Ординатура / Офтальмология / Английские материалы / Optics Learning by Computing with Examples using MATLAB_Dieter Moller_2007

4.Minimum value of Y (σ ) The value of Y (σ ) at the minimum is obtained by differentiation and setting equal to 0. The result is

σ min : −B(n) ·

π

2 · A(n)

σ min 4.286.

Calculation of the corresponding value of Y (σ min)

Y (σ min) −0.013.

For our choice of parameters, Y (σ min) is positive and LSA may not be eliminated.

Application 11.4.

1.Study the π–σ equation and give two examples for elimination of spherical aberration, for a positive and a negative value of x0.

2.Consider a set of lenses all having f 10cm, n 1.5, and radii of curvature

r1 and r2 such that the shape factor σ is between −2 and 2. Plot f s − f depending on σ , where f s is the corrected focal length for spherical aberration. Make sketches of the radii of curvature for values of σ −2, −1, 0, 1, 2 and compare with Jenkins and White (1976, p. 145).

11.5 COMA

So far we have discussed spherical aberration produced by the size of the lens for on-axis points. When the object point is slightly off axis, the resulting aberration is called coma. A new axis appears from the object point through the center of the lens to the center of the image (Figure 11.5). The zones of the lens, indicated by points in Figure 11.5, produce circles instead of image points. Only the center zone produces a point image on the new axis. Larger zones produce circles with larger radii depending on the distance from the new axis. The rings, corresponding to the zones, are arranged like the tail of a comet.

We assume for this discussion of coma that spherical aberration has been eliminated and follow Jenkins and White (1976, p.163). We assume that parallel light is incident on the lens, and the sagittal coma CS is

where ρ is the radius of the largest zone considered, π and σ are defined in the same way as in Eq. (11.24) and β is the angle between the axis of the system and the new axis (Figure 11.6a). For W and G one has

The tangential coma CT is shown in Figure 11.6b and is calculated to be 3CS . The condition for elimination of coma is obtained from Eq. (11.31), when [W σ +Gπ]

424 11. ABERRATION

FIGURE 11.5 Circles of coma of increasing radius corresponding to increasing diameters of the zones of the lens: (a) negative coma; (b) positive coma; (c) zones.

is equal to zero or

In FileFig 11.5, a graph is shown of CT (ρ), calculated depending on the radii of the zones ρ. There are positive and negative values for CT (ρ) depending on the choice of the refractive index and one may choose parameters such that coma is eliminated.

FIGURE 11.6 Coordinates for calculation of coma assuming incident parallel light: (a) Relation of zones of the lens to centers of circles of images. Q0 is on the new axis. The angle between the system axis and the new axis is β. The point Q4 is the center of the circle of the largest zone of the lens; (b) tangential coma CT and sagittal coma CS .

FileFig 11.5 (A5COMAS)

Calculation of coma of the thin lens. It is assumed that spherical aberration is eliminated. Tangential coma is calculated depending on the zone radius.

A5COMAS is only on the CD.

Application 11.5.

1.Choose values of the refractive index n to obtain positive and negative coma.

2.Find an example of a set of parameters for “no coma.”

11.6APLANATIC LENS

One may design a special lens, called an aplanatic lens, which has no coma and no spherical aberration. We choose

426 11. ABERRATION

FIGURE 11.7 Aplanatic lens. Light from the point x0 r1 is not refracted on the first surface of the lens, but on the second. This is shown in (a) for a lens and in (b) for a sphere.

According to Eq. (11.33) there will be no coma, and no spherical aberration because Y 0 (Eq. (11.30)). An example is shown in Figure 11.7. Introduction of σ and π from Eq. (11.34) into the four expressions of Eqs. (11.24) and (11.25) results in

To use it as an aplanatic lens, in addition to the thin lens equation, one has to satisfy the relation between xi and x0:

We see from Figure 11.7a that the emerging rays from x0 are not refracted at the surface of radius r1, but only when arriving at the surface of radius r2. Therefore, we may consider x0 at P as the object point in a medium of refractive index n and spherical surface of radius r2. This is called an aplanatic sphere,

shown in Figure 11.7b. The virtual image point P is obtained by tracing back the ray refracted at the second surface because the image is virtual.

In FileFig 11.6 we show that coma is eliminated for the aplanatic lens.

FileFig 11.6 (A6COMPLANS)

Calculation of coma for the aplanatic lens. The result is that coma is zero.

A6COMPLANS is only on the CD.

Application 11.6.

1.Make a graph of ss(nn) and find back the value for σ at nn 1.5.

2.Get Y from FileFig 9.4 (A4SPHLSIPIS) and using the values from FF6 show that Y 0.

3.Consider a farsighted eye and assume that the focal length is f 2.2 cm and the distance from eye to retina d 2 cm. Design an applanatic lens for correction, that is, bringing the focal point of the eye to the retina.

11.7ASTIGMATISM

We have discussed in Sections 11.3 and 11.5 spherical aberration and coma. For spherical aberration the object points were assumed to be on axis, and for coma slightly off axis. When the object point is farther away from the axis, the image points are no longer in one perpendicular plane with respect to the new axis, as we found for coma. The points appear one behind the other on a new axis from the object point through the center of the lens. There are two planes defined with respect to the new axis, called sagittal (horizontal) and meridional (vertical) planes (see Figures 11.8 and 11.9). Each produces its own image point and between these two points is the circle of least confusion.

The difference of the sagittal and meridional image points on the new axis is called the astigmatic difference, ASD xiH − xiV . Again we follow Jenkins and White (1976, p. 167).

11.7.1 Astigmatism of a Single Spherical Surface

428 11. ABERRATION

φ

φ

FIGURE 11.8 Astigmatism of a single surface. Rays in the meridional (vertical) plane from the primary focus QV , and rays in the sagittal (horizontal) plane form the secondary focus QH . The circle of least confusion is between them.

where r is the radius of curvature of the spherical surface. The angles φ and φ are shown in Figures 11.8a and b for the horizontal and vertical plane, respectively.

In FileFig 11.7 the astigmatic difference ASD is calculated depending on the angle φ. The angle φ may be eliminated using the law of refraction.

11.7.2 Astigmatism of a Thin Lens

For the calculation of the ASD for a thin lens, one has for the sagittal and meridional case each a thin lens equation with a corrected focal length. This is similar to the above discussions of other aberrations. For the sagittal plane one gets

r2 −12, and n 1.3. The first graph shows the calculation of the astigmatic difference ASD depending on the angle φ. The second graph shows for the same lens the dependence on the refractive index n for φ 10 degrees.

FIGURE 11.9 Astigmatism of a thin lens, Meridional plane: All rays from Q meet at primary focus QV . Sagittal plane: All rays from Q meet at point secondary focus QH . The circle of least confusion is indicated by B.

Astigmatism of a single surface. Calculation of astigmatic difference ASD depending on angle φ for a single refracting surface with radius of curvature r.

A7ASTSINS is only on the CD.

Application 11.7.

1.Modify the file for dependence on n for fixed angle φ.

2.Study the ASD for small and large values of r.

FileFig 11.8 (A8ASTISMS)

Astigmatism of a thin lens. Calculation of astigmatic difference ASD for a thin lens of radii of curvature r1 and r2, depending on angle φ or on n.

A8ASTISMS is only on the CD.

Application 11.8.

1.Compare the ASD for lenses with small and large focal lengths.

2.Compare the ASD for a lense with a corresponding lens having one plane surface.

430 11. ABERRATION

FIGURE 11.10 Chromatic aberration: all rays from P1 with shorter wavelength (blue light) converge to point Pblue; all rays with longer wavelength (red light) converge to Pred.

11.8CHROMATIC ABERRATION AND THE ACHROMATIC DOUBLET

The aberrations discussed so far, spherical aberration, coma, and astigmatism, are called monochromatic aberrations. One assumes that monochromatic light is incident on the lens and therefore the refractive index of the law of refraction is a constant. If the incident light is not monochromatic and the constant of the law of refraction is different for different wavelengths, one speaks of chromatic aberrations.

In Figure 11.10 positive chromatic aberration is shown. This means the image point of the shorter wavelength light (blue) is closer to the lens than the image point of the longer wavelength light. The reverse case is called negative chromatic aberration. Chromatic aberration can be compensated by using two lenses. One has positive and the other negative chromatic aberration. Such a double lens is called an achromatic doublet.

An achromatic doublet is made of two lenses in contact. In Chapter 1 we found that the focal length of a compound lens is

where r1 and r2 are the radii of curvature of the lens with refractive index n1, and r1 and r2 are the radii of curvature of the lens with n2. Using the abbreviations

The condition to have the focal length independent of the refractive index in the wavelength range λb to λr is

which translates into a condition for the wavelength dependence of the refractive index

where nb stands for the short wavelength limit of the light (blue) and nr for the long wavelength limit of the light (red). One then gets

Introducing the average refractive indices n1 n1D and n2 n2D , where n1D and n2D are the values of n1 and n2 in the middle between nb and nr , we define

V1 (n1b − n1r )/(n1D − 1) and V2 (n2b − n2r )/(n2D − 1). (11.50)

As a result we have

An example is given in FileFig 11.9 for the calculation of the final focal length f , when the refractive indices and the radii of curvature for calculation of the focal length f1 are given.

FileFig 11.9 (A9ACHROMS)

Calculation of chromatic aberration. Calculation of elimination of chromatic aberration. The focal length of an achromatic doublet with t 0 is calculated.

A9ACHROMS is only on the CD.

Application 11.9.

1.Do the calculation for a chosen value of f , which means determine the corresponding r1 and r2 for f1.

2.A doublet of two lenses should have flat surfaces in the middle. The doublet should have the final focal length f 50 cm. Use V1 and V2 of FF9 and calculate r1 and r4.

3.Do the calculation for a chosen wavelength range in the visible and for f 15cm. Use two different materials. The corresponding refractive indices may be obtained from handbooks or Jenkins and White (1976, p. 177, find f1).

432 11. ABERRATION

11.9CHROMATIC ABERRATION AND THE ACHROMATIC DOUBLET WITH SEPARATED LENSES

Chromatic aberration may also be eliminated when using two lenses at distance t. From Chapter 1, Section 1.3, we have for the focal length of such a system

We want to determine t for given focal lengths f1 and f2 and choose refractive indices with their wavelength dependence.

For one lens one has

From V1 (n1b − n1r )/(n1D − 1) of Eq. (11.50), which may now be written for i equal to 1 or 2

The condition to have no chromatic aberration is obtained from (1/f ) 0 and one gets

An example is given in FileFig 11.10 for the calculation of the distance of the two lenses. To calculate Vi we have used the same values as in FileFig 11.9. The two lenses are assumed to have different radii of curvature and the distance t for no chromatic aberration is calculated.