Two sources from one
Two separate lamps never make fringes, because their phases change at random many millions of times a second. Coherent sources must come from one source. The chapter on Young's double slit split one wavefront with two narrow slits. A mirror or a prism can split it too, and this lesson uses both.
Recall what two coherent sources \(S_1\) and \(S_2\), a distance \(d\) apart, do on a screen a distance \(D\) away. At a height \(y\) from the middle of the screen, the path difference is \(yd/D\). Where it is a whole number of wavelengths the fringe is bright, if the two sources are in phase. So the fringes are equally spaced, with the fringe width \[ \beta = \dfrac{\lambda D}{d}. \]
Lloyd's mirror
Place a narrow slit \(S\), lit by one colour, a small height \(h\) above a long flat mirror. Light reaches a screen in two ways: straight from \(S\), and after reflecting at a grazing angle from the mirror. The reflected light seems to come from \(S'\), the image of \(S\), a depth \(h\) below the mirror's plane. So \(S\) and \(S'\) act as two coherent sources, a distance \(d = 2h\) apart. This is Lloyd's mirror.
Take \(h = 0.5\,\text{mm}\), a screen \(D = 1\,\text{m}\) from the slit, and \(\lambda = 600\,\text{nm}\). The sources are \(d = 2h = 1\,\text{mm}\) apart, so \(\beta = 600 \times 10^{-9} \times 1/10^{-3} = 0.6\,\text{mm}\). Halve the height to \(0.25\,\text{mm}\), and \(d\) halves to \(0.5\,\text{mm}\), so the fringe width doubles to \(1.2\,\text{mm}\). Fig 4.1 shows both.
The dark fringe at the edge
Let the mirror reach the screen, meeting it at O. The point O is exactly as far from \(S\) as from \(S'\), so the path difference there is zero. In Young's experiment such a point is bright. Here it is dark. The reflected wave has met a denser medium, glass, so it gained a half wave. At O the two waves arrive turned against each other and cancel.
The half wave shifts the whole pattern by half a fringe. Every place that would be bright in Young's experiment is dark here, and every dark place is bright. Moving the slit closer to the mirror changes \(\beta\), but O stays at equal paths with the same half wave. So the dark fringe stays at O.
In Lloyd's mirror the slit is \(0.4\,\text{mm}\) above the mirror, the screen is \(1.2\,\text{m}\) from the slit and the light has \(\lambda = 500\,\text{nm}\). What is the fringe width, in mm?
In Lloyd's mirror, the point O on the screen is equally far from the slit and from its image. Why is the fringe at O dark?
In 1834 Humphrey Lloyd, a professor at Trinity College in Dublin, tried to make interference fringes with a single mirror. He placed a slit just above a long, flat plate of glass and looked at the light grazing it. The fringes appeared, and the one at the mirror's edge was dark, not bright. Lloyd saw that this was the half wave on reflection, made visible. It was one of the clearest early proofs that reflection from a denser medium turns the wave over.
Fresnel's biprism
A Fresnel biprism is a thin plate of glass with one flat face. Its other face is two very gentle slopes that meet in a straight edge. So it is two thin prisms joined base to base, each with a small refracting angle \(\alpha\), usually a fraction of a degree. A slit \(S\) stands a distance \(a\) in front of it, parallel to the edge.
The chapter on refraction showed that a thin prism of angle \(\alpha\) turns every ray by \(\delta = (\mu - 1)\alpha\) towards its base. The upper half of the biprism has its base at the edge, so it bends its light down. Traced backwards, that light seems to come from \(S_1\), a distance \(a\delta\) above \(S\). The lower half bends its light up, as if from \(S_2\), \(a\delta\) below \(S\).
\[ \delta = (\mu - 1)\alpha \]
Why. each half is a thin prism of angle \(\alpha\); the thin prism formula from the chapter on refraction.
\[ SS_1 = a\delta \]
Why. a ray turned through the small angle \(\delta\) at the biprism, traced back a distance \(a\), lands \(a\delta\) from where it started.
\[ d = S_1S_2 = 2a\delta = 2a(\mu - 1)\alpha \]
Read it. \(S_2\) lies the same distance below \(S\).
The two virtual sources are images of one slit, so they are coherent. Both beams pass through the same thickness of glass near the edge, and neither is reflected. So no half wave separates them, and the centre of the pattern is bright. The screen is \(b\) beyond the biprism, so it is \(D = a + b\) from the sources.
A Fresnel biprism of \(\mu = 1.5\) and refracting angle \(\alpha = 0.01\,\text{rad}\) is \(a = 10\,\text{cm}\) from the slit. The screen is \(b = 90\,\text{cm}\) beyond the biprism and \(\lambda = 600\,\text{nm}\). Find the deviation by each half, the separation of the virtual sources and the fringe width.
Each half is a thin prism: \(\delta = (\mu - 1)\alpha\). The virtual sources are \(a\delta\) above and below the slit.
\(\delta = 0.5 \times 0.01 = 5 \times 10^{-3}\,\text{rad}\), and \(d = 2a\delta = 2 \times 0.1 \times 5 \times 10^{-3} = 10^{-3}\,\text{m}\).
\(D = a + b = 1\,\text{m}\), so \(\beta = \lambda D/d = 600 \times 10^{-9} \times 1/10^{-3} = 6 \times 10^{-4}\,\text{m}\).
Each half deviates the light by \(5 \times 10^{-3}\,\text{rad}\), the sources are \(1\,\text{mm}\) apart and the fringes are \(0.6\,\text{mm}\) wide.
How is \(d\) found in the lab, where the virtual sources cannot be touched? A convex lens is placed between the biprism and the screen. In one position it forms a large image of the two sources, \(d_1\) apart; in the other, a small one, \(d_2\) apart. The two magnifications multiply to 1, as in the displacement method of the chapter on lenses. So \(d = \sqrt{d_1d_2}\): images \(2\,\text{mm}\) and \(0.5\,\text{mm}\) apart give \(d = 1\,\text{mm}\).
A biprism of \(\mu = 1.6\) has refracting angle \(\alpha = 0.01\,\text{rad}\) and is placed \(20\,\text{cm}\) from the slit. What is the separation of the two virtual sources, in mm?
Where the fringes can be seen
Fringes need two waves. Where only one beam reaches the screen, it is lit evenly, with no fringes. So fringes appear only where the two beams overlap, the shaded regions of Fig 4.1 and Fig 4.2.
In Lloyd's mirror the reflected light reaches only the part of the screen above the mirror's plane. It climbs only as high as the rays from the near end of the mirror. With the mirror running from \(10\,\text{cm}\) in front of the slit to the screen, and \(h = 0.5\,\text{mm}\), that is \(4.5\,\text{mm}\) of screen. It holds about seven fringes. Below the mirror's plane the mirror itself blocks the light, and the screen is dark.
Behind a biprism, the two beams cross at the edge and overlap in a narrow wedge. At the screen the wedge is \(2b\delta\) wide. For the example above, \(2 \times 0.9 \times 5 \times 10^{-3} = 9\,\text{mm}\). With fringes \(0.6\,\text{mm}\) wide, the wedge holds \(15\) fringes. Outside it, each half of the screen is lit by one beam only.
Behind a biprism each half deviates the light by \(4 \times 10^{-3}\,\text{rad}\), and the screen is \(1.2\,\text{m}\) beyond the biprism. The fringe width is \(0.8\,\text{mm}\). How many fringes fit in the overlap region on the screen?
The fringes of a Fresnel biprism are seen only in a limited region of the screen. Explain why, and say what the rest of the screen looks like.
Each half of the biprism bends its light towards the edge, by \(\delta = (\mu - 1)\alpha\). The light from the upper half and the light from the lower half cross behind the edge, and they overlap only in a narrow wedge, \(2b\delta\) wide at the screen. Fringes need both waves, so they appear only there. Outside the wedge, each part of the screen gets light from one half only, so it is lit evenly without fringes.
A radio transmitter on a cliff by the sea sends waves to a ship in two ways: straight, and after reflecting from the calm sea. The sea is a mirror for radio waves of a few metres, and it turns them over, as glass turns light. So at some heights above the water the two waves cancel, and the signal fades there. In the 1940s radio astronomers in Australia put this to use. A single aerial on a cliff watched the Sun rising over the sea, together with its reflection.
- Taking the separation of the sources in Lloyd's mirror as \(h\).The sources are the slit and its image, one \(h\) above the mirror and one \(h\) below: \(d = 2h\).
- Expecting a bright fringe at the mirror's edge, as at the centre of Young's pattern.The reflected wave gains a half wave, so equal paths give a dark fringe.
- Measuring \(D\) for a biprism from the biprism.The virtual sources lie in the plane of the slit, so \(D = a + b\).
- Using the refracting angle in degrees in \(d = 2a(\mu - 1)\alpha\).The thin-prism deviation needs \(\alpha\) in radians here, because \(a\delta\) is an arc.