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# Luminosity of Telescopes

## By Rafael Chamón Cobos

The luminosity of a telescope, I. e., the brightness of its visual field, can be mathematically defined by the ratio between the light flux reaching the eye when the observer looks through the telescope and the light flux reaching the naked eye from the same target.

By applying this definition in the different cases of observing subjects with a telescope (or with a pair of binoculars) it can be proved that the below indicated formulas apply. (The transmission losses of light in the lenses themselves have not been taken into account in these formulas).

We have to distinguish between two types of observed objects:

1. Light emitting punctual objects, like stars
2. Extensive objects, like the Moon, nebulae and all types of terrestrial views with binoculars

In turn, in each one of the two previous cases other two cases have to be considered in relation to the size of the pupils:

1. The exit pupil of the instrument is equal to or greater than the eye pupil (case of 7x50, 10x70 binoculars, etc.)
2. The exit pupil of the instrument is smaller than the eye pupil (common case of telescopes of high magnification)

The formulas for the four combined cases of observation are depicted in the following table:

 LUMINOSITY OF TELESCOPES (Quotient of light fluxes entering the eye, with and without instrument) A) EXIT PUPIL >= EYE PUPIL EXIT PUPIL < EYE PUPIL 1) Watching PUNCTUAL OBJECTS = (MAGNIFICATION)² = (APERTURE / EYE PUPIL)² 2) Watching EXTENSIVE OBJECTS = 1 = (EXIT PUPIL / EYE PUPIL)²

## DISCUSSION

Refer to the above table

### 1-A) PUNCTUAL OBJECTS - EXIT PUPIL EQUAL TO OR GREATER THAN EYE PUPIL

In this case the luminosity is equal to the square of the magnification. In practical terms, the more magnification, the more stars we will see.

LUMINOSITY = (MAGNIFICATION)²

Common values of magnification in this case are 7x, 8x or 10x. Therefore, luminosity can reach values till 10² = 100, which is equivalent to an increment of 6 stellar magnitudes in all observed stars.

### 1- PUNCTUAL OBJECTS – EXIT PUPIL SMALLER THAN EYE PUPIL

In this case the exit pupil restricts the entrance of rays in the eye pupil and the luminosity is lower than in case 1-A) by a factor equal to the square of the quotient of pupils:

LUMINOSITY = (MAGNIFICATION)² * (EXIT PUPIL/EYE PUPIL)²

By applying the relationship MAGNIFICATION = APERTURE/EXIT PUPIL, we get:

LUMINOSITY = (APERTURE / EYE PUPIL)²

The luminosity is the square of the quotient of aperture of the telescope and the eye pupil. The larger aperture, the fainter stars we will see. Since there is no limit for the aperture of a telescope, the luminosity in this case can reach high values if the aperture is large. For this reason, large aperture telescopes enable us to see stars of much higher magnitude than those which are visible with naked eye.

The formula also shows that in this case the luminosity is proportional to the square of the telescope’s aperture with independence of the magnification, i. e., for the same observer, two telescopes of same aperture and different magnification give the same brightness of the observed stars. This means that a higher magnification will not make visible fainter stars (i. e., of higher magnitude). However, the telescope with higher magnification gives a better visual perception of the stars due to other reasons:

• Stars appear more separated each other
• The sky gets darker.

Since the brightness of the stars remain and the sky gets darker, the contrast between stars and sky is increased. The sky gets darker because it can be considered as an extensive object with a certain brightness of its own, due to the atmospheric contamination. To the sky applies then the case 2-B (see table) and a higher magnification means a smaller exit pupil, therefore a lower luminosity of the sky.

### 2-A) EXTENSIVE OBJECTS - EXIT PUPIL EQUAL TO OR GREATER THAN EYE PUPIL

In this case the luminosity of the telescope or binocular is equal to 1 and the field is seen with same brightness as without telescope.

LUMINOSITY = 1

For instance, during a sunny day the eye pupil is small, say 2mm in diameter. By observing terrestrial targets trough a 10x50 binocular, which has an exit pupil of 5mm, the light flux entering the eye is not restricted by the instrument, therefore the luminosity of the observed field is the same as that of the observation with naked eye.

### 2- EXTENSIVE OBJECTS – EXIT PUPIL SMALLER THAN EYE PUPIL

In this case the luminosity is lower than in case 2-A) by a factor equal to the square of the quotient of pupils. Therefore, the luminosity is less than 1 and the field is always seen darker as without telescope.

LUMINOSITY = (EXIT PUPIL / EYE PUPIL)²

For example, during the twilight the eye pupil opens, let's say, till a diameter of 7mm. In this case, a 10x50 binocular darkens the observed field, because the exit pupil of the instrument -5mm- restricts the light flux entering the eye. According to the formula, the luminosity is then (5mm/7mm)² = 0,51

## CONCLUSIONS

1. For stars gazing, the luminosity of the telescope is solely determined by its aperture. The luminosity can be very high (much >1). The more aperture, the higher stellar magnitude of the observed stars.
2. For terrestrial observation, the luminosity is solely determined by the exit pupil of the instrument. The luminosity is =1 if exit pupil is greater than eye pupil and <1 if exit pupil is smaller than eye pupil.

Source: "Introducción al estudio de los instrumentos ópticos" by Pedro Jiménez-Landi Martínez, Editorial UCM