The Workbook

Diffraction is the bending of light round the edge of an obstacle or opening into the region of the geometric shadow. The bending angle is about \(\dfrac{\lambda}{a}\). For a window a metre wide this is about \(10^{-6}\) rad, far too small to see. For a slit \(0.1\) mm wide it is about \(5 \times 10^{-3}\) rad, which gives a pattern millimetres wide on a screen a metre away.

1. Drop the perpendicular \(AN\) on the ray from \(B\). The ray from \(B\) travels the extra distance \(BN = a\sin\theta\).

2. Let \(a\sin\theta = \lambda\). Divide the slit into two halves. A point in the upper half and the point \(\dfrac{a}{2}\) below it send rays to \(P\) with path difference \(\dfrac{a}{2}\sin\theta = \dfrac{\lambda}{2}\). Such a pair cancels. Every point in the upper half has a partner in the lower half, so the whole slit cancels and \(P\) is dark.

Minima: \(a\sin\theta = n\lambda\) with \(n = 1, 2, 3, \ldots\) (not \(n = 0\), which is the central maximum). Secondary maxima lie roughly midway between minima, \(a\sin\theta \approx (2n + 1)\dfrac{\lambda}{2}\).

1. \(\sin\theta = \dfrac{\lambda}{a} = \dfrac{500 \times 10^{-9}}{0.2 \times 10^{-3}} = 2.5 \times 10^{-3}\), so \(\theta \approx 2.5 \times 10^{-3}\) rad.

2. \(y = D\theta = 2 \times 2.5 \times 10^{-3} = 5 \times 10^{-3}\) m \(= 5\) mm.

3. The central maximum runs from the first minimum on one side to the first on the other: width \(= 2y = 10\) mm.

Width \(= \dfrac{2\lambda D}{a}\). (i) Doubling \(a\) halves the width: a wider slit gives a narrower pattern. (ii) Doubling \(\lambda\) doubles the width. (iii) Red light has the longer wavelength, so its central maximum is wider than that of blue.

\(\sin\theta = \dfrac{\lambda}{a} = \dfrac{500}{1000} = 0.5\), so \(\theta = 30^\circ\). With \(a = 0.4\) μm, \(\dfrac{\lambda}{a} = 1.25 > 1\) and no angle satisfies \(a\sin\theta = \lambda\). There is no minimum at all. The slit acts almost like a line source and light spreads over the whole half space in front of it.

1. The first minima are at \(\pm 3\) mm, so the central maximum is \(6\) mm wide.

2. \(y_1 = \dfrac{\lambda D}{a}\), so \(a = \dfrac{\lambda D}{y_1} = \dfrac{600 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} = 3 \times 10^{-4}\) m \(= 0.3\) mm.

1. \(y_n = \dfrac{n\lambda D}{a}\), so \(y_2 = 2 \times 3 = 6\) mm.

2. The first secondary maximum is at \(a\sin\theta \approx \dfrac{3\lambda}{2}\), that is \(y \approx \dfrac{3}{2} \times 3 = 4.5\) mm.

At the first secondary maximum \(a\sin\theta = \dfrac{3\lambda}{2}\). Divide the slit into three parts, each with an edge-to-edge path difference of \(\dfrac{\lambda}{2}\). Two of the parts cancel each other pair by pair; only the third part sends light to the screen. Its amplitude is about one third of the full slit's, so the intensity is roughly one ninth. The exact calculation gives about \(\dfrac{1}{22}\), or \(4.5\%\), of the central intensity.

The central maximum extends from \(\sin\theta = -\dfrac{\lambda}{a}\) to \(+\dfrac{\lambda}{a}\), an angular width of \(\dfrac{2\lambda}{a}\). Each secondary maximum lies between consecutive minima, a width of \(\dfrac{\lambda}{a}\). The central maximum is twice as wide as any secondary maximum.

In water the wavelength becomes \(\dfrac{\lambda}{\mu}\), so the width \(\dfrac{2\lambda D}{a}\) becomes \(\dfrac{10}{4/3} = 7.5\) mm.

\(a = \dfrac{\lambda D}{y_1} = \dfrac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}} = 2.5 \times 10^{-4}\) m \(= 0.25\) mm.

(1) Interference fringes have equal intensity; diffraction maxima fall off quickly away from the centre. (2) Interference fringes are all of equal width; in diffraction the central maximum is twice as wide as the others. (3) Interference is the superposition of waves from two (or more) separate coherent sources. Diffraction is the superposition of wavelets from different parts of one wavefront.

1. Interference maxima are at \(d\sin\theta = n\lambda\) and the first diffraction minimum at \(a\sin\theta = \lambda\). The order that falls on the diffraction minimum is \(n = \dfrac{d}{a} = 5\). Orders \(0, \pm 1, \pm 2, \pm 3, \pm 4\) lie inside, so \(9\) bright fringes.

2. Whenever \(\dfrac{d}{a}\) is an integer, the orders \(n = \dfrac{d}{a}, \dfrac{2d}{a}, \ldots\) coincide with diffraction minima and are missing: the \(5\)th, \(10\)th, \(15\)th, and so on.

Each slit has a finite width and produces its own single slit diffraction pattern. The interference fringes are formed inside that pattern, so their intensity follows the diffraction envelope. The envelope is greatest at the centre and falls off on both sides.

Fringe width \(\beta = \dfrac{\lambda D}{d} = \dfrac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 1\) mm. Central diffraction maximum \(= \dfrac{2\lambda D}{a} = \dfrac{2 \times 500 \times 10^{-9} \times 1}{0.1 \times 10^{-3}} = 10\) mm. Ten fringe widths fit into it, in agreement with the \(9\) bright fringes of Q14.

Two point sources are just resolved when their diffraction patterns overlap in a set way. The central maximum of one must fall on the first minimum of the other. For a circular aperture of diameter \(D\) the first minimum is at \(\sin\theta = \dfrac{1.22\lambda}{D}\), so the least angular separation is \(\theta_{\min} = \dfrac{1.22\lambda}{D}\).

\(\theta_{\min} = \dfrac{1.22\lambda}{D} = \dfrac{1.22 \times 500 \times 10^{-9}}{0.122} = 5 \times 10^{-6}\) rad. Resolving power \(= \dfrac{1}{\theta_{\min}} = 2 \times 10^{5}\) rad\(^{-1}\).

\(\dfrac{1.22\lambda}{D} \le 10^{-5}\) gives \(D \ge \dfrac{1.22 \times 500 \times 10^{-9}}{10^{-5}} = 6.1 \times 10^{-2}\) m \(= 6.1\) cm.

First, a large objective collects more light, in proportion to its area, so faint stars become visible. Second, the limit of resolution \(\dfrac{1.22\lambda}{D}\) falls as \(D\) grows, so close stars can be seen as separate and finer detail is visible.

The smallest distance two points can be apart and still be seen separately is \(d_{\min} = \dfrac{1.22\lambda}{2\mu\sin\theta}\). Here \(d_{\min} = \dfrac{1.22 \times 500}{2 \times 1.22} = 250\) nm. Resolving power \(= \dfrac{1}{d_{\min}} = \dfrac{1}{250 \times 10^{-9}} = 4 \times 10^{6}\) m\(^{-1}\).

Resolving power \(= \dfrac{2\mu\sin\theta}{1.22\lambda}\). It rises with a shorter wavelength (blue or ultraviolet light). It also rises with a larger refractive index between object and objective (oil immersion). A larger cone angle \(\theta\) (wide objective close to the object) helps too. Electrons of a few kilovolts have a de Broglie wavelength of about \(0.01\) nm, tens of thousands of times shorter than visible light. So the electron microscope resolves far finer detail.

1. \(\theta_{\min} = \dfrac{1.22\lambda}{D} = \dfrac{1.22 \times 600 \times 10^{-9}}{2.44 \times 10^{-3}} = 3 \times 10^{-4}\) rad.

2. The headlights subtend \(\dfrac{1.5}{L}\) at the eye. They are just resolved when \(\dfrac{1.5}{L} = 3 \times 10^{-4}\), so \(L = 5000\) m \(= 5\) km.

Radio: \(\theta_{\min} = \dfrac{1.22 \times 1}{100} = 1.22 \times 10^{-2}\) rad, about \(0.7^\circ\). Optical: \(\theta_{\min} = \dfrac{1.22 \times 500 \times 10^{-9}}{0.1} = 6.1 \times 10^{-6}\) rad. The ratio is \(2000\). Even a huge dish resolves far more coarsely than a small optical telescope, because its wavelength is a million times longer.

A wave is polarized when its vibrations are confined to one direction in the plane at right angles to the direction of travel. Only a transverse wave has vibrations across the ray, so only a transverse wave can be polarized. Light is transverse, with \(\vec E\) perpendicular to the ray. Sound in air is longitudinal; its vibrations are along the ray and there is no direction across the ray to select.

In unpolarized light \(\vec E\) is always perpendicular to the ray, but its direction in that plane changes at random many million times a second. On average every direction is equally present. In plane polarized light \(\vec E\) vibrates in one fixed plane containing the ray. An ordinary lamp has countless atoms emitting independently, so its light is unpolarized.

Let the incident \(\vec E\) make angle \(\theta\) with the polaroid axis. Only the component \(E\cos\theta\) passes, so the transmitted intensity is \(I_0\cos^2\theta\). In unpolarized light \(\theta\) takes all values equally, and the average of \(\cos^2\theta\) over a full turn is \(\dfrac12\). So the transmitted intensity is \(\dfrac{I_0}{2}\).

No. Whatever the orientation of the polaroid, the incident light has equal amounts of \(\vec E\) along every direction, so the transmitted intensity stays \(\dfrac{I_0}{2}\). A constant intensity on rotating the polaroid shows that the incident light is unpolarized.

1. The polariser passes half of unpolarized light: \(I_1 = \dfrac{8}{2} = 4\) W/m\(^2\), now plane polarized along the axis of \(P\).

2. Malus's law: \(I_2 = I_1\cos^2 60^\circ = 4 \times \dfrac14 = 1\) W/m\(^2\).

Rotate the polaroid through a full turn and watch the transmitted intensity. If it stays constant the light is unpolarized. If it falls to zero twice in a turn, the light is plane polarized. If it rises and falls but never reaches zero, the light is partially polarized.

Malus's law: let plane polarized light of intensity \(I_0\) fall on an analyser whose axis makes angle \(\theta\) with the plane of polarization. The transmitted intensity is \(I = I_0\cos^2\theta\). It is greatest, \(I_0\), at \(\theta = 0\) (axes parallel) and zero at \(\theta = 90^\circ\) (crossed).

1. \(I_0\cos^2 30^\circ = \dfrac34 I_0\).

2. \(I_0\cos^2 45^\circ = \dfrac12 I_0\).

3. \(I_0\cos^2 60^\circ = \dfrac14 I_0\).

4. \(I_0\cos^2 90^\circ = 0\).

1. \(\cos^2\theta = \dfrac14\), \(\cos\theta = \dfrac12\), \(\theta = 60^\circ\).

2. \(\cos^2\theta = \dfrac12\), \(\cos\theta = \dfrac{1}{\sqrt2}\), \(\theta = 45^\circ\).

The transmitted amplitude is \(E_0\cos\theta\). For half the amplitude \(\cos\theta = \dfrac12\), so \(\theta = 60^\circ\). The intensity is proportional to the square of the amplitude, so it is \(\dfrac14 I_0\).

1. After \(P_1\): \(\dfrac{I_0}{2}\), polarized vertically. After \(P_2\) at \(45^\circ\): \(\dfrac{I_0}{2}\cos^2 45^\circ = \dfrac{I_0}{4}\), polarized along \(P_2\). After \(P_3\), which is \(45^\circ\) from \(P_2\): \(\dfrac{I_0}{4}\cos^2 45^\circ = \dfrac{I_0}{8}\).

2. Without \(P_2\) the light polarized by \(P_1\) meets \(P_3\) at \(90^\circ\) and \(\cos^2 90^\circ = 0\). Nothing comes out.

After the first polaroid: \(\dfrac{I_0}{2}\). After the middle one at \(\theta\): \(\dfrac{I_0}{2}\cos^2\theta\). The last is at \(90^\circ - \theta\) to the middle one: \(I = \dfrac{I_0}{2}\cos^2\theta\sin^2\theta = \dfrac{I_0}{8}\sin^2 2\theta\). This is greatest when \(\sin 2\theta = 1\), that is \(\theta = 45^\circ\), giving \(\dfrac{I_0}{8}\).

1. The curve is Malus's law with \(I_{\max} = 4\) W/m\(^2\) at \(\theta = 0\): \(I = 4\cos^2\theta\).

2. At \(60^\circ\): \(4 \times \dfrac14 = 1\) W/m\(^2\). \(I = 2\) when \(\cos^2\theta = \dfrac12\), at \(\theta = 45^\circ\) (and again at \(135^\circ\)).

\(I = I_0\cos^2\theta\) is greatest at \(\theta = 0\) and again at \(\theta = 180^\circ\), since \(\cos^2\theta\) repeats every half turn. So there are two maxima in each rotation. At \(5\) turns per second the intensity reaches its maximum \(10\) times a second.

Brewster's law: the reflected light is completely plane polarized when the tangent of the angle of incidence equals the refractive index, \(\tan i_p = \mu\). For \(\mu = \sqrt3\): \(\tan i_p = \sqrt3\), so \(i_p = 60^\circ\).

Snell's law: \(\sin i_p = \mu\sin r\). Brewster's law: \(\mu = \tan i_p = \dfrac{\sin i_p}{\cos i_p}\). Substituting, \(\sin i_p = \dfrac{\sin i_p}{\cos i_p}\sin r\), so \(\sin r = \cos i_p\) and \(r = 90^\circ - i_p\). The reflected ray makes angle \(i_p\) with the normal and the refracted ray makes \(r\) on the other side. The angle between them is \(180^\circ - i_p - r = 90^\circ\).

\(\mu = \tan i_p = \tan 53^\circ = \dfrac43\). The angle of refraction is \(r = 90^\circ - i_p = 37^\circ\). Check: \(\sin 53^\circ = 0.8 = \dfrac43 \times 0.6 = \mu\sin 37^\circ\).

At the polarizing angle the reflected light is completely plane polarized. Its \(\vec E\) is perpendicular to the plane of incidence (the plane containing the incident ray and the normal), that is parallel to the surface. The refracted light is only partially polarized. It contains all of the component with \(\vec E\) in the plane of incidence and part of the other component.

For glass to air the relative index is \(\dfrac{1}{\sqrt3}\), so \(\tan i_p' = \dfrac{1}{\sqrt3}\) and \(i_p' = 30^\circ\). For air to glass \(i_p = 60^\circ\). The two polarizing angles add to \(90^\circ\): each is the angle of refraction for the other.

\(\sin C = \dfrac{1}{\mu}\) gives \(\mu = \dfrac{1}{\sin 45^\circ} = \sqrt2\). Then \(\tan i_p = \sqrt2\), so \(i_p = \tan^{-1}\sqrt2 \approx 54.7^\circ\). In general \(\tan i_p = \mu = \dfrac{1}{\sin C}\).

Air molecules are much smaller than the wavelength, and for such scatterers the scattered intensity is proportional to \(\dfrac{1}{\lambda^4}\). Blue light, of shorter wavelength, is scattered several times more strongly than red, so the scattered light of the sky is blue. The molecule scatters by acting as an oscillating electric dipole driven by the incident \(\vec E\). A dipole sends no radiation along its own axis. So light scattered at \(90^\circ\) can only carry the \(\vec E\) component perpendicular to the plane of scattering, and it is plane polarized. At other angles it is partly polarized.

1. The light reaches the observer after scattering through exactly \(90^\circ\), so it is completely plane polarized.

2. The incident \(\vec E\) is perpendicular to the sunlight, that is in the vertical plane at right angles to the Sun's direction. Only the component perpendicular to the scattering plane (the vertical plane containing the Sun, the molecule and the observer) is radiated downward. So \(\vec E\) is horizontal and at right angles to the direction of the Sun.

Reflection near the polarizing angle favours the component of \(\vec E\) perpendicular to the plane of incidence. For a horizontal road or lake that component is horizontal. Polaroid sunglasses are made with a vertical transmission axis, which blocks the horizontal \(\vec E\) of the glare while passing most of the ordinary light.

The polaroid always passes half of the unpolarized part, \(2\) W/m\(^2\). The polarized part passes \(2\cos^2\theta\), between \(2\) and \(0\). Greatest: \(2 + 2 = 4\) W/m\(^2\), when the axis is along the polarization. Least: \(2 + 0 = 2\) W/m\(^2\), when the axis is across it.