A Slab in One Beam, and White Light

Optical path

Two waves that start in phase can fall out of step in two ways. One may travel farther, as in the last two lessons. Or one may travel the same distance, but through glass, where it is slower. To compare the two effects, count every path as the distance light would cover in air in the same time. For a thickness \(t\) of a medium of refractive index \(\mu\), that distance is the optical path \(\mu t\).

Take a glass slab \(2\,\text{mm}\) thick, of \(\mu = 1.5\). Its optical path is \(1.5 \times 2 = 3\,\text{mm}\). Light of \(600\,\text{nm}\) in air has a wavelength of \(600/1.5 = 400\,\text{nm}\) in the glass, so \(2\,\text{mm}/400\,\text{nm} = 5000\) waves fit inside it. The same \(2\,\text{mm}\) of air holds only about \(3333\) waves. The glass holds \(1667\) extra waves, which is the extra path \((\mu - 1)t = 1\,\text{mm}\) divided by \(600\,\text{nm}\).

The time tells the same story. In the glass, light travels at \(c/\mu = 2 \times 10^8\,\text{m/s}\), so it crosses \(2\,\text{mm}\) in \(2 \times 10^{-3}/(2 \times 10^8)\) \(= 10^{-11}\,\text{s}\). In that time, light in air would cover \(3 \times 10^8 \times 10^{-11}\) \(= 3\,\text{mm}\), exactly the optical path. The frequency does not change in the slab; only the speed and the wavelength fall.

A slab over one slit

Put a thin, transparent sheet of thickness \(t\) and index \(\mu\) over the slit \(S_1\) in Young's experiment. The light from \(S_1\) now has an extra optical path \((\mu - 1)t\). At a point P at height \(y\) above O, the geometric path from \(S_2\) is longer by \(yd/D\). The slab makes the path from \(S_1\) longer by \((\mu - 1)t\).

The central fringe moves to \(y_0\), towards the covered slit. The reason is simple. The slab slows the light from \(S_1\), so equal optical paths need a point closer to \(S_1\), where the geometric path from \(S_1\) is shorter. Every other fringe keeps its place relative to the centre. The positions become \(y_n = y_0 + n\lambda D/d\), so the whole pattern slides, and the fringe width \(\beta = \lambda D/d\) is unchanged.

The second form counts the shift in fringes: the pattern moves by \((\mu - 1)t/\lambda\) fringe widths, one fringe for every wavelength of extra path. Take \(t = 6\,\mu\text{m}\), \(\mu = 1.5\), \(d = 1\,\text{mm}\), \(D = 1\,\text{m}\) and \(\lambda = 600\,\text{nm}\). The shift is \(0.5 \times 6 \times 10^{-6} \times 1/10^{-3}\) \(= 3 \times 10^{-3}\,\text{m}\) \(= 3\,\text{mm}\). With \(\beta = 0.6\,\text{mm}\), that is five fringe widths, and indeed \((\mu - 1)t/\lambda\) \(= 3\,\mu\text{m}/0.6\,\mu\text{m}\) \(= 5\).

Try the slider of the figure above in \(600\,\text{nm}\) light. At \(t = 1.2\,\mu\text{m}\) the centre has moved exactly one fringe width, and the new pattern looks the same as the old one. At \(t = 1.8\,\mu\text{m}\) the shift is one and a half fringe widths. The point O, where the old centre was, is now dark.

Other questions run the formula backwards. If a slab of \(\mu = 1.6\) moves the centre by \(1.2\,\text{mm}\), with \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\), then \(t\) \(= y_0d/((\mu - 1)D)\) \(= 1.2 \times 10^{-3} \times 10^{-3}/0.6\) \(= 2\,\mu\text{m}\). To bring the centre to the old fourth bright fringe with \(600\,\text{nm}\) light, the slab of \(\mu = 1.5\) must add \(4\lambda\). Its thickness is \(t\) \(= 4 \times 600\,\text{nm}/0.5\) \(= 4.8\,\mu\text{m}\).

When both slits are covered, only the difference matters. Slabs \(4\,\mu\text{m}\) thick of \(\mu_1 = 1.5\) over \(S_1\) and \(\mu_2 = 1.6\) over \(S_2\) add \(2.0\,\mu\text{m}\) and \(2.4\,\mu\text{m}\). The net extra path is \(0.4\,\mu\text{m}\) on the \(S_2\) side. So the pattern moves \(0.4 \times 10^{-6} \times 1/10^{-3}\) \(= 0.4\,\text{mm}\) towards \(S_2\).

A slab under water

The whole apparatus can also be put in water, as the last lesson did. A glass slab of \(\mu_g = 1.5\) then replaces \(t\) of water, not \(t\) of air. Its extra optical path is \((\mu_g - \mu_w)t\), while the geometric path difference, measured in the water, is \(\mu_w yd/D\). Equal optical paths give \(y_0 = (\mu_g/\mu_w - 1)tD/d\).

White-light fringes

Now replace the sodium lamp by white light, a mixture of every wavelength from \(400\) to \(700\,\text{nm}\). At the centre O the path difference is zero for every wavelength, so every colour is bright there. Together they make a white central fringe. Away from O each colour has its own fringe width, \(\beta = \lambda D/d\), and the fringes of the different colours no longer fall together.

Violet has the shortest wavelength, so its first bright fringe is closest to the centre. Red, the longest, is farthest out. Each coloured fringe beside the white centre is violet on its inner edge and red on its outer edge. After a few fringes the colours of neighbouring orders overlap so much that the screen looks evenly white again. Only a handful of coloured fringes can be seen.

At a single point P the question turns round: which colours are bright there? With \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\), a point \(1.2\,\text{mm}\) from O has a path difference of \(1.2\,\mu\text{m}\). A colour is bright if \(1200\,\text{nm} = n\lambda\), that is \(\lambda\) \(= 1200, 600, 400, 300, \dots\,\text{nm}\). Only \(600\,\text{nm}\) and \(400\,\text{nm}\) are visible.

Missing colours follow in the same way. At a point where the path difference is \(1.05\,\mu\text{m}\), a colour is dark if \(1050\,\text{nm} = (2n - 1)\lambda/2\), that is \(\lambda\) \(= 2100, 700, 420, 300, \dots\,\text{nm}\). The visible colours missing there are \(700\,\text{nm}\) and \(420\,\text{nm}\), the deep red and the violet. The point looks green.

The one fringe that cannot be mistaken

With a sodium lamp, every bright fringe looks the same. Put a slab over one slit, and the pattern jumps by several fringe widths while you watch. Afterwards it looks exactly as before. You cannot tell which fringe is now the centre, so you cannot count the shift. Only the fraction of a fringe beyond a whole number could be seen.

White light solves this. Only the zero-order fringe is white; every other fringe is coloured. The white fringe is the fringe of zero optical path difference, so the slab carries it to \(y_0 = (\mu - 1)tD/d\). A slab of \(\mu = 1.5\) and \(t = 6\,\mu\text{m}\), with \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\), carries it \(3\,\text{mm}\) towards \(S_1\). The coloured fringes move with it. Try the white-light chip of the slab figure above.