A Quick Rough Sketch

When the power and space layout has been completed, a rough sketch, drawn to scale, is a worthwhile investment of your time. The components can be drawn as single lens elements to start, either as planoconvex or as equiconvex shapes (or planoconcave or equicon-cave). The process is simple because the radius of a planoconvex or planoconcave lens is equal to the focal length multiplied by (n1).

Thus for a glass lens with an index of 1.5, the radius is just half the focal length. If the index is higher, the radius is longer; for an index of 2.0, the radius equals the focal length, and for germanium (at an index of 4.0), the radius is three times the focal length. For an equiconvex, we neglect the thickness and simply use a value of 2(n1)(EFL) for the radius. This works out to the radius equal to the focal length for an index of 1.5, equal to twice the focal length for an index of 2.0, and six times the focal length for germanium.

An achromatic doublet can easily be drawn in a planoconvex (or planoconcave) shape. The positive element (of the planoconvex dou-blet) is drawn as an equiconvex lens with radii the same as the radius of the planoconvex singlet cited above, e.g., half of the doublet focal length for glass elements. This is a crude approximation, but it

suf-fices for our purposes. Several examples of these sketches are shown in Fig. 5.4.

The components should be drawn at the diameters necessary to pass the rays of the system. A paraxial raytrace [using Eqs. (1.19) and (1.20)] of the axial marginal ray and of the principal ray will yield y

n = 1.5

Plano-convex Equi-convex Achromat

n = 2.0

Focal length

Focal length

n = 4.0

Focal length

Figure 5.4 Sample quick sketches of singlets and doublets for several values of the index of refraction. All lenses are drawn to the same focal length and diameter for easy comparison. For the planoconvex lenses (both singlets and doublets) the

sur-and y_pfor each component. Then the diameter necessary to pass the axial bundle is simply equal to 2y. The diameter needed to pass the oblique bundle is equal to 2(y_p V^•y), where V is the vignetting factor for the oblique bundle. The largest of these diameters is the one to use. The rays (axial and vignetted oblique) should be drawn into the system sketch in order to get a better picture of the way the system functions and also to avoid impossibilities. Section 1.9, and specifical-ly Eqs. (1.23) through (1.26), can be used to generate the oblique bun-dle ray data from the data of the axial and principal rays.

If the lenses look too fat to you, the odds are that later on the lens designer will have to split them up in order to get a good-quality image. You can get a feel for what the system may look like by split-ting them yourself; simply draw the lenses with radii increased by a factor equal to the number of elements you split them into. If you need to split a lens in two in order to make it look reasonable, double the radii of your initial sketch. If you need to split it into three, triple the radii, etc. What you will wind up with is a crude, simple represen-tation of what the system might look like. The following paragraphs and the next chapter can help you to decide what sort of a lens design (i.e., singlet, doublet, anastigmat, etc.) will be needed for each compo-nent.

For each component, calculate the relative aperture (f-number) in two ways, one by dividing its focal length by the diameter determined two paragraphs above and the other by using the diameter indicated by the axial marginal ray. The latter is an indication of the speed (f-number) at which it must form a decent image; the former indicates the diameter at which it must be constructed.

Similarly, we need to estimate the angular field that the component will have to cover. Again, we can get two values. The first is simply the angle that a ray from the edge of the object (the object for the com-ponent, not the object for the entire system) through the center of the component makes with the axis; this field angle has to do with the construction, vignetting, etc. Note that this is not the slope of the principal ray, because the principal ray may not go through the center of the component. The second field of view is determined by dividing the smaller of the object or image heights (for the component) by the component focal length.

As an example, consider the erector (component b) of the lens-erect-ing telescope discussed in Sec. 5.3, and shown in Fig. 5.3. Here the objective lens (component a) forms an image of the field, which is then reimaged by the erector. The (first) half-field of view is the height of the image formed by the objective, divided by the object dis-tance s for the relay. The other (second) half-field is the same image

height divided by the component focal length f_b. When the object is at infinity, both field values are the same, but for a system with finite conjugate distances, as is the case for this erector lens, the second field will be larger than the first. This is the field angle to use in deciding what type of lens design form is apt to be able to cover your system’s field of view. The reason for this is that the basic field curva-ture of a lens, called the Petzval curvacurva-ture, is determined by the image size, rather than the field angle, and the second field angle described above gives a measure of this relationship which is closer to what most people describe as the “coverage” angle of a lens system.

These field angles and f-numbers can be used to get a preliminary idea of what sort of optics will be needed to make your system work, perhaps by referring to Fig. 6.13.