IPhO 1981 · Theory problem 3

Radio interferometer on the sea shore

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A detector of radiowaves in a radioastronomical observatory is placed on the sea beach at height \(h = 2\) m above the sea level. After the rise of a star, radiating electromagnetic waves of wavelength \(\lambda = 21\) cm, above the horizont the detector registers series of alternating maxima and minima. The registered signal is proportional to the intensity of the detected waves. The detector registers waves with electric vector, vibrating in a direction parallel to the sea surface.

  1. Determine the angle between the star and the horizont in the moment when the detector registers maxima and minima (in general form).

  2. Does the signal decrease or increase just after the rise of the star?

  3. Determine the signal ratio of the first maximum to the next minimum. At reflection of the electromagnetic wave on the water surface, the ratio of the intensities of the electric field of the reflected (\(E_r\)) and incident (\(E_i\)) wave follows the low:

\[ \frac{E_r}{E_i}=\frac{n-\cos\varphi}{n+\cos\varphi}, \]

where \(n\) is the refraction index and \(\varphi\) is the incident angle of the wave. For the surface “air-water” for \(\lambda = 21\) cm, the refraction index \(n = 9\).

  1. Does the ratio of the intensities of consecutive maxima and minima increase or decrease with rising of the star?

Assume that the sea surface is flat.

Official English text of the problem.

Answers

  1. Writing \(\alpha\) for the star’s elevation above the horizon, the usual interference-fringe positions are

\[ \boxed{\alpha_m^{\max}\simeq \arcsin\!\left(\frac{(2m+1)\lambda}{4h}\right)},\qquad \boxed{\alpha_m^{\min}\simeq \arcsin\!\left(\frac{m\lambda}{2h}\right)}. \]

Here \(m=0,1,2,\ldots\), restricted to arguments not exceeding \(1\). The first maximum is at approximately \(1.50^\circ\), and the following minimum at \(3.01^\circ\). The small correction from angle-dependent reflection is quantified below.

  1. The signal increases immediately after the star rises, starting from a minimum at the horizon.

  2. The signal ratio is

\[ \boxed{\frac{S_{\text{first maximum}}}{S_{\text{next minimum}}} \simeq 2.96\times10^4}. \]

  1. The ratio of successive maximum signals to minimum signals decreases as the star rises.

Solution

1. Treat the distant star as a steady source of plane waves, and the flat sea as a coherent specular reflector. Let \(\alpha\) be the elevation of the star and \(I_0\) the intensity of the direct wave alone.

The reflected path can be calculated by replacing the detector, at height \(h\), with its mirror image at depth \(h\). The separation between detector and image is \(2h\); its projection along the direction toward the star gives the path difference

\[ \Delta=2h\sin\alpha. \]

The detected electric field is parallel to the sea and perpendicular to the plane of incidence. Reflection from air into the optically denser water reverses this electric field, introducing a phase change of \(\pi\).

Let \(r\) denote the positive reflected-to-incident field-amplitude ratio, and define the propagation phase difference

\[ \delta=\frac{2\pi\Delta}{\lambda} =\frac{4\pi h}{\lambda}\sin\alpha. \]

Including the reflection phase reversal, the resultant intensity is

\[ \boxed{\frac{I}{I_0} =1+r^2-2r\cos\delta =(1-r)^2+4r\sin^2\!\left(\frac{\delta}{2}\right)}. \]

Treating \(r\) as locally constant across a fringe, maxima occur when \(\cos\delta=-1\), whereas minima occur when \(\cos\delta=1\). Thus

\[ 2h\sin\alpha_m^{\max} =\left(m+\frac12\right)\lambda, \qquad 2h\sin\alpha_m^{\min}=m\lambda, \]

which gives the formulas in the Answers section. Numerically,

\[ \sin\alpha_m^{\max}=0.02625(2m+1), \qquad \sin\alpha_m^{\min}=0.05250m. \]

The horizon, \(m=0\), is a minimum.

For completeness, the supplied angle-dependent reflection coefficient causes small shifts from these phase-alignment positions. Since the incidence angle is measured from the surface normal,

\[ \varphi=\frac{\pi}{2}-\alpha, \qquad \cos\varphi=\sin\alpha. \]

Set \(s=\sin\alpha\) and \(K=4\pi h/\lambda\). Then

\[ r(s)=\frac{n-s}{n+s}, \qquad r'(s)=-\frac{2n}{(n+s)^2}. \]

Differentiating the full intensity gives the exact condition for interior extrema:

\[ \boxed{ r'(s)\bigl[r(s)-\cos(Ks)\bigr] +Kr(s)\sin(Ks)=0. } \]

The phase-alignment formulas give \(1.50419^\circ\) and \(3.00941^\circ\) for the first maximum and following minimum. Keeping the variation of \(r\) gives \(1.50241^\circ\) and \(3.00940^\circ\), respectively. Thus the simple formulas give both angles correctly at the precision reported above.

2. At the horizon,

\[ \alpha=0,\qquad \delta=0,\qquad r=1. \]

The direct and reflected fields therefore cancel:

\[ I(0)=0. \]

As the star rises, the propagation phase difference becomes nonzero and the cancellation is lost. More explicitly, for small positive \(\alpha\), measured in radians,

\[ r=1-\frac{2\alpha}{n}+O(\alpha^2), \qquad \delta=K\alpha+O(\alpha^3). \]

Consequently,

\[ \frac{I}{I_0} =\left(K^2+\frac{4}{n^2}\right)\alpha^2+O(\alpha^3). \]

The leading coefficient is positive, so the signal initially increases quadratically with elevation.

3. The given ratio \(E_r/E_i\) is a ratio of electric-field amplitudes. The measured signal is proportional to the square of the resultant field.

At the first maximum and following minimum, respectively,

\[ s_1=\frac{\lambda}{4h}=0.02625, \qquad s_2=\frac{\lambda}{2h}=0.05250. \]

Their reflection coefficients are different:

\[ r_1=\frac{9-0.02625}{9+0.02625} \simeq0.994184, \]

\[ r_2=\frac{9-0.05250}{9+0.05250} \simeq0.988401. \]

At a maximum the field amplitudes add, while at a minimum they subtract. Hence

\[ I_{\max,1}=I_0(1+r_1)^2, \qquad I_{\min,1}=I_0(1-r_2)^2. \]

Using

\[ 1+r_1=\frac{2n}{n+s_1}, \qquad 1-r_2=\frac{2s_2}{n+s_2}, \]

the requested ratio becomes

\[ \begin{aligned} R &=\left[\frac{n(n+s_2)}{s_2(n+s_1)}\right]^2\\ &=\left[ \frac{9(9+0.05250)} {0.05250(9+0.02625)} \right]^2\\ &\simeq 29558.9. \end{aligned} \]

Therefore,

\[ \boxed{R\simeq2.96\times10^4}. \]

Using the strict extrema from part 1 changes this ratio by less than \(10^{-5}\) relatively, leaving the reported result unchanged.

4. As the star rises, \(\sin\alpha\) increases and the reflection amplitude decreases:

\[ \frac{dr}{d\alpha} =-\frac{2n\cos\alpha}{(n+\sin\alpha)^2}<0 \qquad \left(0<\alpha<\frac{\pi}{2}\right). \]

The constructive-interference envelope is

\[ I_{\mathrm{upper}}=I_0(1+r)^2, \]

which decreases as \(r\) decreases. The destructive-interference envelope is

\[ I_{\mathrm{lower}}=I_0(1-r)^2, \]

which increases. Thus later maxima become weaker while later minima become stronger, so the maximum-to-minimum signal ratio decreases.

The solution was checked against the official solution.