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IPhO 1986, Теория — Задача 1. Q1

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IPhO 1986 — Theory

Условие

Plane wavefronts (three vertical lines) incident on a screen with two slits separated by d; from each slit a ray leaves at angle θ to the dashed normal. — 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,\ \pm 1,\ \pm 2\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,\ \text{and } d = 1.2\times 10^{-6}\ \mathrm{m}.

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

Отговори

Решение

Answers Question 1

(i) Vector Diagram

Vector diagram: vector OP of length a along the horizontal, vector PQ of length a at angle φ to the dashed extension, resultant A from O to Q (double arrow) at angle β to OP; the parallelogram completed by lines from O and P to a top vertex.

Two slits separated by d on a vertical screen; parallel rays leave at angle θ, a perpendicular dropped from the upper slit to the lower ray shows the path difference, with angle θ marked twice.

If the phase of the light from the first slit is zero, the phase from second slit is

\phi = \frac{2\pi}{\lambda} d\sin\theta

Adding the two waves with phase difference $\phi$ where $\xi = 2\pi\left(ft - \frac{x}{\lambda}\right)$ ,

a\cos(\xi + \phi) + a\cos(\xi) = 2a\cos(\phi/2)(\xi + \phi/2)

a\cos(\xi + \phi) + a\cos(\xi) = 2a\cos\beta\{\cos(\xi + \beta)\}

This is a wave of amplitude $A = 2a\cos\beta$ and phase $\beta$ . From vector diagram, in isosceles triangle OPQ,

\beta = \frac{1}{2}\phi = \frac{\pi}{\lambda} d\sin\theta \qquad (NB\quad \phi = 2\beta)

and

A = 2a\cos\beta.

Thus the sum of the two waves can be obtained by the addition of two vectors of amplitude a and angular directions 0 and $\phi$ .

(ii) Each slit in diffraction grating produces a wave of amplitude a with phase $2\beta$ relative to previous slit wave. The vector diagram consists of a 'regular' polygon with sides of constant length $a$ and with constant angles between adjacent sides.
Let O be the centre of circumscribing circle passing through the vertices of the polygon. Then radial lines such as OS have length R and bisect the internal angles of the polygon. Figure 1.2.

Polygon of equal vectors a from T through S towards Z, each turned by φ relative to the previous one; radii R drawn from the circumcentre O to every vertex. — Figure 1.2

O\hat{S}T = O\hat{T}S = \frac{1}{2}(180 - \phi)

\text{and}\quad T\hat{O}S = \phi

In the triangle $TOS$ , for example

a = 2R\sin(\phi/2) = 2R\sin\beta \ \text{ as }\ (\phi = 2\beta)

\therefore R = \frac{a}{2\sin\beta} \qquad (1)

As the polygon has $N$ faces then:

T\hat{O}Z = N(T\hat{O}Z) = N\phi = 2N\beta

Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by

2R\sin N\beta\ .

Hence form (1) this amplitude is

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

Resultant phase is

\begin{aligned} &= Z\hat{T}S \\ &= O\hat{T}S - O\hat{T}Z \\ &\left(90 - \frac{\phi}{2}\right) - \frac{1}{2}(180 - N\phi) \\ &-\frac{1}{2}(N-1)\phi \\ &= (N-1)\beta \end{aligned}

Sketch against β: a sin Nβ as a rapid small oscillation about the β axis, and 1/sin β as U-shaped branches alternating above and below the axis between dashed vertical asymptotes; arrows label the curves 1/sin β and a sin Nβ.

(iii)

Intensity $I = \dfrac{a^2\sin^2 N\beta}{\sin^2\beta}$

Intensity I against β: tall principal maxima at 0, π, 2π and 3π with small subsidiary maxima between them.

(iv) For the principle maxima $\beta = \pi p$ where $p = 0 \pm 1 \pm 2........$

I_{\max} = a^2\left(\frac{N\beta'}{\beta'}\right) = N^2a^2 \qquad \beta' = 0\ \text{ and } \beta = \pi p + \beta'

(v) Adjacent max. estimate $I_1$ :

\sin^2 N\beta = 1,\quad \beta = 2\pi p \mp \frac{3\pi}{2N}\ \text{ i.e }\ \beta = \pm\frac{3\pi}{2N}

$\left[\beta = \pi p \pm \dfrac{\pi}{2N}\right]$ does not give a maximum as can be observed from the graph.

I_1 = a^2\frac{1}{\frac{3\pi^2}{2n}} = \frac{a^2N^2}{23}\ \text{ for } N \gg 1

Adjacent zero intensity occurs for $\beta = \pi\rho \pm \dfrac{\pi}{N}$ i.e. $\delta = \pm\dfrac{\pi}{N}$

For phase differences much greater than $\delta$ , $\mathrm{I} = \mathrm{a}^2\left(\dfrac{\sin N\beta}{\sin\beta}\right) = a^2$ .

(vi)

\beta = n\pi \quad \text{for a principle maximum}

\text{i.e. } \frac{\pi}{\lambda} d\sin\theta = n\pi \qquad n = 0,\pm1,\pm2..........

Differentiating w.r.t, $\lambda$

d\cos\theta\Delta\theta = n\Delta\lambda

\Delta\theta = \frac{n\Delta\lambda}{d\cos\theta}

Substituting $\lambda = 589.0\mathrm{nm}$ , $\lambda + \Delta\lambda = 589.6\ \mathrm{nm}$ . n = 2 and $d = 1.2\times 10^{-6}\mathrm{m}$ .

\Delta\theta = \frac{n\Delta\lambda}{d\sqrt{1 - \left(\frac{n\lambda}{d}\right)^2}} \quad \text{as } \sin\theta = \frac{n\lambda}{d} \text{ and } \cos\theta = \sqrt{1 - \left(\frac{n\lambda}{d}\right)^2}

\Rightarrow \Delta\theta = 5.2\times 10^{-3}\ rads\ \text{ or }\ 0.30^{0}

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf