Where the path difference is fixed
Every bright fringe is a set of points where the path difference from the two sources has one value, \(n\lambda\). So the shape of a fringe is the shape of the set of points P with \(S_2P - S_1P\) fixed. In a plane containing both sources, that set is a hyperbola with \(S_1\) and \(S_2\) as its two foci. The zero curve is the straight line midway between them.
Space is three-dimensional, so turn each curve of Fig 5.1 about the line \(S_1S_2\). It sweeps out a surface, a hyperboloid of revolution. The bright places in space are a family of such surfaces, one inside the other. A screen shows the curves where these surfaces cut it.
In Young's experiment the screen is parallel to \(S_1S_2\). It cuts each hyperboloid in a hyperbola, so the fringes are hyperbolas. But a screen a metre away shows only the few millimetres near the centre, where the hyperbolas are nearly straight. For slits \(1\,\text{mm}\) apart and \(500\,\text{nm}\) light, the tenth fringe bends by about \(60\,\text{nm}\) across a centimetre of its length. No eye can see that, so the fringes look like straight lines.
The fringes of Young's double slit are hyperbolas, yet near the centre of the screen they look like straight lines. Explain.
A bright fringe is where the path difference has a fixed value, and those points lie on a hyperboloid with the slits as foci. The screen is parallel to the line of the slits and far from them compared with their separation. Near the centre it cuts each hyperboloid far from its bend, where the hyperbola is almost a straight line. Over the few millimetres of the visible pattern the curve is too small to see, so the fringes look straight and parallel to the slits.
Sources one behind the other
Now put the two sources on the axis, one behind the other, and hold the screen across the axis. The hyperboloids are symmetric about that axis, so the screen cuts each of them in a circle. The fringes are concentric rings round the axis.
At the centre of the screen, the path difference is the whole separation \(d\) of the sources. Take \(d = 1\,\text{mm}\) and \(\lambda = 500\,\text{nm}\). The order of the centre is \(d/\lambda = 10^{-3}/(500 \times 10^{-9}) = 2000\), a whole number, so the centre is bright. The centre does not have order zero here, unlike Young's pattern.
Away from the centre, a point seen at a small angle \(\theta\) from the axis has path difference \(d\cos\theta \approx d(1 - \theta^2/2)\). It falls below \(d\) as \(\theta^2\) grows. The next bright ring, of order \(1999\), is at \(\theta = \sqrt{2\lambda/d} = 0.032\,\text{rad}\), about \(3.2\,\text{cm}\) out on a screen a metre away. The rings crowd outward, like Newton's rings, and for the same reason: the path changes as the square of the distance.
Two coherent sources, \(0.5\,\text{mm}\) apart, lie one behind the other on an axis, and a screen stands across the axis. The light has \(\lambda = 500\,\text{nm}\). What is seen at the centre of the screen?
A Michelson interferometer splits one beam in two and sends the halves to two mirrors. Seen from the eyepiece, the two images of the source lie one behind the other on the axis, so the fringes are rings. Move one mirror by half a wavelength, and one ring shrinks into the centre. The LIGO detectors use the same idea with arms \(4\,\text{km}\) long. They sense gravitational waves that change an arm by far less than the width of a proton. One is being built in Hingoli district in Maharashtra, LIGO-India.
A beam that arrives at a slant
In Young's experiment the light usually falls straight on the slits, so the wave reaches both slits in step. Now tilt the beam upward by a small angle \(\alpha\). Its wavefronts are square to the beam, so they tilt too. A wavefront now reaches the lower slit \(S_2\) first and the upper slit \(S_1\) a little later.
\[ \Delta_{\text{before}} = d\sin\alpha \]
Why. the wavefront that touches \(S_2\) is still a distance \(d\sin\alpha\) from \(S_1\), measured along the beam; the wave at \(S_1\) lags by that path.
\[ \Delta_{\text{after}} = S_2P - S_1P = \dfrac{yd}{D} \]
Why. for a point P at height \(y\) on the screen, the path from \(S_2\) is longer, as in Young's experiment.
\[ d\sin\alpha = \dfrac{y_0d}{D} \]
Why. the central fringe is where the total paths through the two slits are equal, so the lag before \(S_1\) is paid back after it.
\[ y_0 = D\sin\alpha \approx D\alpha \]
Read it. divide by \(d\); for a small angle, \(\sin\alpha \approx \alpha\).
The central fringe moves up, the way the beam travels, by \(D\sin\alpha\). It does not depend on \(d\) or on \(\lambda\). The fringe width \(\lambda D/d\) does not change, so the whole pattern slides as one. Take \(\alpha = 2 \times 10^{-3}\,\text{rad}\), \(D = 1\,\text{m}\), \(d = 1\,\text{mm}\) and \(\lambda = 500\,\text{nm}\). Then \(y_0 = 2\,\text{mm}\), and \(\beta = 0.5\,\text{mm}\), so the centre has moved four fringe widths.
The slab chip of Fig 5.3 shows how to undo the shift. The tilt makes the wave at \(S_1\) lag by \(d\sin\alpha = 10^{-3} \times 2 \times 10^{-3} = 2\,\mu\text{m}\). A glass slab of thickness \(t\) adds an extra optical path \((\mu - 1)t\) to the light that crosses it. Put it before \(S_2\) and make \((\mu - 1)t = 2\,\mu\text{m}\); with \(\mu = 1.5\), \(t = 4\,\mu\text{m}\). Both slits then carry the same delay, and the central fringe returns to O.
A parallel beam falls on two slits, tilted upward at a small angle \(\alpha\) to the axis. Which way does the central fringe move?
A parallel beam falls on a double slit at \(1 \times 10^{-3}\,\text{rad}\) to the axis. The screen is \(1.5\,\text{m}\) from the slits. How far does the central fringe move, in mm?
Slits \(0.8\,\text{mm}\) apart are lit by a beam at \(1.5 \times 10^{-3}\,\text{rad}\) to the axis. What thickness of glass of \(\mu = 1.6\), placed before the slit the beam reaches first, brings the central fringe back to O, in µm?
A source off the axis
A point source \(S\) behaves the same way. Let it be a distance \(a\) before the slits and a height \(s\) above the axis. It is now nearer to \(S_1\) than to \(S_2\), by about \(sd/a\). So the wave reaches \(S_1\) first, and the wave at \(S_2\) lags. The balance point on the screen must favour \(S_2\), so it lies below O.
The central fringe lies where the whole optical path from the source to the screen is the same through both slits. A tilted beam, a source off the axis or a slab adds an extra path before the slits. The fringe then moves to where the paths after the slits make up the difference.
A point source is \(0.5\,\text{m}\) before a double slit, on its axis, and the screen is \(1\,\text{m}\) beyond the slits. The source is moved \(1\,\text{mm}\) above the axis. How far does the central fringe move, and which way?
The source is now nearer \(S_1\), by \(sd/a\). The central fringe is where the path from \(S_2\) to the screen is shorter by the same amount.
\(\dfrac{y_0d}{D} = \dfrac{sd}{a}\), so \(y_0 = \dfrac{sD}{a}\).
\(y_0 = \dfrac{1\,\text{mm} \times 1}{0.5} = 2\,\text{mm}\).
The central fringe moves \(2\,\text{mm}\) below O, to the side opposite the source. Raising the source lowers the pattern.
A point source is \(0.4\,\text{m}\) before a double slit and the screen is \(1.2\,\text{m}\) beyond it. The source is moved \(2\,\text{mm}\) below the axis. How far does the central fringe move, in mm?
- Moving the central fringe towards the side of a raised source.A raised source reaches the upper slit first, so the balance point lies nearer the lower slit. The fringe moves to the opposite side.
- Moving the fringe against the direction of a tilted beam.A beam heading upward delays the upper slit, and the fringe moves up, the way the beam travels.
- Thinking a tilt or a raised source changes the fringe width.\(\beta = \lambda D/d\) holds as before. The whole pattern slides without changing its spacing.
- Giving the centre of circular fringes the order zero.When the sources lie one behind the other, the centre has the full path difference \(d\), of order \(d/\lambda\).