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Solutions to Problems in Chapter 3: Geometrical Optics

[Figure and equation numbers in the answers are preceded by "iii" so as to avoid confusion with figure and equation numbers in the text.] 3.1 The foci of a mirror in the form of an ellipsoid of revolution are conjugate points (Problem 2.5). What is the magnification produced, in terms of the eccentricity of the ellipsoid? (Tricky) In fact, the image formed from a finite object is so severely aberrated that its magnification can not be defined. If we consider the system as a symmetrical one (rays passing through the region around A in Fig. iii.1), then its magnification is clearly unity. If we consider the region around B as an approximately spherical mirror, then the magnification is (1 + e)/(1 - e), where e (a2 - b2 )1/2 /a is the eccentricity of the ellipsoid; similarly, the region around C gives (1 - e)/(1 + e).

Fig. iii.1. Imaging by an ellipsoid of revolution.

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Fig. iii.2. Periscope design. The rays designated by full lines enter and leave parallel to the axis, and those designated by broken lines are at 30 to the axis.

3.2 In order to use a microscope to observe an inaccessible specimen, one can introduce a relay lens between the specimen and the objective, so that the microscope looks at a real image of the specimen. Draw a ray diagram of the system, and find the influence of the relay lens on the exit pupil and the field of view. For ray-tracing we represent the eyepiece by a single lens. The full NA of the objective will only be used if the relay lens subtends a cone of rays at least as steep as the objective can accept. This is of course difficult to achieve if the objective has high NA, and so the relay lens becomes the aperture stop of the system and reduces its effective NA. Its image in the combination of objective and eyepiece is now the exit pupil. Since the relay lens is outside the focus of the objective, a real image of it is formed between the objective and the eyepiece, and then the exit pupil is further away from the eyepiece than in the original microscope. The field of view is limited by the eyepiece, and is unaffected by the relay lens according to Gaussian optics. In practice, it is difficult to find long focal length lenses of good enough quality not to degrade the performance of a well-designed objective at off-axis points, and so the relay lens generally reduces both the resolution and the usable field of view of the microscope. 3.3 Design a periscope having a length of 2 m and a tube diameter of 0.1 m. The field of view must be a cone of semi-angle 30 . The periscope needs several relay and field lenses. Use paraxial optics only. We neglect the mirrors at the two ends (Fig. iii.2). All the light entering the tube must leave through the other end, so these are entrance and exit

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pupils. There has to be a lens at the entrance to permit all the light up to 30 to enter. Its focal length, which we shall call 2f , is determined by the requirement that a marginal ray at the edge of the angular field of view focusses within the tube; this gives 2f 8 cm. The focal plane is the place for a field lens, of focal length f , which creates an intermediate pupil of diameter equal to the entrance pupil. In the plane of this pupil we put a relay lens, also of focal length f , and so on. For the given dimensions, and 2f = 8 cm, the total number of lenses is 26; the figure shows the principle for a shorter periscope. 3.4 A Galilean telescope has an objective lens with a long positive focal length and an eyepiece with a short negative focal length.

1. What is the distance between the two lenses, when the telescope is focussed on infinity and the image appears at infinity? 2. Is the image upright or inverted? 3. Where is the exit pupil? 4. What determines the field of view in this type of telescope? 5. Why are Galilean telescopes rarely used except as toys?

(a) The distance between the lenses is the algebraic sum of the focal lengths (see Fig. 3.10), which is shorter than the focal length of the objective lens. (b) The image is upright. (c) The exit pupil, being the image of the objective lens in the (diverging) eyepiece, is between the two lenses. (d) The field of view is determined by the size of the objective lens, since the eye can not be placed at the exit pupil. (e) Galilean telescopes are rarely used because their field of view is limited, and also because a reticle or cross-hairs can not be placed at the intermediate image position, which is behind the observer's eye. However, a Galilean telescope is shorter than an astronomical telescope using the same objective, and the image is upright. 3.5 Two converging lenses are separated by a distance a little greater than the sum of their focal lengths. Show that this combination produces a real image of a distant source, but that the focal length is negative! How can you explain this surprising fact physically? We have to locate the principal planes of the combination. When we do this, we find that H is to the right of F2 , which explains why the focal length is negative; see Fig. iii.3.

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Fig. iii.3. Two converging lenses have a negative focal length.

3.6 A compound lens consists of two positive thin lenses L1 and L2 , with focal lengths 90 mm and 30 mm and apertures 60 mm and 20 mm respectively. L1 L2 = 50 mm. Between the lenses, in the plane 30 mm from L1 there is an axial aperture with diameter 10 mm. Where is the aperture stop, for a given axial object 120 mm in front of L1 ? Find also the positions of the entrance and exit pupils. The axial aperture is the aperture stop. The entrance pupil is 45 mm to the right of L1 and the exit pupil 60 mm to the left of L2 . They are both virtual images of the aperture. 3.7 The following is a useful method of finding the refractive index of a transparent material in the form of a parallel-sided plate with thickness d. A microscope is focussed on an object. The plate is inserted between the object and the microscope objective, and the microscope is re-focussed. The distance that the microscope moves in refocussing is measured. Find the relationship between this distance, the refractive index and d. Estimate the accuracy of the method. For Gaussian optics, the displacement is x = (1 - n-1 )d. For large apertures, spherical aberration becomes significant. As in problem 2.6 the depth of focus is given by x = 2f 2 /D2 , where D is the diameter of the lens, from which the accuracy of determining n can be deduced. 3.8 A planar object is imaged by a thin lens. The object lies in a plane which is not normal to the optical axis of the lens. Show that the image lies in a plane which is also inclined to the optical axis, and that the object and image planes intersect in the plane of the lens. This is called the Scheimpflug construction, and is important in the design of cameras for architectural and aerial photography. Show that the image of a rectangular object is distorted into a trapezoidal shape. We will solve this algebraically. For a thin lens, v -1 - u-1 = f -1 . Now

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Fig. iii.4. Scheimpflug construction.

we define the object position in terms of its height h1 as u = u0 + h1 , and the corresponding image position will be v = v0 + h2 , where the object and image intersect the axis at u0 and v0 and their planes are tilted at and to the axis (Fig. iii.4). Using the small-angle approximation, we have:

1 1 1 1 1 = - = - f v u v0 (1 + h2 /v0 ) u0 (1 + h1 /u0 ) 1 h2 1 h1 (1 - )- (1 - ) v0 v0 u0 u0 h2 h1 1 = - 2 + 2 . f v0 u0 (iii.1) (iii.2) (iii.3)

Thus

h1 h2 = 2 . 2 v0 u0 (iii.4)

Now h2 /h1 = v0 /u0 from which (iii.4) gives / = v0 /u0 . Remembering that v0 and u0 have opposite signs, this means that the object and image planes intersect in the plane of the lens. 3.9 Within the limitations of Gaussian optics, is it possible to replace a glass sphere of any refractive index by a single thin lens ? If the thin lens is symmetrical and made of glass with the same refractive index as the sphere, what are its radii of curvature? The matrix for a glass sphere with radius R and refractive index n is

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Fig. iii.5. Retro-reflection (a) by an eye, (b) by a glass sphere. The latter is obtained for non-paraxial rays if 2 < n < 2.

a c

b d

=

2/n - 1 -2(n - 1)/Rn

2R/n 2/n - 1

(iii.5)

Then, we find that the distance from the first vertex to the first principal plane is (a - 1)/c = R and that from the second vertex to the second principal plane is (1 - d)/c) = -R. Thus, the two principal planes coincide and lie in the diametrical plane of the sphere, z = 0. As a result, the sphere can be replaced by a single thin lens. For a biconvex lens with radii ±R0 the value of c = -1/f = 2(1 - n)/R0 and equating this to c in (iii.5) we find that R0 = nR. 3.10 Why do eyes reflect incident light back in the direction of the source (the "red eye" phenomenon in flash photography)? How are "cat's eye" reflectors on roads and road signs constructed? Fig. iii.5(a) shows the rays trace for parallel light incident on an eye. The light is focussed onto the retina. Any reflected light returns parallel to the incident direction and therefore appears as a bright source. This is called a "retro-reflector". The colour red indicated that the retinal cells reflect more red light than other colours; for cats, the relevant colour is green. "Cat's eye" reflectors on the road are an approximation to this situation. For a sphere of glass, the front surface acts as a lens with a lot of spherical aberration. Paraxial incident rays focus onto the back surface when the refractive index n = 2, but rays further from the axis focus closer to the front and are not retro-reflected. When 2 < n < 2 rays at some distance from the axis (i.e. angle of incidence ^ between 0 and i i 90 ) are retro-relected. From Fig. iii.5(b) you can see that ^ has to satisfy n sin(^/2) = sin ^, which is satisfied by an annulus of the incident light; i i enough to give retro-reflection of part of the incident light. For example, for n = 1.51, retro-reflection occurs for rays at ^ = 80 (shown emphasized i in the figure).

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Fig. iii.6. Relationship between focal length and f-number, and the separation between the zoom lens components.

3.11 A zoom lens consists of two lenses, with focal lengths 100 mm and -20 mm respectively. Plot a graph showing the effective focal length and f -number of the combination, as a function of the distance between the two lenses. From (3.52) we see that for two lenses with focal lengths f1 and f2 separated by l,

-c = 1 1 1 l = + - . feff f1 f2 f1 f2 (iii.6)

Fig. iii.6 shows the focal length calculated from (iii.6). We assume the positive lens to be the aperture stop and take its diameter as, say, 20 mm. The f -number is inversely proportional to the angle made at the focus by an edge ray from infinity going through the aperture stop. It is therefore given here by f /20, its value is shown on the right hand scale. 3.12 A glass shell with refractive index 1.5 has equal radii of curvature on both sides ( one is convex, the other concave). The radii are both 100 mm and the thickness is 1.5 mm. (a) Without carrying out any calculation, decide whether the shell acts as a lens with positive or negative optical power. (b) Find its focal length and principal planes. (a) See Fig. iii.7. A ray entering parallel to the axis will be refracted towards it, and therefore meets the second surface when it is closer to the axis. It is therefore deflected back by a smaller angle. It is useful to look at the prism formed by the relevant tangent planes at the points of intersection (shown by broken lines) to see the ray deviation, which is always in the direction of the base of the prism. The lens therefore has

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Fig. iii.7. Ray trace for a glass shell with equal radii of curvature on both sides.

positive power. (b) The focal length is 40 m and H1 and H2 are 200 mm to the left of the two vertices, respectively. 3.13 The glass shell of the previous problem now has two concentric surfaces, the outer one having radius 100mm. Answer the same questions about the new shell. (a) A ray entering parallel to the axis is refracted towards it. It now meets the second surface at a point where normal is at a larger angle (use the prism construction again) and therefore is refracted back at a larger angle than the incident; thus the shell behaves as a divergent lens. (b) The focal length is -19.7 m and the principal planes coincide at distance 100 mm to the right of the first surface. Like the sphere (problem 3.9) the system is equivalent to a single thin lens at its centre. 3.14 Write a computer program based on the Gaussian matrices to find the cardinal points of any paraxial optical system defined by coaxial spherical interfaces between regions of given refractive indices, and/or thin lenses. Use it to check the results of problems 3.4-3.6. A sample program is descibed in the software section of the web-site. 3.15 Show that in a symmetrical imaging system with unit paraxial magnification the distortion must be zero. Each light ray through the system is reversible. On the one hand, the distortion must double if we go back from the image through the lens to the object. On the other hand, the reversal must eliminate the distortion. Thus it must be zero. 3.16 Design a lens of the type shown in Fig. 3.26(b) with n = 2 and f = . What is m when O is at the aplanatic point? Explain physically why the lens magnifies, even though its effective focal length is infinite.

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Fig. iii.8. Ray trace for an aplanatic lens, imaging O to I. The ray in the lower half shows why the system is afocal.

A lens with r1 = -0.5, t = 1, r2 = -1 will have f = and the aplanatic point of the second surface is at the centre of the first one (Fig. iii.8. The lens has its principal planes at . The magnification M = (v - vp )/(u - up ) can then be finite. To avoid infinities, consider the case where r2 is replaced by -0.999 in the above lens. Then f = 499.5 and H1 and H2 are at +250 and +500 with respect to their vertices. Since the object is at -0.5 and the image at -3, the magnification is approximately 2. Actually, imaging by an afocal system is quite common; consider a simple telescope, Fig. 3.10(a), looking at an object at a finite distance. The image, which can be real or virtual, is also at a finite distance. 3.17 An observer sees a object through a thick glass window. Design a paraxial optical system which can be placed before the window so that the observer sees the object in its actual position, as if the window were not there. (This problem was posed as a challenge to optical designers by the journal "Applied Optics" some decades ago, but is quite easily solved using matrix optics when you decide exactly what you require of the principal points.) We define two planes outside the thick window, on opposite sides, separated by distance d. Then, if the complete system including the window is between these planes, and it does nothing, then the matrix between these planes must be the translation matrix in air:

1 0 d 1 (iii.7)

Clearly the system is afocal (feff ), and every entering ray exits along the same line as it enters. One symmetrical system which satisfies this requirement is shown in Fig. iii.9(a), consisting of three lenses. Two rays are shown, one inclined to the axis and one parallel to it. Then, the central

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Fig. iii.9. Cancelling the effect of a thick window: (a) Principle using a three-lens system; (b) Distributing the components symmetrically on both sides.

Fig. iii.10. A symmetrical system with H1 to the right of H2 .

diverging lens can be thickened sufficiently to include the window (Fig. iii.9(b)). Practically, one would build two identical units each consisting of a converging lens and a plano-concave lens; then the plane surfaces would be place in contact with the two sides of the window. 3.18 Can you find a thin lens combination which has positive effective focal length but has the principal point H1 to the right of H2 ? One way to do this is to design a symmetrical zoom lens. Then, since H2 is to the left of the system, by symmetry, H1 must be to its right. Fig. iii.10 is an example.

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Fig. iii.11. Ray diagram for a slide projector.

3.19 A slide projector has a powerful lamp, a condenser lens, a slideholder and an imaging lens. Draw a ray diagram for rays forming an image of a point on the slide on a distant screen, and determine where are the entrance and exit pupils of this system. The ray diagram and pupils are shown in Fig. iii.11 3.20 Discuss the ray optics involved in the formation of a rainbow (both first and second orders) by refraction and reflection of sunlight by spherical water drops. There are also "supernumerary bows", which occur next to the blue edge of the rainbow, when the raindrops are uniform in size. These can be explained by interference. If the water drops were replaced by an assembly of ZnS spheres (n = 2.32), at what angles would the rainbows appear, and how many would there be? [n.b. This problem can not be solved analytically, but needs numerical calculations.] With the help of Chapter 5, explain why light from the rainbow is polarized. The general topic of the rainbow is too complicated to do more than summarize it here, and the reader is referred to "The Rainbow Bridge", R. L. Lee and A. B. Fraser, Penn. State University Press (2001). When the deviation of an incident ray, expressed as a function of the angle of incidence (i.e. impact parameter) has a minimum, a stronly reflected ray is seen. Because the refractive index is a function of wavelength, this angle varies with the colour and a rainbow is seen, of order m which is the number of internal reflections suffered by the ray. For water, n = 1.33, the rays with m = 1 and m = 2 have minima, and are reflected opposite the sum (/2 < < 3/2). However, it follows from the arguments in problem 3.10 above that when n > 2 the picture of first and second-order bows turns out to change dramatically and the results are shown in Fig. iii.13(b) suggesting that the 3rd and 4th order rainbows would be opposite the Sun.

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Fig. iii.12. Formation of rainbows: (a) Ray diagram for a ray with impact parameter b; (b) deviation as a function of angle of incidence for n = 2.32. The minima represent rainbows. Deviations between 1-3, 5-7 ...×/2 (shaded) are opposite the sun.

A ray parallel to the optical axis is incident on a spherical drop of radius R with refractive index n at distance b from the axis. This defines the ray's angle of incidence at the drop surface as ^ = sin-1 (b/R). Then i Snell's law gives the angle of refraction r . From Fig. ??(a) it is easy to see ^ that after m reflections at the drop-air interface, the angle of deviation of the exiting ray is

m (^) = 2(^ - r ) + m( - 2^) i i ^ r = m + 2^ - 2(m + 1)^. i r (iii.8) (iii.9)

The rainbow arises when this function has an extreme value. For n = 1.33 (water), there are extreme values for all m > 0. For a rainbow to be visible, it has to be sought opposite to the sun, implying /2 < < 3/2, 5/2 < < 7/2 etc. Plotting out the values of (iii.9) numerically for 0 < ^ < /2 i o gives visible water rainbows at 138 = (180 - 42) and 230 = (180 + 50)o for m = 1 and 2 respectively, giving bows at 42o and 50o respectively. The colours arise because these angles are dependent on n. The next visible rainbow is for m = 5 at (540 - 49)o . However, the intensity of a rainbow is determined the mth power of the reflection coefficient at an air-water interface. This is typically a few percent (it depends on ^ and i the polarization) and therefore even the second order rainbow is difficult to see, let alone the fifth order one. Looking at the reflection coefficient (see §5.4) it is clear that the angles ^ are closer to the Brewster angle (49o for water) than to normal incidence, i and so the polarization is reflected much better than the . Thus the rainbow is quite strongly polarized tangentially with respect to the anti-

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Fig. iii.13. Aplanatic spheres used as a solar collector.

Sun point. Supernumerary bows occur because, at an angle close to the extremum, there are two rays at different b which have the same deviation. These can interfere, the path difference depending on and R. If R is well-defined, the resultant interference oscillations in the intensity as a function of are visible. When the refractive index is n = 2.32, the third and fourth orders have extrema in the region 5/2 < < 7/2 at 486o = (540 + 54)o and 621o = (630 - 9)o (Fig. ??(b)). The reflection coefficient is rather larger for this refractive index, so but the bows are still quite strongly polarized. 3.21 The aplanatic surfaces of a sphere are spherical, and therefore the edges of a circular source can be imaged with no aberrations using a spherical immersion lens. Discuss how this remark could be developed into a design for an ideal solar concentrator, with the addition of a single lens (as in the microscope objective in §3.8). How could you overcome the problem of chromatic dispersion in this system? Fig. iii.13 shows a cross-section of a glass sphere with the two conjugate aplanatic spheres. Now the Sun is imaged by a collector lens so that the edges of the image lie on the outer aplanatic sphere, and then the edges of the aplanatic image within the sphere have no spherical aberration; thus a solar cell can be tailored to this image with minimum losses. This system is very efficient, particularly because the solar cell is "immersed" in the sphere. This idea can be used to improve the efficiency of an electronic camera, when an aplanatic microlens is attached to each pixel. A problem is that the solar cell or pixels have to be round in order to use it

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most efficiently. Chromatic dispersion could be addressed by designing a suitable doublet for the first (conventional) lens.

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