## Abstract

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1. It is possible to obtain slightly greater film illumination on the axis of the optical system by displacing the film a short distance toward the exit pupil. However, the illumination of the film is then not uniform and the actual gain is small.
2. If the system is not free from spherical aberration, there will be a small deflection of the spot as the mirror rotates. However, with a perfect image forming system which we shall assume, there can be no displacement of the spot when the mirror is imaged on the film.
3. Since the more logical position of the lens is between the source and the mirror, this is equivalent to making the angle subtended at the film by the lens greater than that subtended by the mirror.
4. The diaphragm opening may then be considered as the source and the illumination remains unaltered if the actual source is of sufficient size. It must, of course, be large enough to avoid being either a field or an aperture stop of the system.
5. The average photographic material used for making negatives is capable of resolving at least 25 lines per millimeter under favorable conditions. The choice of 0.1 mm for the width of the line is consequently conservative.
6. There is a saving in film not only because of the decrease in its width but because the speed of the film will be decreased proportionately also. Thus the area of film required is about proportional to the square of the deflection. These advantages usually more than compensate for the necessity of enlarging the records.
7. When the exit pupil is circular and of radius ρ and is situated a distance p from the film, the illumination on the film isI=πB(ρ2p2+ρ2)For small values of ρ with respect to p this becomesI=πBρ2p2=ωBwhere ω is the solid angle subtended at the film by the exit pupil. For small values of ω, this result is independent of the shape of the exit pupil. In the present case, the exit pupil in one plane does not coincide with the exit pupil in the plane at right angles to it. The effective value of ω, however, isL4·w3V2·V1as given by equation (12).

#### Other (7)

It is possible to obtain slightly greater film illumination on the axis of the optical system by displacing the film a short distance toward the exit pupil. However, the illumination of the film is then not uniform and the actual gain is small.

If the system is not free from spherical aberration, there will be a small deflection of the spot as the mirror rotates. However, with a perfect image forming system which we shall assume, there can be no displacement of the spot when the mirror is imaged on the film.

Since the more logical position of the lens is between the source and the mirror, this is equivalent to making the angle subtended at the film by the lens greater than that subtended by the mirror.

The diaphragm opening may then be considered as the source and the illumination remains unaltered if the actual source is of sufficient size. It must, of course, be large enough to avoid being either a field or an aperture stop of the system.

The average photographic material used for making negatives is capable of resolving at least 25 lines per millimeter under favorable conditions. The choice of 0.1 mm for the width of the line is consequently conservative.

There is a saving in film not only because of the decrease in its width but because the speed of the film will be decreased proportionately also. Thus the area of film required is about proportional to the square of the deflection. These advantages usually more than compensate for the necessity of enlarging the records.

When the exit pupil is circular and of radius ρ and is situated a distance p from the film, the illumination on the film isI=πB(ρ2p2+ρ2)For small values of ρ with respect to p this becomesI=πBρ2p2=ωBwhere ω is the solid angle subtended at the film by the exit pupil. For small values of ω, this result is independent of the shape of the exit pupil. In the present case, the exit pupil in one plane does not coincide with the exit pupil in the plane at right angles to it. The effective value of ω, however, isL4·w3V2·V1as given by equation (12).

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### Equations (25)

$w 4 = w 3 · v 3 u 3 l 4 = l 3 · v 3 u 3$
$I = B · w 4 I 4 ( v 2 - v 3 ) 2 = B · v 3 2 u 3 2 · w 3 l 3 ( v 2 - v 3 ) 2$
$D = 2 α ( u 3 - u 2 ) · v 2 u 2 = 2 α v 2 ( u 3 u 2 - 1 ) = 2 α v 3 u 3 ( 1 v 2 - v 3 )$
$1 u 2 + 1 v 2 = 1 u 3 + 1 v 3$
$w 5 = w 2 v / u$
$l 5 = l 2 v / u$
$2 f = 1 u + 1 v$
$I = B · w 3 l 3 v 2$
$D = 2 · α · v$
$w 5 ≧ 2 λ v w 3$
$l 5 ≧ 2 λ v l 3$
$2 f = 1 u 1 + 1 v 1$
$w 5 = w 2 v 1 / u 1$
$l 5 = l 3 v 2 / u 2$
$I = B · l 4 v 2 · w 3 v 1$
$D = 2 · α · v 1$
$s 1 = d D d t = 2 · v · d α d t$
$E = I · w 5 s 1 = I · l 5 s 2$
$E ≧ σ$
$I · w 5 σ · s 1 ≧ 1$
$B σ · w 3 l 3 v 2 · w 5 s 1 ≧ 1.$
$B σ · l 4 v 2 · w 3 v 1 · w 6 s 1 ≧ 1$
$I=πB(ρ2p2+ρ2)$
$I=πBρ2p2=ωB$
$L4·w3V2·V1$