IPhO 1979, Теория и експеримент — Задача 3

• 1) $2.6 \times 10^{-7}$ rad

  δφ ≈ λ/D ≈ 2.6·10⁻⁷ rad ≈ 0.05″

• 2) $1 \times 10^{-12}$

  K₀ = (dD/2λL)⁴ ≈ 10⁻¹² (order-of-magnitude estimate)

• 3) No: only N = 12 photons reach the pupil, N < n, so the pulse cannot be perceived with a naked eye.

• 4) $2 \times 10^{6}$

  β = K₀/K ≈ 2·10⁶

1. The beam divergence angle $\delta\varphi$ caused by diffraction defines the accuracy of the telescope optical axis installation:

2. The part $K_1$ of the light energy of a laser, directed to a reflector, may be found by the ratio of the area of $S_1$ reflector ( $S_1 = \pi d^2/4$ ) versus the area $S_2$ of the light spot on the Moon ($S_2 = \pi r^2$, where $r = L\,\delta\varphi \approx L\lambda/D$, $L$ – the distance from the Earth to the Moon)

The reflected light beam diverges as well and forms a light spot with the radius $R$ on the Earth’s surface:

That’s why the part $K_2$ of the reflected energy, which got into the telescope, makes

The part $K_0$ of the laser energy, that got into the telescope after having been reflected by the reflector on the Moon, equals

3. The pupil of a naked eye receives as less a part of the light flux compared to a telescope, as the area of the pupil $S_e$ is less than the area of the telescope mirror $S_t$:

So the number of photons $N$ getting into the pupil of the eye is equal

Since $N < n$, one can not perceive the reflected pulse with a naked eye.

4. In the absence of a reflector $\alpha$ =10% of the laser energy, that got onto the Moon, are dispersed by the lunar surface within a solid angle $\Omega_1 = 2\pi$ steradian.

The solid angle in which one can see the telescope mirror from the Moon, constitutes

That is why the part $K$ of the energy gets into the telescope and it is equal

Thus, the gain $\beta$, which is obtained through the use of the reflector is equal

Note. The result obtained is only evaluative as the light flux is unevenly distributed inside the angle of diffraction.