Читать книгу Optical Engineering Science - Stephen Rolt - Страница 41

Figure 2.3 shows a simplified illustration of an early type of camera lens, the Cooke triplet.

By convention, image space is assumed to be on the left-hand side of the illustration. All lenses are assumed to have no tangible thickness (thin lens approximation) and the axial origin lies at the first lens. Positive axial displacement is to the right.

Figure 2.3 Cooke triplet.

1 Position and Size of Exit PupilIt is easiest, first of all, to calculate the position of the exit pupil, as this is the stop imaged by a single lens (the third lens) of focal length 32.8 mm. The position of the aperture stop, the object in this instance, is 6.4 mm to the left of this lens. The distance, v, of the exit pupil from the third lens is therefore given by:Thus, the exit pupil is 7.95 mm to the left of the third lens and 8.05 mm from the origin. The magnification is given by (minus) the ratio of image and object distances and so it is easy to calculate the size of the exit pupil:

2 Cardinal Points of the LensThe distance between the first and second lenses is 6.8 mm and between the second and third lenses is 9.2 mm. By convention, we retrace dummy rays −16 mm back to the origin at the first lens, so that all matrix ray tracing formulae are referred to a common origin. The matrix for the system is given below:To calculate the position of the exit pupil we need to know the focal length of the system and the positions of the two focal points. Following the matrix relations set out in Chapter 1, we can calculate the following:Focal length: 52.3 mmLocation of First Focal Point: −41.2 mmLocation of Second Focal Point: 57.7 mmAll distances are referenced to the axial origin at the first lens. There is, of course, a single effective focal length as both object and image spaces are considered to lie within media of the same refractive index.

3 Position and Size of the Entrance PupilThe imaged pupil or exit pupil lies in image space, 8.05 mm from the origin. This is 49.65 mm to the left of the second focal point. In applying Newton's formula, the second focal distance, x2 is then equal to −49.65. We can now calculate the first focal distance to determine the position of the entrance pupil.The object or entrance pupil therefore lies 55.1 mm to the right of the first focal point and 13.9 mm (−41.2 + 55.1) to the right of the first lens.The location of the entrance pupil expressed as an object distance is 52.3 − 55.1 or −2.8 mm. Similarly the location of the exit pupil expressed as an image distance is equal to −49.65 + 52.3 or +2.65 mm. The magnification (image/object) is, in this instance equal to 2.65/2.8 or 0.946. Therefore we have:The diameter of the entrance pupil is, therefore, 15.1 mmSo, in summary we have:

Entrance Pupil Axial Location: 13.9 mm	Entrance Pupil Diameter: 15.1 mm
Exit Pupil Axial Location: 8.05 mm	Exit Pupil Diameter: 14.3 mm

Figure 2.4 Optical system with a telecentric output.