IPhO 1979 · Theory problem 3

Laser location of the Moon

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During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser (\(\lambda=0{,}69\,\mu\mathrm{m}\)) was directed to the Moon’s surface by the telescope with a diameter of the mirror \(D=2{,}6\,\mathrm{m}\). The reflector on the Moon’s surface reflected the light backward as an ideal mirror with the diameter \(d=20\,\mathrm{cm}\). The reflected light was then collected by the same telescope and focused at the photodetector.

1) What must the accuracy to direct the telescope optical axis be in this experiment?

2) What part of emitted laser energy can be detected after reflection on the Moon, if we neglect the light loses in the Earth’s atmosphere?

3) Can we see a reflected light pulse with naked eye if the energy of single laser pulse \(E=1\,\mathrm{J}\) and the threshold sensitivity of eye is equal \(n=100\) light quantum?

4) Suppose the Moon’s surface reflects \(\alpha=10\%\) of the incident light in the spatial angle \(2\pi\) steradian, estimate the advantage of a using reflector.

The distance from the Earth to the Moon is \(L=380000\,\mathrm{km}\). The diameter of pupil of the eye is \(d_p=5\,\mathrm{mm}\). Plank constant is \(h=6.6\cdot10^{-34}\,\mathrm{J\cdot s}\).

Official English text of the problem.

Answers

The numerical results below assume uniform coherent illumination of the circular telescope aperture, a centered reflector, and ideal collection.

  1. Pointing accuracy of order \[ \boxed{\delta\theta\lesssim 3\times10^{-7}\,\mathrm{rad}\approx0.06''}. \]

  2. The fraction collected by the telescope is \[ \boxed{\eta=\left(\frac{\pi Dd}{4\lambda L}\right)^4 \approx5.9\times10^{-12}}. \]

  3. No. The unaided eye receives approximately \[ \boxed{N_{\rm eye}\approx76\ \text{photons}<100}. \]

  4. Relative to diffuse reflection from the entire illuminated lunar patch, the reflector increases the received energy by approximately \[ \boxed{G\approx1.0\times10^7}. \]

The elementary approximation of uniformly illuminated diffraction spots gives \(\eta\approx10^{-12}\), \(N_{\rm eye}\approx12\), and \(G\approx1.7\times10^6\). The distinction comes from the diffraction intensity profile, as explained below.

Solution

We take \[ \lambda=6.9\times10^{-7}\,\mathrm m,\qquad L=3.8\times10^8\,\mathrm m,\qquad c=3.0\times10^8\,\mathrm{m\,s^{-1}}. \] We neglect atmospheric absorption, optical losses, and detector inefficiency. The laser fills the telescope aperture coherently, and the reflector is centered in the outgoing beam.

1) Required pointing accuracy

A finite aperture produces diffraction. For an aperture of diameter \(a\), the characteristic angular spread is \[ \theta\sim\frac{\lambda}{a}. \] For a uniformly illuminated circular aperture, the first dark ring occurs at \[ \theta_1=1.22\frac{\lambda}{a}. \]

For the transmitting telescope, \[ \frac{\lambda}{D} =\frac{6.9\times10^{-7}}{2.6} =2.65\times10^{-7}\,\mathrm{rad} =0.0547''. \] The first-dark-ring angle is \(3.24\times10^{-7}\,\mathrm{rad}=0.0668''\).

The corresponding central diffraction spot on the Moon has radius \[ r_1=L\theta_1\approx123\,\mathrm m, \] much larger than the reflector’s radius \(0.10\,\mathrm m\).

A pointing error \(\delta\theta\) moves the beam center by \(L\delta\theta\). To keep the reflector in the central diffraction spot, this displacement must be smaller than roughly \(10^2\,\mathrm m\). Thus \[ \boxed{\delta\theta\lesssim3\times10^{-7}\,\mathrm{rad} \approx0.06''}. \] For nearly maximal return, the pointing error should be appreciably smaller than this characteristic width.

2) Fraction of laser energy collected after reflection

Diffraction occurs twice: first at the transmitting telescope and then at the lunar reflector.

To retain the circular-aperture numerical factor, consider a uniformly illuminated aperture of area \(A\). In the far field, the contributions from all parts of the aperture arrive in phase on its axis. The Huygens–Fresnel formula gives the axial field amplitude, apart from an overall phase, as \[ u(0,L)=\frac{u_0A}{i\lambda L}. \] Because the transmitted energy is proportional to \(|u_0|^2A\), the axial energy per unit area, or fluence, is \[ \boxed{\mathcal F(0,L)=\frac{EA}{\lambda^2L^2}}. \] Consequently, a small receiving aperture of area \(B\), centered on the axis, intercepts the fraction \[ \frac{E_B}{E}\approx\frac{AB}{\lambda^2L^2}. \]

Here the telescope and reflector areas are \[ A_D=\frac{\pi D^2}{4}=5.31\,\mathrm{m^2}, \qquad A_d=\frac{\pi d^2}{4}=3.14\times10^{-2}\,\mathrm{m^2}. \] The reflector is tiny compared with the outgoing diffraction spot, so the incident fluence is essentially constant across it. The fraction reaching it is therefore \[ f_{\rm out} =\frac{A_DA_d}{\lambda^2L^2} =2.43\times10^{-6}. \]

The illuminated reflector acts as a coherent aperture of diameter \(d\). Its returning diffraction pattern has first-dark-ring radius \[ r_{\rm return}=1.22\frac{\lambda L}{d} \approx1.60\times10^3\,\mathrm m. \] The telescope is tiny compared with this pattern, so the same central-fluence calculation applies: \[ f_{\rm back} =\frac{A_dA_D}{\lambda^2L^2} =2.43\times10^{-6}. \]

Multiplying the two fractions, \[ \eta=f_{\rm out}f_{\rm back} =\left(\frac{A_DA_d}{\lambda^2L^2}\right)^2 =\left(\frac{\pi Dd}{4\lambda L}\right)^4. \] Thus \[ \boxed{\eta=5.9\times10^{-12}}. \]

3) Visibility to the naked eye

The returning fluence is essentially uniform over both the telescope aperture and the much smaller pupil. The energy entering an unaided eye is therefore \[ E_{\rm eye} =E\eta\frac{A_p}{A_D} =E\eta\left(\frac{d_p}{D}\right)^2. \] Substitution gives \[ E_{\rm eye} =(1)(5.89\times10^{-12}) \left(\frac{0.005}{2.6}\right)^2 =2.18\times10^{-17}\,\mathrm J. \]

The energy of one photon is \[ \varepsilon_\gamma=\frac{hc}{\lambda} =\frac{(6.6\times10^{-34})(3.0\times10^8)} {6.9\times10^{-7}} =2.87\times10^{-19}\,\mathrm J. \] Hence \[ N_{\rm eye} =\frac{E_{\rm eye}}{\varepsilon_\gamma} \approx75.9. \]

Since \[ \boxed{N_{\rm eye}\approx76<100}, \] the pulse does not meet the stated naked-eye visibility threshold, even under these ideal assumptions.

Uniform, in-phase illumination maximizes the axial fluence for a fixed aperture and transmitted energy. Thus an illumination profile giving a lower central intensity would not improve this conclusion.

4) Advantage over reflection from the lunar surface

For the requested estimate, suppose the diffusely reflected energy \(\alpha E\) is distributed uniformly over \(2\pi\) steradians. The receiving telescope subtends the solid angle \[ \Omega_D\approx\frac{A_D}{L^2}. \] The fraction of the emitted energy collected from the diffuse surface is \[ \eta_{\rm diffuse} =\alpha\frac{\Omega_D}{2\pi} =\frac{\alpha A_D}{2\pi L^2} =\frac{\alpha D^2}{8L^2}. \] Numerically, \[ \eta_{\rm diffuse} =\frac{0.10(2.6)^2}{8(3.8\times10^8)^2} =5.85\times10^{-19}. \] This includes reflection from the whole illuminated lunar patch.

The reflector’s advantage is therefore \[ G=\frac{\eta}{\eta_{\rm diffuse}} =\frac{5.89\times10^{-12}}{5.85\times10^{-19}} \approx1.01\times10^7. \] Thus \[ \boxed{G\approx10^7}. \]

Elementary diffraction-spot approximation

A simpler estimate treats each beam as uniformly spread over a disk of radius \[ r\approx\frac{\lambda L}{a}, \] where \(a\) is the emitting aperture diameter. It gives \[ r_{\rm Moon}\approx101\,\mathrm m,\qquad r_{\rm Earth}\approx1311\,\mathrm m. \] Taking the intercepted energy fractions as area ratios then yields \[ \eta_{\rm spot} \approx \frac{(d/2)^2}{r_{\rm Moon}^2} \frac{(D/2)^2}{r_{\rm Earth}^2} = \left(\frac{Dd}{2\lambda L}\right)^4 =9.7\times10^{-13}. \] The corresponding results are \[ N_{\rm eye,spot}\approx12,\qquad G_{\rm spot}\approx1.7\times10^6. \]

This is a useful elementary estimate. The circular-aperture calculation gives a larger return because the actual diffraction pattern is centrally peaked, and both receiving apertures sample its central region.

The solution was checked against the official solution.