The intensity across the screen
The last lesson found where the fringes are. It did not say how the brightness changes between them. At a height \(y\) the path difference is \(yd/D\), so the phase difference is \(\phi = (2\pi/\lambda)(yd/D)\). Since \(\beta = \lambda D/d\), this is simply \(\phi = 2\pi y/\beta\). Each fringe width of height adds one full cycle of phase.
For two equal slits, each giving an intensity \(I_0\) alone, the third lesson found \(I = 4I_0\cos^2(\phi/2)\). Put in \(\phi = 2\pi y/\beta\), and write \(I_{\max} = 4I_0\), the intensity at the centre.
The intensity is \(I_{\max}\) at \(y = 0, \pm\beta, \pm2\beta\) and zero at \(y = \pm\beta/2, \pm3\beta/2\). The fringes are not sharp stripes. The brightness rises and falls smoothly, like the square of a cosine, as Fig 5.1 shows.
Read Fig 5.1 at a few points, with \(\beta = 0.6\,\text{mm}\). At \(y = 0.15\,\text{mm}\), \(y/\beta = \tfrac14\) and \(I\) \(= I_{\max}\cos^2 45°\) \(= \tfrac12 I_{\max}\). At \(y = 0.2\,\text{mm}\), \(y/\beta = \tfrac13\) and \(I\) \(= I_{\max}\cos^2 60°\) \(= \tfrac14 I_{\max}\). At \(y = 0.1\,\text{mm}\), \(y/\beta = \tfrac16\) and \(I\) \(= I_{\max}\cos^2 30°\) \(= \tfrac34 I_{\max}\). The intensity falls to half at \(\beta/4\), a quarter of a fringe width from the centre of a bright fringe.
The fringe width is \(0.6\,\text{mm}\). What is the intensity, as a fraction of \(I_{\max}\), at a point \(0.2\,\text{mm}\) from the central maximum?
A graph of the intensity is often all an experiment gives. Suppose neighbouring maxima on such a graph are \(0.5\,\text{mm}\) apart, for \(500\,\text{nm}\) light and a screen \(1\,\text{m}\) away. The fringe width is \(0.5\,\text{mm}\), and the slit separation is \(d\) \(= \lambda D/\beta\) \(= 500 \times 10^{-9} \times 1/(0.5 \times 10^{-3})\) \(= 10^{-3}\,\text{m}\) \(= 1\,\text{mm}\).
The fringe width is \(0.6\,\text{mm}\). How many bright fringes lie within \(6\,\text{mm}\) on either side of the centre, counting the central fringe?
Two colours on one screen
Light of two wavelengths through the same slits makes two patterns at once. Both are bright at the centre, where the path difference is zero for every wavelength. Away from the centre the fringe widths differ, so the two sets of fringes drift out of step. A bright fringe of one colour falls on a bright fringe of the other where their heights agree.
Take \(600\,\text{nm}\) and \(400\,\text{nm}\) with \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\). The fringe widths are \(0.6\,\text{mm}\) and \(0.4\,\text{mm}\). The condition \(600n_1 = 400n_2\) is \(3n_1 = 2n_2\), whose smallest solution is \(n_1 = 2\), \(n_2 = 3\). So the second bright fringe of one falls on the third of the other, at \(2 \times 0.6 = 1.2\,\text{mm}\) from the centre.
Light of \(480\,\text{nm}\) and \(600\,\text{nm}\) falls on the same double slit. The \(n\)th bright fringe of \(600\,\text{nm}\) falls on the \((n + 1)\)th bright fringe of \(480\,\text{nm}\). Find \(n\), and the height of this fringe for \(d = 1\,\text{mm}\), \(D = 1\,\text{m}\).
The two bright fringes are at the same height, so their path differences are equal.
\(600n = 480(n + 1)\).
\(120n = 480\), so \(n = 4\). The height is \(4 \times 0.6 = 2.4\,\text{mm}\).
The fourth bright fringe of \(600\,\text{nm}\) falls on the fifth of \(480\,\text{nm}\), \(2.4\,\text{mm}\) from the centre.
Light of \(500\,\text{nm}\) and \(600\,\text{nm}\) falls on two slits \(1\,\text{mm}\) apart, with the screen \(1\,\text{m}\) away. How far from the centre do bright fringes of the two colours first fall together again, in mm?
A sodium lamp, the yellow street light of Indian towns, gives two wavelengths very close together, \(589.0\,\text{nm}\) and \(589.6\,\text{nm}\). Near the centre their fringes lie almost exactly on top of each other. Each order moves them \(0.6\,\text{nm}\) of path further out of step. After about \(589.3/(2 \times 0.6) \approx 491\) orders, a bright fringe of one falls on a dark fringe of the other, and the fringes fade. Watching where fringes fade and return is one way to measure how close two wavelengths are.
The experiment under water
Now fill the space between the slits and the screen with water, \(\mu = \tfrac43\). The frequency of the light cannot change, but its speed and its wavelength both fall by \(\mu\). The path difference is still \(yd/D\), but it must now be counted in the shorter wavelength \(\lambda/\mu\). So the fringe width becomes \(\beta' = (\lambda/\mu)D/d = \beta/\mu\).
A fringe width of \(0.6\,\text{mm}\) in air becomes \(0.6 \div \tfrac43 = 0.45\,\text{mm}\) under water. The fringes crowd together, but the pattern does not shift: the centre is still bright, because the two paths are still equal there. In a liquid of \(\mu = 1.5\), light of \(600\,\text{nm}\) has a wavelength of \(400\,\text{nm}\). With \(d = 1\,\text{mm}\) and \(D = 1\,\text{m}\) the fringes are then \(0.4\,\text{mm}\) apart, and the angular fringe width is \(4 \times 10^{-4}\,\text{rad}\).
When a double slit apparatus is placed in a liquid, the fringe width falls from \(0.6\,\text{mm}\) to \(0.45\,\text{mm}\). What is the refractive index of the liquid?
When the slits are not equal
The formula \(I = I_{\max}\cos^2(\pi y/\beta)\) assumed two equal slits. Suppose a grey filter over \(S_2\) cuts its intensity to a quarter. The amplitude from \(S_2\) is then half that from \(S_1\), since intensity goes as the square of the amplitude. The bright fringes get \((a + a/2)^2 = \tfrac94a^2\), and the dark ones \((a - a/2)^2 = \tfrac14a^2\).
The ratio is \(I_{\max}/I_{\min} = (3/2)^2/(1/2)^2 = 9\). The dark fringes are no longer dark; the pattern loses contrast. The positions of the fringes do not change, because they depend only on the path difference. Narrow slits of unequal width act the same way: each slit's intensity grows in proportion to its width. Widths in the ratio \(4 : 1\) give intensities \(4 : 1\), amplitudes \(2 : 1\) and \(I_{\max}/I_{\min} = 9\).
In a double slit experiment with equal slits, the intensity at the centre O is \(4I_0\). The slit \(S_2\) is now closed completely. What happens to the fringes, and what is the intensity at O?
With one slit there is no second wave to interfere with, so the fringes disappear. The screen is lit almost evenly by the one slit.
Each slit alone gave \(I_0\) at O. With both slits in phase the amplitudes added, \(2a\), which gave \(4I_0\).
With \(S_2\) closed only the amplitude \(a\) is left, so the intensity at O is \(I_0\), a quarter of what it was.
A single slit is not quite even across the screen. Its light spreads by diffraction, and it is brightest opposite the slit and fades far out. The chapter on diffraction and polarisation studies that pattern. Over the few millimetres of Young's fringes, it is nearly even.
A filter over one slit of a double slit reduces the intensity from that slit to one quarter of its earlier value. What is \(I_{\max}/I_{\min}\) of the new pattern?
One slit of a double slit is made narrower than the other. Explain what happens to the bright fringes, the dark fringes and the positions of the fringes.
The narrower slit sends less light, so the two waves reaching any point have unequal amplitudes. At the bright fringes the amplitudes still add, but to less than twice the larger one. At the dark fringes they do not cancel completely, so the minima are grey instead of dark, and the contrast falls. The positions of the fringes depend only on the path difference, \(y = n\lambda D/d\), so they do not move and the fringe width is unchanged.
- Treating the fringes as sharp stripes, all of one brightness.The intensity varies smoothly as \(\cos^2(\pi y/\beta)\); it is half the maximum a quarter of a fringe width from each bright centre.
- Dividing the fringe width by \(\mu\) and also moving the central fringe when the apparatus is put in water.Only the wavelength changes. The fringes crowd by \(\mu\), but the centre, with equal paths, stays at O.
- Taking \(I_{\max}/I_{\min}\) as the ratio of the two intensities.Take square roots to get the amplitudes, add and subtract them, then square the ratio.
- Expecting the fringes to move when one slit is dimmed.Positions depend only on the path difference. Dimming one slit greys the dark fringes but does not move them.