The Workbook

Light may reflect from the surface of a denser medium, one of higher \(\mu\). The reflected wave then has its phase changed by \(\pi\), equivalent to an extra path of \(\dfrac{\lambda}{2}\). Reflection from a rarer medium gives no phase change. For a soap film in air the top surface is air to soap, denser, so that reflection gets the \(\pi\) shift. The bottom surface is soap to air, rarer, so no shift.

The wave reflected from the bottom travels an extra optical path \(2\mu t\cos r\) inside the film. The top reflection alone has the \(\pi\) shift, so the effective path difference is \(2\mu t\cos r \pm \dfrac{\lambda}{2}\). At normal incidence \(\cos r = 1\). Bright: \(2\mu t = (2n - 1)\dfrac{\lambda}{2}\). Dark: \(2\mu t = n\lambda\).

1. \(2\mu t = 2 \times 1.5 \times 400 = 1200\) nm.

2. Bright when \(1200 = (2n - 1)\dfrac{\lambda}{2}\), so \(\lambda = \dfrac{2400}{2n - 1} = 2400, 800, 480, 343, \ldots\) nm. The visible one is \(480\) nm.

3. Dark when \(1200 = n\lambda\), so \(\lambda = 1200, 600, 400, 300, \ldots\) nm. Missing in the visible range: \(600\) nm and \(400\) nm.

Bright: \(2\mu t = (2n - 1)\dfrac{\lambda}{2}\); the least \(t\) has \(n = 1\): \(t = \dfrac{\lambda}{4\mu} = \dfrac{600}{6} = 100\) nm. Dark: \(2\mu t = n\lambda\); with \(n = 1\), \(t = \dfrac{\lambda}{2\mu} = 200\) nm.

When \(t \ll \lambda\) the path \(2\mu t\) is almost zero, so the only difference between the two reflected waves is the \(\pi\) shift of the top reflection. They cancel and the film looks black. Neither transmitted wave has a \(\pi\) shift, so in transmitted light the same part looks bright. The reflected and transmitted patterns are complementary.

\(2\mu t\cos r = 2 \times 1.5 \times 500 \times 0.6 = 900\) nm. Bright: \(\lambda = \dfrac{1800}{2n - 1} = 1800, 600, 360, \ldots\) nm, so \(600\) nm is strongly reflected. Dark: \(\lambda = \dfrac{900}{n} = 900, 450, 300, \ldots\) nm, so \(450\) nm is missing.

Air to coating and coating to glass are both rarer to denser, so both reflections have the \(\pi\) shift and it cancels. The two reflected waves cancel when \(2\mu_c t = \dfrac{\lambda}{2}\). \(t = \dfrac{\lambda}{4\mu_c} = \dfrac{552}{4 \times 1.38} = 100\) nm.

Let the centre of curvature be \(C\), a distance \(R\) above \(O\). The point on the lens surface at horizontal distance \(r\) is at height \(t\) above the plate. In the right triangle with \(C\): \(r^2 + (R - t)^2 = R^2\), so \(r^2 = 2Rt - t^2\). For \(t \ll R\) the \(t^2\) term is negligible and \(r^2 = 2Rt\), giving \(t = \dfrac{r^2}{2R}\).

At the point of contact the air film has zero thickness, so the geometric path difference is zero. But the reflection at the lower surface (air to glass) has the \(\pi\) shift and the upper one (glass to air) does not. The two reflected waves are in antiphase and the centre is dark. In transmitted light there is no net phase shift, so the centre is bright.

For the air film \(\mu = 1\) and normal incidence. Dark rings (counting the \(\pi\) shift) need \(2t = n\lambda\). With \(t = \dfrac{r^2}{2R}\): \(\dfrac{r_n^2}{R} = n\lambda\), so \(r_n = \sqrt{n\lambda R}\). For \(n = 5\): \(r_5^2 = 5 \times 500 \times 10^{-9} \times 1.6 = 4 \times 10^{-6}\) m\(^2\), so \(r_5 = 2 \times 10^{-3}\) m \(= 2\) mm. Diameter \(= 4\) mm.

\(D_n^2 = 4n\lambda R\), so \(D_{20}^2 - D_5^2 = 4(20 - 5)\lambda R = 60\lambda R\). \(\lambda = \dfrac{(8^2 - 4^2) \times 10^{-6}}{60 \times 1.6} = \dfrac{48 \times 10^{-6}}{96} = 5 \times 10^{-7}\) m \(= 500\) nm.

1. Consecutive dark rings differ by a fixed step in thickness, \(\dfrac{\lambda}{2}\). Since \(t \propto r^2\), equal steps in \(t\) need smaller and smaller steps in \(r\) as \(r\) grows: \(r_n \propto \sqrt n\). So the rings crowd together outward.

2. With water the condition becomes \(2\mu t = n\lambda\), so \(r_n^2 = \dfrac{n\lambda R}{\mu}\). Every radius shrinks by the factor \(\dfrac{1}{\sqrt\mu} = \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt3}{2} \approx 0.87\).

1. Light reaching the screen directly from \(S\) interferes with light reflected from the mirror. The reflected light appears to come from the image \(S'\), at \(h\) below the mirror plane. The two coherent sources are \(S\) and \(S'\), separated by \(d = 2h = 1\) mm.

2. \(\beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9} \times 1}{10^{-3}} = 0.6\) mm.

At \(O\) the paths from \(S\) and \(S'\) are equal. But the reflected wave has an extra phase of \(\pi\) from the mirror (air to glass). So the two waves cancel and the fringe at \(O\) is dark, unlike the bright centre of a YDSE. Fringes appear only where both direct and reflected light arrive, which is above the mirror plane. Below it only direct light reaches the screen.

1. Each half of the biprism is a thin prism and deviates light by \(\delta = (\mu - 1)\alpha\). The virtual source is displaced by \(a\delta\), so \(d = 2a(\mu - 1)\alpha = 2 \times 0.1 \times 0.5 \times 0.01 = 10^{-3}\) m \(= 1\) mm.

2. The screen is \(D = a + b = 1\) m from the virtual sources. \(\beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9} \times 1}{10^{-3}} = 0.6\) mm.

\(\delta = (\mu - 1)\alpha = 0.5 \times 0.01 = 5 \times 10^{-3}\) rad. The upper half bends its light downward and the lower half bends its light upward. The two refracted beams cross and overlap only in a wedge-shaped region behind the biprism. Fringes need both beams, so they are seen only where the beams overlap on the screen.

\(d = 2h = 0.5\) mm, half the earlier value, so \(\beta = \dfrac{\lambda D}{d}\) doubles to \(1.2\) mm. The point \(O\) still has equal paths and the same \(\pi\) shift, so the dark fringe stays at \(O\).

The set of points with \(S_2P - S_1P = \) constant is a hyperboloid of revolution with the two sources as foci and \(S_1S_2\) as axis. A screen parallel to \(S_1S_2\) cuts these surfaces in hyperbolae, so the fringes are hyperbolae. Near the centre they are almost straight.

1. The screen is perpendicular to the axis \(S_1S_2\), and the hyperboloids are symmetric about that axis. The screen cuts them in circles, so the fringes are concentric circles about the axis.

2. At the centre the path difference is the full separation, \(1\) mm. \(n = \dfrac{10^{-3}}{500 \times 10^{-9}} = 2000\), an integer, so the centre is bright, of order \(2000\).

The fringes are hyperbolae with the slits as foci. Close to the central axis the hyperbolae have a very large radius of curvature. Over the few millimetres of the visible pattern they do not bend noticeably. So the fringes look like straight lines parallel to the slits.

1. The beam travels upward, so its wavefront, perpendicular to the beam, reaches the lower slit \(S_2\) first. The wave at \(S_1\) lags by the extra path \(d\sin\alpha\).

2. For a point at height \(y\), the path from \(S_1\) to the screen is shorter by \(\dfrac{yd}{D}\). Total path difference \(= d\sin\alpha - \dfrac{yd}{D}\). It is zero at \(\dfrac{y_0 d}{D} = d\sin\alpha\), so \(y_0 = D\sin\alpha \approx D\alpha\). The central fringe moves upward, in the direction the beam is travelling.

\(y_0 = D\alpha = 1 \times 2 \times 10^{-3} = 2\) mm. \(\beta = \dfrac{\lambda D}{d} = \dfrac{500 \times 10^{-9}}{10^{-3}} = 0.5\) mm. So the shift is \(\dfrac{2}{0.5} = 4\) fringe widths.

The tilt makes the wave at \(S_1\) lag by \(d\sin\alpha = 10^{-3} \times 2 \times 10^{-3} = 2\) μm. To cancel this the path from \(S_2\) must be lengthened by the same amount, so the slab goes before \(S_2\). \((\mu - 1)t = 2\) μm gives \(t = \dfrac{2}{0.5} = 4\) μm.

With \(S\) at height \(s\) above the axis at distance \(a\) from the slits, \(S\) is closer to \(S_1\) by about \(\dfrac{sd}{a}\). The wave at \(S_2\) now lags, so the central fringe moves to the \(S_2\) side, below \(O\). Its position: \(\dfrac{y_0 d}{D} = \dfrac{sd}{a}\), so \(y_0 = \dfrac{sD}{a} = \dfrac{1 \times 1}{0.5} = 2\) mm below \(O\). The pattern moves opposite to the source.