IAO 2011, Теория, Групи α и β — α/β-2. Планетариуми

Група α — осветеност на купола

Source gap: the top of the scanned page is cut off. The surviving calculation begins with the following displayed formula; the missing opening is not reconstructed.

where $\Gamma$ is the magnification coefficient of the projecting system, $\Gamma=R/r$, $r$ is the distance from the hole in the foil to the projecting lens. By the lens formula, $1/R+1/r=1/F$, but since $R\ll F$, one may use $\Gamma\approx R/F$.

Accordingly, the area of the illuminated surface of the dome which gives the star $0^{\mathrm m}$ is equal to

After slides-foils have been removed the whole dome will be illuminated, i.e. a hemisphere of radius $R=5\,\mathrm m$ will shine, the area of shining surface is

Thus, the shining area will be

In terms of stellar magnitudes the difference is

Thus, the stellar magnitude of the dome will be $0^{\mathrm m}-19.25^{\mathrm m}=-19.25^{\mathrm m}$. It gives an illuminance 1000 times less, than the illuminance outdoors at clear sunny day, and in 400 times larger than illuminance from full Moon. It corresponds approximately to indoors illuminance at daytime. Of course, it is possible to read without any problem.

We should note that the solution does not require the values of limiting stellar magnitude $6^{\mathrm m}$, the number of projecting systems for the hemisphere and the diameter of the dome of the planetarium.

Transcription note: the source's inequality $R\ll F$ is retained, although its numerical approximation uses the opposite scale relation.

Група β — 2.1.

The visible angular size of the “stars” should be less than the resolution of an eye, that is the linear size (diameter) of images of these “stars” on the dome would not exceed $L_0=\alpha\cdot R$, where $\alpha$ is the resolving power of a human eye in the dark (nearby $50''\approx2.5\cdot10^{-4}\,\mathrm{rad}$), $R$ is the radius of the hall of the planetarium. In our case $L_0=\alpha\cdot R\approx2.5\cdot10^{-4}\cdot5\,\mathrm m\approx1.25\,\mathrm{mm}$.

The size of the image of one star appearing on the dome with the optical system is defined by two parameters:

a) The first parameter is geometrical, it is defined by the optical increase of the size of a star at its display on the dome. If the size of a star on a slide is $l_0=0.1\,\mathrm{mm}$, the size of the image is calculated by the formula of increase of the objective (lens): $L=\Gamma\cdot l_0=l_0\cdot R/r$, where $r$ is the distance from the mentioned hole in a foil to the projecting lens. According to the formula for a lens $1/R+1/r=1/F$. In our case the increasing factor should not exceed $\Gamma_0=L_0/l_0$, whence we find that the focal length of the system should be not less than

It is the first condition. The condition is quite feasible.

b) The second parameter is diffraction. It is defined by the size of the Airy disc, the angular size of a divergence of the rays from a dot source (the source being near to focus of a lens) is equal to $\lambda/D$, where $\lambda$ is the wavelength (of order 500 nm or $5\cdot10^{-7}\,\mathrm m$), and $D$ is the diameter of the lens of the projecting optical system. Thus the size of the image of a dot source on the dome of radius $R$ is $R\cdot\lambda/D$. So we need a condition

So, the second condition: the diameter of the objective should be not less than 2 mm. The condition is quite feasible as well.

By the way, we may notice that this value is easy to point out at once: it is the diameter of a pupil exposure that corresponds to the resolution of a human eye in the conditions of day light.

Formal answer: two parameters:

1. Focal length of the system $F\ge l_0/\alpha\approx40\,\mathrm{cm}$.

Source gap: the continuation of the formal list is missing at the top of the next scanned page. The second condition, $D\ge2\,\mathrm{mm}$, is present in the preceding paragraph.

Група β — 2.2.

Source gap: the top of the scanned page is cut off. The surviving calculation begins with the following displayed formula; the missing opening is not reconstructed.

where $\Gamma$ is the magnification coefficient of the projecting system, $\Gamma=R/r$, $r$ is the distance from the hole in the foil to the projecting lens. By the lens formula, $1/R+1/r=1/F$, but since $R\ll F$, one may use $\Gamma\approx R/F$.

Accordingly, the area of the illuminated surface of the dome which gives the star $0^{\mathrm m}$ is equal to

After slides-foils have been removed the whole dome will be illuminated, i.e. a hemisphere of radius $R=5\,\mathrm m$ will shine, the area of shining surface is

Thus, the shining area will be

In terms of stellar magnitudes the difference is

Thus, the stellar magnitude of the dome will be $0^{\mathrm m}-20.25^{\mathrm m}=-20.25^{\mathrm m}$. It gives an illuminance 400 times less, than the illuminance outdoors at clear sunny day, and in 1000 times larger than illuminance from full Moon. It corresponds approximately to indoors illuminance at daytime. Of course, it is possible to read without any problem.

We should note that the solution does not require the values of limiting stellar magnitude $6^{\mathrm m}$, the number of projecting systems for the hemisphere and the diameter of the dome of the planetarium.

Transcription note: the source's inequality $R\ll F$ is retained, although its numerical approximation uses the opposite scale relation.