IPhO 1986 · Theory problem 1

Two slits and a diffraction grating: vector addition of amplitudes

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Figure 1

Figure 1.1

A plane monochromatic light wave, wavelength \(\lambda\) and frequency \(f\), is incident normally on two identical narrow slits, separated by a distance \(d\), as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance \(x\) in a direction \(\theta\) at time \(t\), by

\[ y=a\cos[2\pi(ft-x/\lambda)] \]

where the amplitude \(a\) is the same for both waves. (Assume \(x\) is much larger than \(d\)).

(i) Show that the two waves observed at an angle \(\theta\) to a normal to the slits, have a resultant amplitude \(A\) which can be obtained by adding two vectors, each having magnitude \(a\), and each with an associated direction determined by the phase of the light wave.

Verify geometrically, from the vector diagram, that

\[ A=2a\cos\theta \]

where

\[ \beta=\frac{\pi}{\lambda}d\sin\theta \]

(ii) The double slit is replaced by a diffraction grating with \(N\) equally spaced slits, adjacent slits being separated by a distance \(d\). Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude \(a\), form a part of a regular polygon with vertices on a circle of radius \(R\) given by

\[ R=\frac{a}{2\sin\beta}, \]

Deduce that the resultant amplitude is

\[ \frac{a\sin N\beta}{\sin\beta} \]

and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.

(iii) Sketch, in the same graph, \(\sin N\beta\) and \((1/\sin\beta)\) as a function of \(\beta\). On a separate graph show how the intensity of the resultant wave varies as a function of \(\beta\).

(iv) Determine the intensities of the principal intensity maxima.

(v) Show that the number of principal maxima cannot exceed

\[ \left(\frac{2d}{\lambda}+1\right) \]

(vi) Show that two wavelengths \(\lambda\) and \(\lambda+\delta\lambda\), where \(\delta\lambda\ll\lambda\), produce principal maxima with an angular separation given by

\[ \Delta\theta=\frac{n\Delta\lambda}{d\cos\theta} \qquad\text{where }n=0,\ \pm1,\ \pm2\ldots\text{etc} \]

Calculate this angular separation for the sodium D lines for which

\[ \lambda=589.0\,\mathrm{nm},\quad \lambda+\Delta\lambda=589.6\,\mathrm{nm},\quad n=2,\quad\text{and }d=1.2\times10^{-6}\,\mathrm{m}. \]

\[ \left[ \text{reminder:}\quad \cos A+\cos B = 2\cos\left(\frac{A+B}{2}\right)\cdot \cos\left(\frac{A-B}{2}\right) \right] \]

Official English text of the problem.

Answers

Let \(I_1\) be the intensity from one slit alone, and let \[ \beta=\frac{\pi d\sin\theta}{\lambda}. \] A signed amplitude coefficient is denoted by \(A_{\mathrm s}\); its absolute value is the physical amplitude.

  • (i) The phase difference is \(2\beta\), giving \[ \boxed{A_{\mathrm s}=2a\cos\beta,\qquad A=2a|\cos\beta|.} \] The printed \(\cos\theta\) should be \(\cos\beta\).

  • (ii) \[ \boxed{R=\frac{a}{2\sin\beta}\quad(0<\beta<\pi),\qquad A_{\mathrm s}=a\frac{\sin N\beta}{\sin\beta}.} \] Generally, \(R=a/(2|\sin\beta|)\). For slit phases increasing by \(2\beta\), the resultant phase relative to the first edge slit is \((N-1)\beta\) with the signed coefficient above. Using a positive amplitude adds \(\pi\) wherever that coefficient is negative. At \(\beta=m\pi\), the physical amplitude is \(Na\).

  • (iii) \(\sin N\beta\) has amplitude \(1\) and period \(2\pi/N\); \(1/\sin\beta\) has vertical asymptotes at \(\beta=m\pi\). The intensity is \[ \boxed{I(\beta)=I_1\left(\frac{\sin N\beta}{\sin\beta}\right)^2.} \] It is even and has period \(\pi\), with principal peaks at \(m\pi\). Between consecutive principal peaks are \(N-1\) zeros and \(N-2\) secondary maxima.

  • (iv) \[ \boxed{I_{\mathrm{principal}}=N^2I_1,\qquad d\sin\theta=n\lambda.} \]

  • (v) Counting grazing orders as limiting directions, \[ \boxed{M=2\left\lfloor\frac d\lambda\right\rfloor+1 \leq\frac{2d}{\lambda}+1.} \]

  • (vi) To first order, \[ \boxed{\Delta\theta\simeq\frac{n\Delta\lambda}{d\cos\theta}.} \] The angular separation is the magnitude of this signed shift. For the specified sodium lines in second order, \[ \boxed{\Delta\theta\simeq5.2\times10^{-3}\,\mathrm{rad} \simeq0.30^\circ.} \]

Solution

(i)

Normal incidence gives the same incident phase at both slits. The emerging waves are coherent, so their instantaneous displacements add.

In the far field, the rays reaching the observation point are effectively parallel. The difference between the two path lengths is therefore \[ \Delta x=d\sin\theta. \] Consequently, the phase difference has magnitude \[ \delta=\frac{2\pi\Delta x}{\lambda} =\frac{2\pi d\sin\theta}{\lambda} =2\beta. \]

Choose the reference slit so that the other wave advances in phase by \(2\beta\). Writing the reference phase as \(\psi\), the two waves are \[ y_0=a\cos\psi,\qquad y_1=a\cos(\psi+2\beta). \] The given identity gives \[ y_0+y_1 =2a\cos\beta\cos(\psi+\beta). \]

To interpret this geometrically, represent each wave by a vector of length \(a\), whose direction equals its phase. Its projection on a fixed axis is the instantaneous wave displacement. Both vectors rotate with angular speed \(2\pi f\), maintaining their angular separation \(2\beta\). Since projections add, their vector sum represents the resultant wave.

For the vector diagram, draw the two arrows from a common origin, making angles \(-\beta\) and \(+\beta\) with their phase bisector. Their components perpendicular to this bisector cancel; their components along it add: \[ A_{\mathrm s}=a\cos\beta+a\cos\beta=2a\cos\beta. \] Thus \[ \boxed{A=2a|\cos\beta|}. \] A negative signed coefficient simply represents a reversal of the resultant vector, or an additional phase of \(\pi\).

The \(\theta\) in the printed amplitude formula must therefore be replaced by \(\beta\).

(ii)

Apply the same far-field approximation to the whole grating, neglecting amplitude variation and wavefront curvature across it. Label the slits from one edge so that their phases are \[ \psi,\ \psi+2\beta,\ \ldots,\ \psi+2(N-1)\beta. \]

Draw the \(N\) amplitude vectors head to tail. Every vector has length \(a\), and each turns through the same angle \(2\beta\) relative to its predecessor. Their endpoints consequently form an equal-chord polygonal chain on a circle.

For the diagram, label the successive endpoints \(P_0,P_1,\ldots,P_N\) and the circle centre \(O\). Each side \(P_jP_{j+1}\) has length \(a\) and subtends a directed central angle \(2\beta\). Bisecting the corresponding isosceles triangle gives, for \(0<\beta<\pi\), \[ \frac a2=R\sin\beta. \] Hence \[ \boxed{R=\frac{a}{2\sin\beta}}. \] For arbitrary \(\beta\), the positive radius is \(a/(2|\sin\beta|)\). The chain need not close after the \(N\) vectors.

The resultant vector runs directly from \(P_0\) to \(P_N\). These endpoints are separated by a total directed central angle \(2N\beta\), so its length is \[ A_N=2R|\sin N\beta| =a\left|\frac{\sin N\beta}{\sin\beta}\right|. \] Thus the requested signed coefficient is \[ \boxed{A_{\mathrm s}=a\frac{\sin N\beta}{\sin\beta}}. \]

The phase follows from the chord directions. The first small chord has a midpoint radius at angular position \(\beta\), while the resultant chord has a midpoint radius at \(N\beta\). Each chord is perpendicular to its midpoint radius, so their angular difference is \[ N\beta-\beta=(N-1)\beta, \] up to a reversal of the resultant.

Equivalently, the complete resultant wave is \[ \boxed{ y_{\mathrm{tot}} = a\frac{\sin N\beta}{\sin\beta} \cos\!\left[\psi+(N-1)\beta\right]. } \] Therefore, with the chosen edge convention, the phase relative to that edge slit is \((N-1)\beta\) when the signed coefficient is retained. With a nonnegative amplitude, it is \[ \phi= \begin{cases} (N-1)\beta,&\sin N\beta/\sin\beta>0,\\[2mm] (N-1)\beta+\pi,&\sin N\beta/\sin\beta<0, \end{cases} \qquad(\bmod\,2\pi). \] Referencing the opposite edge reverses the relative phase modulo \(2\pi\). At a zero resultant, phase is undefined.

When \(\beta=m\pi\), adjacent phases differ by \(2m\pi\). All vectors point in the same direction, giving \(A_N=Na\) and zero relative phase modulo \(2\pi\). The circle construction then has infinite limiting radius.

(iii)

The time-averaged intensity is proportional to the square of the physical amplitude. Since the one-slit amplitude is \(a\), \[ \frac{I}{I_1}=\frac{A_N^2}{a^2}, \] and hence \[ \boxed{\frac{I(\beta)}{I_1} =\left(\frac{\sin N\beta}{\sin\beta}\right)^2}. \]

For the first graph, use common horizontal \(\beta\) and vertical function-value axes:

  • \(\sin N\beta\) oscillates between \(-1\) and \(+1\), with period \(2\pi/N\). Its zeros are at \(\beta=k\pi/N\), and its extrema occur at \(\beta=(2k+1)\pi/(2N)\), with values \((-1)^k\).
  • \(1/\sin\beta\) has period \(2\pi\), no zeros, and vertical asymptotes at \(\beta=m\pi\). Between \(0\) and \(\pi\), it falls from \(+\infty\) to its minimum \(1\) at \(\pi/2\), then rises to \(+\infty\). The next branch, between \(\pi\) and \(2\pi\), is negative, reaching \(-1\) at \(3\pi/2\) and tending to \(-\infty\) at both ends. These branches repeat.

Although the reciprocal sine diverges at \(m\pi\), the product with \(\sin N\beta\) has a finite limiting magnitude \(N\).

For the intensity graph, plot \(I/I_1\). Its principal features are:

Position \(I/I_1\)
\(\beta=m\pi\) \(N^2\)
\(\beta=m\pi+r\pi/N,\quad r=1,\ldots,N-1\) \(0\)
\(\beta=m\pi+\pi/2\), \(N\) even \(0\)
\(\beta=m\pi+\pi/2\), \(N\) odd \(1\)

The curve is nonnegative, even, and periodic with period \(\pi\). Each principal peak is symmetric about \(\beta=m\pi\). Its nearest zeros are \[ \beta=m\pi\pm\frac{\pi}{N}, \] so its width between these zeros is \(2\pi/N\).

Between neighboring principal peaks there are \(N-1\) zeros, with \(N-2\) smaller secondary peaks between successive zeros. Their exact positions satisfy \[ \boxed{N\cot(N\beta)=\cot\beta}, \] obtained by differentiating the intensity away from its zeros. Away from the asymptotes, the curve lies beneath the envelope \[ I=\frac{I_1}{\sin^2\beta}. \]

(iv)

The largest possible resultant of \(N\) vectors of length \(a\) has length \(Na\). This occurs when all vectors point in the same direction: \[ 2\beta=2n\pi. \] Thus the principal maxima satisfy \[ \beta=n\pi \quad\Longleftrightarrow\quad \boxed{d\sin\theta=n\lambda}. \]

At these directions, \[ A_N=Na, \] so \[ \boxed{I_{\mathrm{principal}}=N^2I_1}. \] Under the stated equal-amplitude model, all principal maxima have this intensity.

The same result follows by taking the limit: \[ \lim_{\beta\to n\pi} \frac{\sin N\beta}{\sin\beta} =N(-1)^{n(N-1)}, \] whose square is \(N^2\).

(v)

A principal maximum of order \(n\) requires \[ \sin\theta=\frac{n\lambda}{d}. \] Since \(|\sin\theta|\leq1\), \[ |n|\leq\frac d\lambda. \]

The allowed integer orders therefore run from \[ -\left\lfloor\frac d\lambda\right\rfloor \quad\text{to}\quad +\left\lfloor\frac d\lambda\right\rfloor. \] Counting the zero order gives \[ M=2\left\lfloor\frac d\lambda\right\rfloor+1. \] Because \(\lfloor u\rfloor\leq u\), \[ \boxed{M\leq\frac{2d}{\lambda}+1}. \]

This count includes any orders exactly at grazing incidence, \(\theta=\pm90^\circ\), as limiting directions. Excluding such directions can only reduce the number.

(vi)

Consider the principal maxima of the same order \(n\) for the two wavelengths. Writing \(\Delta\lambda=\delta\lambda\), their grating equations are \[ d\sin\theta=n\lambda, \] \[ d\sin(\theta+\Delta\theta) =n(\lambda+\Delta\lambda). \]

For a small angular shift, measured in radians, \[ \sin(\theta+\Delta\theta) \simeq\sin\theta+\Delta\theta\cos\theta. \] Subtracting the first grating equation from the second gives \[ d\cos\theta\,\Delta\theta\simeq n\Delta\lambda, \] hence \[ \boxed{\Delta\theta\simeq\frac{n\Delta\lambda}{d\cos\theta}}. \] This is a first-order result for existing, nongrazing maxima. For negative \(n\), the signed shift is negative; the angular separation is its absolute value. The zero-order maxima coincide.

For the sodium lines, \[ \Delta\lambda=(589.6-589.0)\,\mathrm{nm} =0.6\times10^{-9}\,\mathrm m. \] At \(\lambda=589.0\,\mathrm{nm}\) in second order, \[ \sin\theta =\frac{2(589.0\times10^{-9})}{1.2\times10^{-6}} =0.9816667, \] so \[ \theta\simeq79.01^\circ,\qquad \cos\theta\simeq0.19061. \] Therefore, \[ \Delta\theta \simeq \frac{2(0.6\times10^{-9})} {(1.2\times10^{-6})(0.19061)} =5.246\times10^{-3}\,\mathrm{rad}. \] To sensible precision, \[ \boxed{\Delta\theta\simeq5.2\times10^{-3}\,\mathrm{rad} \simeq0.30^\circ}. \]

As a numerical check, subtracting the two angles from the unexpanded grating equation gives \(5.32\times10^{-3}\,\mathrm{rad}\), about \(1.4\%\) above the first-order estimate.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.