,\text{m}\) and \(\lambda = 600\,\text{nm}\). A point P on the screen is \(1.5\,\text{mm}\) above O. Is P bright or dark, and which fringe is it?
P is at \(y = 1.5\,\text{mm}\); the slits are \(1\,\text{mm}\) apart and the screen \(1\,\text{m}\) away.
\(\Delta x = yd/D\), then compare \(\Delta x\) with \(\lambda\).
\(\Delta x\) \(= 1.5 \times 10^{-3} \times 10^{-3}/1\) \(= 1.5\,\mu\text{m}\), and \(1.5\,\mu\text{m}/0.6\,\mu\text{m} = 2.5\). So \(\Delta x\) \(= 5\lambda/2\) \(= (2 \times 3 - 1)\lambda/2\).
P is dark: it lies on the third dark fringe.
Why is the fringe at the centre O always bright, whatever the wavelength?
The fringe width
The distance from one bright fringe to the next is \(y_{n+1} - y_n = \lambda D/d\). The distance from one dark fringe to the next is the same. This is the fringe width \(\beta\). A bright fringe and the dark fringes beside it are each half a fringe width apart.
Take \(\lambda = 600\,\text{nm}\), \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\). Then \(\beta\) \(= 600 \times 10^{-9} \times 1/10^{-3}\) \(= 6 \times 10^{-4}\,\text{m}\) \(= 0.6\,\text{mm}\). The fringes are small but easy to see with a magnifying eyepiece. Doubling \(d\) halves \(\beta\) to \(0.3\,\text{mm}\). Doubling \(D\) instead doubles \(\beta\) to \(1.2\,\text{mm}\). Blue light of \(450\,\text{nm}\) in place of \(600\,\text{nm}\) gives \(0.6 \times 450/600 = 0.45\,\text{mm}\): shorter waves give closer fringes.
Try the sliders of the figure above. Change \(D\) alone and \(\beta\) changes, but the second readout, \(\beta/D\), does not move. This is the angular fringe width, the angle that one fringe subtends at the slits: \(\theta_\beta = \lambda/d\). For \(600\,\text{nm}\) and \(1\,\text{mm}\) it is \(6 \times 10^{-4}\,\text{rad}\), about \(0.03°\). Moving the screen spreads the fringes on it, but their angle stays fixed.
The first and the sixth bright fringes on the same side of O are \(3.0\,\text{mm}\) apart. Find the fringe width, and the distance of the fourth dark fringe from O.
From the first to the sixth bright fringe there are five fringe widths, so \(\beta = 3.0/5 = 0.6\,\text{mm}\).
The \(n\)th dark fringe is at \((2n - 1)\beta/2\). For \(n = 4\), \(y = \tfrac72 \times 0.6 = 2.1\,\text{mm}\).
The fringe width is \(0.6\,\text{mm}\). How far apart are the third bright fringe and the second dark fringe when they lie on opposite sides of O?
Light of \(500\,\text{nm}\) falls on two slits \(0.5\,\text{mm}\) apart. The screen is \(1.5\,\text{m}\) from the slits. What is the fringe width, in mm?
Measuring a wavelength
Young's fringes are the simplest way to measure the wavelength of light. Measure \(\beta\), \(D\) and \(d\), and \(\lambda = \beta d/D\). With \(d = 1.5\,\text{mm}\), \(D = 1\,\text{m}\) and a measured \(\beta = 0.4\,\text{mm}\), \(\lambda\) \(= 0.4 \times 10^{-3} \times 1.5 \times 10^{-3}/1\) \(= 6 \times 10^{-7}\,\text{m}\) \(= 600\,\text{nm}\).
It is hard to measure \(D\) to the slits exactly, so a better way moves the screen. Moving the screen \(0.5\,\text{m}\) farther raises \(\beta\) by \(\lambda\Delta D/d\). If \(\beta\) grows by \(0.3\,\text{mm}\) with slits \(1\,\text{mm}\) apart, then \(\lambda\) \(= 0.3 \times 10^{-3} \times 10^{-3}/0.5\) \(= 6 \times 10^{-7}\,\text{m}\), again \(600\,\text{nm}\). Plotting \(\beta\) against \(D\) for several positions gives a straight line through the origin, of slope \(\lambda/d\).
A student plots the fringe width \(\beta\) against the screen distance \(D\) and gets a straight line through the origin, with \(\beta = 0.5\,\text{mm}\) at \(D = 1\,\text{m}\). The slits are \(1\,\text{mm}\) apart. What is the wavelength, in nm?
Now return to the blackened slide. A green laser pointer gives \(532\,\text{nm}\). The two scratches are \(0.25\,\text{mm}\) apart, and the wall is \(2\,\text{m}\) away. The fringe width is \(\beta\) \(= 532 \times 10^{-9} \times 2/(0.25 \times 10^{-3})\) \(\approx 4.3 \times 10^{-3}\,\text{m}\). The spots on the wall are about \(4.3\,\text{mm}\) apart. Scratch the lines closer together, and the spots spread out.
- Counting the first bright fringe as the one at O.The fringe at O is the zero-order fringe. The first bright fringe is one fringe width away, at \(\lambda D/d\).
- Putting the \(n\)th dark fringe at \(n\beta\) or at \(n\beta/2\).Dark fringes lie half-way between bright ones: \(\beta/2, 3\beta/2, 5\beta/2\), that is \((2n - 1)\beta/2\).
- Mixing millimetres, metres and nanometres in \(\beta = \lambda D/d\).Convert every length to metres first, then convert \(\beta\) to mm at the end.
- Expecting the angular fringe width to grow when the screen moves back.\(\theta_\beta = \lambda/d\) has no \(D\) in it. The fringes on the screen widen, but the angle each one subtends stays the same.