The Workbook

Every point on a wavefront acts as a fresh source of secondary wavelets. These spread out in all directions with the speed of the wave in that medium. The new wavefront at a later time is the surface that touches all these wavelets on their forward side, their common envelope.

(i) Spheres centred on the source. (ii) Cylinders with the line as axis. (iii) Planes: a small part of a sphere of enormous radius is flat.

1. In the time \(t\) that the wave takes to travel from \(B\) to \(C\), the wavelet from \(A\) spreads a distance \(vt\). So \(AD = BC = vt\).

2. Triangles \(ABC\) and \(ADC\) have the common side \(AC\), \(AD = BC\), and right angles at \(B\) and \(D\). They are congruent, so \(\angle BAC = \angle DCA\). These angles equal the angles of incidence and reflection, so \(i = r\).

1. In triangle \(ABC\), \(\sin i = \dfrac{BC}{AC} = \dfrac{v_1t}{AC}\). In triangle \(ADC\), \(\sin r = \dfrac{AD}{AC} = \dfrac{v_2t}{AC}\).

2. Dividing, \(\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}\), a constant for the pair of media. This is Snell's law, and it identifies the refractive index as \(\mu = \dfrac{v_1}{v_2}\), the ratio of the wave speeds.

The emerging wavefront is spherical and converges on the focus. Light travels more slowly in glass, and the centre of the lens is thickest, so the middle of the wavefront is delayed most. The edges get ahead and the plane wavefront bends into a sphere curving in towards the axis.

\(v = \dfrac{c}{\mu} = 2 \times 10^{8}\) m/s. \(\lambda' = \dfrac{\lambda}{\mu} = 400\) nm. The frequency is fixed by the source: \(f = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{6 \times 10^{-7}} = 5 \times 10^{14}\) Hz, the same in both media.

Coherent sources have the same frequency and a phase difference that stays constant in time. In a bulb, atoms emit short wave trains independently, so the phase changes at random about \(10^{8}\) times a second. The interference pattern from two bulbs shifts just as fast and only a uniform average is seen.

\(\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x\). Here \(\Delta\phi = \dfrac{2\pi}{600} \times 150 = \dfrac{\pi}{2}\).

(i) \(\Delta\phi = \pi\): crest meets trough, destructive. (ii) \(\Delta\phi = 2\pi\): in step, constructive. (iii) \(\Delta\phi = 3\pi\): destructive again. Odd multiples of \(\dfrac{\lambda}{2}\) give darkness, whole wavelengths give brightness.

\(y = a\sin\omega t + a\sin(\omega t + \phi) = 2a\cos\dfrac{\phi}{2}\sin\left(\omega t + \dfrac{\phi}{2}\right)\), amplitude \(A = 2a\cos\dfrac{\phi}{2}\). For \(\phi = 60^\circ\): \(A = 2a\cos 30^\circ = \sqrt3a\), and \(I \propto A^2 = 3a^2\), so \(I = 3I_0\).

Optical path is the distance light would travel in vacuum in the same time: \(\mu t\). Through the slab: \(1.5 \times 1 = 1.5\) mm. Extra path \((\mu - 1)t = 0.5\) mm \(= 5 \times 10^{-4}\) m, which is \(\dfrac{5 \times 10^{-4}}{6 \times 10^{-7}} \approx 833\) wavelengths.

1. \(\Delta x = 6.2 - 5.0 = 1.2\) μm \(= 1200\) nm \(= 2\lambda\).

2. A whole number of wavelengths: the waves arrive in phase and \(P\) is bright.

\(I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi\). With \(I_1 = I_2 = I_0\): \(I = 2I_0(1 + \cos\phi) = 4I_0\cos^2\dfrac{\phi}{2}\). At \(\phi = 0\): \(4I_0\). At \(\dfrac{\pi}{2}\): \(2I_0\). At \(\pi\): \(0\).

\(\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{a_1 + a_2}{a_1 - a_2}\right)^2 = \left(\dfrac{7}{1}\right)^2 = 49\).

\(I \propto a^2\), so \(\dfrac{a_1}{a_2} = \sqrt9 = 3\). \(\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{3 + 1}{3 - 1}\right)^2 = 4\).

\(\dfrac{a_1 + a_2}{a_1 - a_2} = \sqrt{25} = 5\), so \(a_1 + a_2 = 5a_1 - 5a_2\), giving \(4a_1 = 6a_2\) and \(\dfrac{a_1}{a_2} = \dfrac{3}{2}\). Intensities: \(\dfrac{9}{4}\).

1. \(I = 4I_0\cos^2\dfrac{\phi}{2}\): maxima \(4I_0\) at \(\phi = 0, 2\pi, 4\pi\) and zeros at \(\pi, 3\pi\).

2. \(\phi = \dfrac{2\pi}{3}\): \(I = 4I_0\cos^2 60^\circ = 4I_0 \times \dfrac{1}{4} = I_0\).

\(I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi\). Over the pattern \(\phi\) takes all values equally and \(\cos\phi\) averages to zero, so \(\langle I\rangle = I_1 + I_2\), exactly the sum of the two separate intensities. Interference only redistributes the energy from the dark places to the bright ones.

\(I_{\max} = \left(\sqrt{4I_0} + \sqrt{I_0}\right)^2 = (2 + 1)^2I_0 = 9I_0\). \(I_{\min} = (2 - 1)^2I_0 = I_0\). The dark fringes are not fully dark because the amplitudes are unequal.

\(\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c} = \dfrac{3 \times 10^{6}}{3 \times 10^{8}} = 0.01\). \(\Delta\lambda = 5\) nm, towards longer wavelengths for a receding source: the line is seen at \(505\) nm, a red shift.

1. The wavelength is longer, a red shift, so the galaxy is receding.

2. \(v = c\dfrac{\Delta\lambda}{\lambda} = 3 \times 10^{8} \times \dfrac{3}{600} = 1.5 \times 10^{6}\) m/s.

\(\Delta\lambda = \lambda\dfrac{v}{c} = 500 \times \dfrac{6 \times 10^{5}}{3 \times 10^{8}} = 1\) nm. The source approaches, so the wavelength shortens to \(499\) nm: a blue shift of \(1\) nm.

The exact formula involves relativistic time dilation and terms in \(\dfrac{v^2}{c^2}\). For \(v \ll c\) these are negligible and the simple first-order result \(\dfrac{\Delta\lambda}{\lambda} = \dfrac{v}{c}\) holds. For speeds comparable with \(c\) it would be badly wrong.

Intensity is proportional to the square of the resultant amplitude. In phase the amplitudes add, \(A_{\max} = a_1 + a_2\); in antiphase they subtract, \(A_{\min} = a_1 - a_2\). So \(\dfrac{I_{\max}}{I_{\min}} = \dfrac{(a_1 + a_2)^2}{(a_1 - a_2)^2}\).

Light from a narrow slit \(S\) falls on two close, parallel slits \(S_1\) and \(S_2\). The two slits act as coherent sources and their light overlaps on a distant screen. The screen shows alternate bright and dark bands, called fringes, parallel to the slits and equally spaced.

Both \(S_1\) and \(S_2\) receive light from the same wavefront from \(S\). So any random change of phase in the source appears at both slits together. The phase difference between them stays constant: they are coherent. Two separate lamps change phase independently and at random. The fringes then shift millions of times a second and the screen looks uniformly lit.

1. Drop the perpendicular \(S_1N\) on \(S_2P\). For a distant screen \(S_1P \approx NP\), so the path difference is \(S_2N = d\sin\theta\).

2. For small \(\theta\), \(\sin\theta \approx \tan\theta = \dfrac{y}{D}\). So \(\Delta x = d\sin\theta = \dfrac{yd}{D}\).

Bright: the waves arrive in phase, \(\Delta x = n\lambda\) with \(n = 0, 1, 2, \ldots\). Dark: they arrive in antiphase, \(\Delta x = (2n - 1)\dfrac{\lambda}{2}\) with \(n = 1, 2, 3, \ldots\).

\(O\) is equidistant from the two slits, so the path difference there is zero for every wavelength. The waves arrive in phase and the fringe is bright. It is the zero-order fringe, \(n = 0\).

\(\beta = \dfrac{\lambda D}{d} = \dfrac{600 \times 10^{-9} \times 1}{10^{-3}} = 6 \times 10^{-4}\) m \(= 0.6\) mm.

1. \(\beta \propto \dfrac{1}{d}\), so doubling \(d\) halves \(\beta\): \(0.3\) mm.

2. \(\beta \propto D\), so doubling \(D\) doubles \(\beta\): \(1.2\) mm.

\(\beta \propto \lambda\) for the same \(d\) and \(D\). \(\beta' = 0.6 \times \dfrac{450}{600} = 0.45\) mm. A shorter wavelength gives narrower fringes, so they come closer.

The angular fringe width is the angle that one fringe subtends at the slits, \(\theta_0 = \dfrac{\beta}{D} = \dfrac{\lambda}{d}\). Here \(\theta_0 = \dfrac{600 \times 10^{-9}}{10^{-3}} = 6 \times 10^{-4}\) rad. It does not contain \(D\), so moving the screen leaves it unchanged.

1. From the first to the sixth bright fringe there are \(5\) fringe widths. \(\beta = \dfrac{3.0}{5} = 0.6\) mm.

2. The \(n\)th dark fringe is at \((2n - 1)\dfrac{\beta}{2}\). For \(n = 4\): \(\dfrac{7}{2} \times 0.6 = 2.1\) mm.

\(\lambda = \dfrac{\beta d}{D} = \dfrac{0.4 \times 10^{-3} \times 1.5 \times 10^{-3}}{1} = 6 \times 10^{-7}\) m \(= 600\) nm.

1. \(y_3 = 3\beta = 1.8\) mm.

2. \(y = (2 \times 3 - 1)\dfrac{\beta}{2} = \dfrac{5}{2} \times 0.6 = 1.5\) mm.

1. Third bright at \(3\beta = 1.8\) mm; second dark at \(\dfrac{3}{2}\beta = 0.9\) mm. Same side: \(1.8 - 0.9 = 0.9\) mm.

2. Opposite sides: \(1.8 + 0.9 = 2.7\) mm.

1. \(\Delta x = \dfrac{yd}{D} = \dfrac{1.5 \times 10^{-3} \times 10^{-3}}{1} = 1.5 \times 10^{-6}\) m \(= 1.5\) μm.

2. \(\dfrac{\Delta x}{\lambda} = \dfrac{1.5}{0.6} = 2.5\), so \(\Delta x = \dfrac{5\lambda}{2} = (2 \times 3 - 1)\dfrac{\lambda}{2}\). \(P\) is dark; it lies on the third dark fringe.

The fringe widths are \(0.6\) mm and \(0.4\) mm. Bright fringes coincide where \(0.6\,n_1 = 0.4\,n_2\), that is \(3n_1 = 2n_2\). The smallest solution is \(n_1 = 2\), \(n_2 = 3\). Both give \(y = 1.2\) mm.

\(y_n(600) = y_{n+1}(480)\) gives \(600n = 480(n + 1)\), so \(120n = 480\) and \(n = 4\). The fourth bright fringe of \(600\) nm falls on the fifth of \(480\) nm.

Bright fringes are at \(y = 0, \pm\beta, \pm 2\beta, \ldots\). Within \(6\) mm on one side there are \(\dfrac{6}{0.6} = 10\) fringes beyond the centre. Total: \(10 + 10 + 1 = 21\).

\(I = I_{\max}\cos^2\dfrac{\phi}{2}\) with \(\phi = \dfrac{2\pi}{\lambda}\dfrac{yd}{D} = \dfrac{2\pi y}{\beta}\). So \(I = I_{\max}\cos^2\dfrac{\pi y}{\beta}\). It is \(I_{\max}\) at \(y = 0, \pm\beta, \ldots\) and zero midway between.

1. \(\dfrac{y}{\beta} = \dfrac{0.15}{0.6} = \dfrac{1}{4}\): \(I = I_{\max}\cos^2\dfrac{\pi}{4} = \dfrac{1}{2}I_{\max}\).

2. \(\dfrac{y}{\beta} = \dfrac{1}{3}\): \(I = I_{\max}\cos^2\dfrac{\pi}{3} = \dfrac{1}{4}I_{\max}\).

1. Adjacent maxima are at \(0\) and \(0.5\) mm, so \(\beta = 0.5\) mm.

2. \(d = \dfrac{\lambda D}{\beta} = \dfrac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 10^{-3}\) m \(= 1\) mm.

\(\cos^2\dfrac{\pi y}{\beta} = \dfrac{1}{2}\) gives \(\dfrac{\pi y}{\beta} = \dfrac{\pi}{4}\), so \(y = \dfrac{\beta}{4} = 0.15\) mm.

\(\beta = \dfrac{\lambda D}{d}\), so \(\Delta\beta = \dfrac{\lambda\,\Delta D}{d}\). \(\lambda = \dfrac{\Delta\beta \, d}{\Delta D} = \dfrac{0.3 \times 10^{-3} \times 10^{-3}}{0.5} = 6 \times 10^{-7}\) m \(= 600\) nm.

1. Slope \(= \dfrac{\beta}{D} = \dfrac{0.5 \times 10^{-3}}{1} = 5 \times 10^{-4}\).

2. \(\beta = \dfrac{\lambda}{d}D\), so the slope is \(\dfrac{\lambda}{d}\). \(\lambda = 5 \times 10^{-4} \times 10^{-3} = 5 \times 10^{-7}\) m \(= 500\) nm.

The exact condition for the \(n\)th bright fringe is \(d\sin\theta = n\lambda\) and its position is \(y = D\tan\theta\). Near the centre \(\sin\theta \approx \tan\theta\), so \(y \propto n\) and the fringes are equally spaced. Far from the centre \(\tan\theta\) grows faster than \(\sin\theta\). Equal steps in \(\sin\theta\) then give increasing steps in \(y\), so the fringes become wider.

The optical path is the distance light would cover in air (or vacuum) in the time it takes to cross the medium. For a geometric path \(x\) in a medium of index \(\mu\) the optical path is \(\mu x\). Here it is \(1.5 \times 2 = 3\) mm.

1. \(\lambda_g = \dfrac{\lambda}{\mu} = \dfrac{600}{1.5} = 400\) nm.

2. Number of waves \(= \dfrac{t}{\lambda_g} = \dfrac{2 \times 10^{-3}}{400 \times 10^{-9}} = 5000\). The same length of air holds only \(3333\) waves.

Optical paths: glass \(\mu t = 3\) mm, air \(t = 2\) mm. Path difference \(= \mu t - t = (\mu - 1)t = 1\) mm.

Speed in glass \(v = \dfrac{c}{\mu} = 2 \times 10^{8}\) m/s. Time \(= \dfrac{t}{v} = \dfrac{2 \times 10^{-3}}{2 \times 10^{8}} = 10^{-11}\) s. This equals the optical path divided by \(c\).

The frequency is fixed by the source and cannot change at a boundary. The speed falls to \(\dfrac{c}{\mu}\) and the wavelength to \(\dfrac{\lambda}{\mu}\). The beam through the slab gains an extra optical path \((\mu - 1)t\), that is an extra phase \(\dfrac{2\pi}{\lambda}(\mu - 1)t\).

1. The geometric path difference is \(S_2P - S_1P = \dfrac{yd}{D}\). The slab adds \((\mu - 1)t\) to the path from \(S_1\). Net path difference \(\Delta x = \dfrac{yd}{D} - (\mu - 1)t\).

2. The central fringe is where \(\Delta x = 0\): \(\dfrac{y_0 d}{D} = (\mu - 1)t\), so \(y_0 = \dfrac{D(\mu - 1)t}{d}\). It moves towards \(S_1\), the slab side.

Shift \(= \dfrac{D(\mu - 1)t}{d} = \dfrac{1 \times 0.5 \times 6 \times 10^{-6}}{10^{-3}} = 3 \times 10^{-3}\) m \(= 3\) mm. The fringe width is \(\beta = \dfrac{\lambda D}{d} = 0.6\) mm, so the shift is \(\dfrac{3}{0.6} = 5\) fringe widths. Equivalently \(\dfrac{(\mu - 1)t}{\lambda} = \dfrac{3 \times 10^{-6}}{600 \times 10^{-9}} = 5\).

The slab lengthens the optical path from \(S_1\). To restore zero path difference the point must move closer to \(S_1\), so the whole pattern shifts towards \(S_1\). The fringe width \(\beta = \dfrac{\lambda D}{d}\) does not contain \(t\) or \(\mu\), so it stays the same.

Shift \(= \dfrac{D(\mu - 1)t}{d}\), so \(t = \dfrac{\text{shift} \times d}{D(\mu - 1)} = \dfrac{1.2 \times 10^{-3} \times 10^{-3}}{1 \times 0.6} = 2 \times 10^{-6}\) m \(= 2\) μm.

The fourth bright fringe is \(4\beta\) from the centre. The shift is \(4\beta\) when \((\mu - 1)t = 4\lambda\). \(t = \dfrac{4 \times 600 \times 10^{-9}}{0.5} = 4.8 \times 10^{-6}\) m \(= 4.8\) μm.

At \(O\) the geometric paths are equal, so the path difference is just the slab's contribution, \((\mu - 1)t = 0.5 \times 1.8 = 0.9\) μm \(= 1.5\lambda\). An odd multiple of \(\dfrac{\lambda}{2}\) gives a dark point. \(1.5\lambda = (2 \times 2 - 1)\dfrac{\lambda}{2}\), so \(O\) lies on the second dark fringe of the shifted pattern.

1. Extra optical paths: \(S_1\) side \((\mu_1 - 1)t = 2\) μm; \(S_2\) side \((\mu_2 - 1)t = 2.4\) μm. Net extra path \(= 0.4\) μm on the \(S_2\) side.

2. Shift \(= \dfrac{D \times 0.4 \times 10^{-6}}{10^{-3}} = 0.4 \times 10^{-3}\) m \(= 0.4\) mm. The thicker optical path is on the \(S_2\) side, so the pattern moves towards \(S_2\).

In water the wavelength becomes \(\lambda' = \dfrac{\lambda}{\mu}\), so the fringe width becomes \(\beta' = \dfrac{\lambda' D}{d} = \dfrac{\beta}{\mu}\). \(\beta' = \dfrac{0.6}{4/3} = 0.45\) mm. The fringes come closer; the pattern does not shift.

1. \(\lambda' = \dfrac{600}{1.5} = 400\) nm. \(\beta = \dfrac{\lambda' D}{d} = \dfrac{400 \times 10^{-9} \times 1}{10^{-3}} = 0.4\) mm.

2. Angular fringe width \(= \dfrac{\lambda'}{d} = \dfrac{400 \times 10^{-9}}{10^{-3}} = 4 \times 10^{-4}\) rad.

\(\beta' = \dfrac{\beta}{\mu}\), so \(\mu = \dfrac{\beta}{\beta'} = \dfrac{0.6}{0.45} = \dfrac{4}{3}\).

In water the slab replaces \(t\) of water by \(t\) of glass. The extra optical path is \((\mu_g - \mu_w)t = \mu_w\left(\dfrac{\mu_g}{\mu_w} - 1\right)t\). Measured in the water wavelength, the shift is \(\dfrac{D}{d}\left(\dfrac{\mu_g}{\mu_w} - 1\right)t = \dfrac{1}{10^{-3}}(1.125 - 1) \times 6 \times 10^{-6} = 0.75\) mm. In air the same slab shifted the pattern by \(3\) mm, so the shift is one quarter of that.

At \(O\) the two paths are equal, so every wavelength has zero path difference and every colour is bright there. Together they make white. Away from \(O\) the fringe width \(\beta = \dfrac{\lambda D}{d}\) is different for each colour. The bright fringes of the colours no longer fall together and the fringes appear coloured.

Violet has the shortest wavelength, so its first bright fringe is at the smallest distance \(\beta = \dfrac{\lambda D}{d}\) from the centre. Violet lies on the inner edge and red on the outer edge of each coloured fringe.

1. \(\Delta x = \dfrac{yd}{D} = \dfrac{1.2 \times 10^{-3} \times 10^{-3}}{1} = 1.2 \times 10^{-6}\) m \(= 1.2\) μm.

2. Bright when \(\Delta x = n\lambda\), so \(\lambda = \dfrac{1.2}{n}\) μm \(= 1200, 600, 400, 300, \ldots\) nm. In the visible range: \(600\) nm (\(n = 2\)) and \(400\) nm (\(n = 3\)).

Dark when \(\Delta x = (2n - 1)\dfrac{\lambda}{2}\), so \(\lambda = \dfrac{2 \times 1.05}{2n - 1}\) μm \(= 2100, 700, 420, 300, \ldots\) nm. The visible wavelengths missing at that point are \(700\) nm (\(n = 2\)) and \(420\) nm (\(n = 3\)).

The white fringe is the fringe of zero path difference. It moves to \(y_0 = \dfrac{D(\mu - 1)t}{d} = \dfrac{1 \times 0.5 \times 6 \times 10^{-6}}{10^{-3}} = 3\) mm, on the \(S_1\) side of \(O\). The coloured fringes move with it.

With monochromatic light all bright fringes look the same. When the pattern shifts by several fringe widths the eye cannot tell the new central fringe from any other. With white light only the central fringe is white, so its new position is seen at once and the shift can be measured directly.

Intensity is proportional to amplitude squared. If the amplitude from \(S_1\) is \(a\), that from \(S_2\) becomes \(\dfrac{a}{2}\). \(\dfrac{I_{\max}}{I_{\min}} = \dfrac{(a + a/2)^2}{(a - a/2)^2} = \dfrac{(3/2)^2}{(1/2)^2} = 9\). The dark fringes are no longer fully dark.

With one slit there is no second wave to interfere with, so the fringes disappear. The screen is lit almost uniformly; only the broad single-slit diffraction pattern remains. Each slit alone gives \(I_0\) at \(O\); with both slits in phase the amplitude doubled and gave \(4I_0\). Now the intensity at \(O\) is \(I_0\), one quarter of before.