INTRODUCTION THIN LENSES. Introduction. given by the paraxial refraction equation derived last lecture: Thin lenses (19.1) = 1. Double-lens systems

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1 Chapter 9 OPTICAL INSTRUMENTS Introduction Thin lenses Double-lens systems Aberrations Camera Human eye Compound microscope Summary INTRODUCTION Knowledge of geometrical optics, diffraction and interference, allows calculation of the operation and performance of instruments that focus electromagnetic waves. There are many applications to both optical and non optical instruments. This discussion will be limited to three optical systems that are most frequently encountered in the medical field, namely, the camera, the human eye, and the microscope. These illustrate the general features of optics. First it is necessary to complete the discussion of the optics of thin lens systems. Converging thin lens THIN LENSES Consider two refracting convex spherical surfaces as shown in figure. Assume that the object and image are in air with refractive index unity, while the glass lens has refractive index n. At the first surface the image distance is drawn for the case where it is negative. At the first surface the object and image distance are given by the paraxial refraction equation derived last lecture: = (9.) The image of surface serves as the object for surface 2 with object distance 2 = + where is the distance separating the two surfaces. Thus for the second surface we have a paraxial refraction equation: + + = 2 2 The geometry for this pair of refracting surfaces is simplified if the thin lens approximation is assumed that is, =0Then this last refraction equation equals: + = (9.2) 2 2 Combining the paraxial refraction equations for each surface gives: + µ =( ) 2 2 (Thin-lens approximation) Define the focal length as: µ =( ) (Lens-makers equation) 2 Thus the thin lens equation can be written as: + = (Thin-lens equation.) where the subscripts and 2 are dropped. The adopted sign convention has positive on the incident (left-hand) side, positive on the transmission (right-hand) side, and positive for a converging thin lens and negative for a diverging thin lens. Note that if either or are infinite, then the other distance equals the focal length. That is, a parallel ray focusses at the focal point a distance from the lens. Figure 2 shows the focal points for converging and diverging lenses. The power ofalensisdefined as: = (Power of a lens) The unit of lens power is the diopter which is the value of when the focal length is given in meters. Thus a +0 diopter lens is a converging lens with focal length 0, whilea 0 diopter lens is a diverging lens of focal length 0. From figure 3 it is obvious by proportion that the magnification for a converging lens is given by: Figure lens. Refraction at two spherical surfaces of a glass = = (Magnification) 4

2 Figure 2 Focal points for converging (a) and diverging (b) lenses. Figure 4 Image for a converging lens of focal length =333 for (a) =20, and(b) =0. Figure 3 Magnification for a converging lens. Figure 5 A diverging lens system. Example of converging lens Consider the glass has =5, and the lens surfaces have = 0 and 2 = 20 Then using the lens-makers equation we have; = µ =(5 0) 00 =75 diopters 020 Thatis,thefocallengthis =333. Considerthecaseof =20 shown in figure 4a, then substitution in the lens formula gives: 02 + =75 that is, =40 For this case the magnification is = 20 that is the image is inverted. Considerthecasewhere =0 shown in figure 4b, then substitution in the lens formula gives = 40 and magnification =+40that is the image now is erect, not inverted. Example of diverging lens For the diverging lens shown in figure 5, we have a power given by: = µ =(5 0) 05 = that is the focal length is = 2 If the object is at =20 then the thin lens formula gives the image at I where: 02 + = 8333 Thus = 75, that is the image is virtual. The magnification is =+0375that is the image is erect not inverted. DOUBLE-LENS SYSTEMS For multiple lens systems, the final image is found by calculating the image from the first lens and then using this image as the object for the next lens to calculate the second image. This process can be repeated for a series of lenses. Let us consider double-lens systems. Twoconverginglenses The microscope, telescope, magnifying lens, and the use of reading glasses all are examples where one has systems of two converging lenses. For the system shown in figure 6, the thin-lens formula for the first lens of focal length gives: = (Lens ) because is negative. Using the image of lens as the object of lens 2, and taking into account the separation 42

3 Figure 6 Double-lens system d between the lenses, gives: ( + ) + = (Lens 2) 2 2 Using these two equations allows calculation of the final image. Moreover, the total magnification of the system is the product of the separate magnifications: = 2 Note the special case when the separation between thelensesiszero, =0Then adding the two lens equations gives: + = That is, the power of a compound pair of lenses with zero separation is Figure 7 Barrel distortion for a wide angle lens. = + 2 (Compound lenses with zero separation) Thus the powers are additive if the separation between the lenses is zero. Contact lenses are an example of compound lenses with essentially zero separation between the lens. The eye needs to have an effective power of about 50 diopters to focus far objects on the retina. For the myopic, short-sighted, eye the power of the lens in the eye is too powerful so a diverging lens with negative power, e.g. 0 diopters, is used to reduce the power of the compound system. For the long-sighted eye, the eye lens is not powerful enough so a converging lens, with positive power, is used to increase the power of the compound system. ABERRATIONS The lens and mirror equations were derived assuming the paraxial approximation, that is, all angles are small enough to assume that sin = This breaks down if the radius of the lens or mirror is large compared with the object or image distances, or if the object is far from the axis of the lens system. When the smallangle approximation fails, the rays through different parts of the lens do not focus at the same point giving a blurred image. Spherical aberration is due the Figure 8 Chromatic aberration is the different focal lengths for different frequencies due to the variation of refractive index on frequency. break down caused by the size of the lens while barrel or pin-cushion distortion is due to large diplacement of theobjectfromtheaxisofthelens. Comaandastigmatism are combinations of these two effects. Barrel distortion is observed when using a wide-angle camera lens as shown in figure 7. Chromatic aberration is caused by the fact that the refractive index is frequency dependent. Thus the focal length of a thin lens is frequency dependent causing different colors to focus at different image distances as shown in figure 8. It is possible to make a compound lens of two types of glass that have different dependences of refractive index on frequency. With such an achromatic doublet, illustrated in figure 9, it is possible to arrange that the focal length is the same at two frequencies. Note that chromatic aberration is not a problem for reflective imaging systems. 43

4 Figure 9 Achromatic doublet. be assumed that ' The performance of the camera can be broken into five main elements; lens speed, shutter speed, resolution, depth of focus, and field of view. Camera design is a compromise between the conflicting requirements of these five aspects. Lens speed The light flux collected by the camera lens is proportional to the area of the lens aperture, that is 2 4 Since the image size is proportional to, then the image area is proportional to 2 Thus the power ³ 2 density of the image is Define the Fnumberof the lens, as = (F-number of a lens.) Thus the image power density is: Figure 0 The main elements of a camera. CAMERA The camera is a relatively simple optical system and yet it demonstrates the many compromises that must be made is design and use of optical systems. Figure 0 shows a schematic of the main elements of the camera. The lens is shown in figure is an achromatic doublet to correct for chromatic aberration. Most camera lens comprise several lenses to minimize aberrations in order to obtain a good resolution with a wide field of view. The iris of the lens has an aperture of diameter D. The object distance normally is thus the image distance is For example, a typical 35mm camera lens has a focal length of =50. Thus =50 for = and =53 when =2 Therefore in the subsequent discussion it will Image power density 2 The brightness of the final image is proportional to the energy density which is the power density times the time the shutter is open. Note that for an expensive camera lens =22 is the smallest aperture and =0 the largest aperture. The ratio of image power densities is almost 500 for these extreme cases. Moreover, one can vary the shutter speed by a factor of 0 3 increasing the dynamic range of usable light intensity to nearly 0 6 Shutter speed Clearly, the shutter speed must be sufficiently small to ensure that the image does not move significantly while the shutter is open. Unfortunately, the shorter the shutter speed, the less light that is collected. A fast film, and small F number are required to compensate for use of a fast shutter speed. Resolution There are two limits to resolution, diffraction and granularity of the film emulsion or the CCD digital sensor. Diffraction sets an angular resolution of =22 For an image distance of = this gives the spatial width between the first minima as: =244 (Diffraction spatial resolution.) Figure The optics of a camera. This implies that the best diffraction resolution is obtained with small F numbers. Typically, for = 500the spatial resolution is =22. The emulsion of a slow fine-grain film has a resolution of about 0 which limits the resolution in a 35mm camera for 0 Faster speed films, that is, those with a higher ASA number have poorer spatial 44

5 Figure 2 Diffraction spatial width. Figure 4 Illustration of the depth of focus of a camera. Figure 3 Depth of focus resolution. Digital cameras have a poorer resolution due to the limited number of pixels used to record the picture, that is, spatial resolution of 35 for a typical camera. Depth of focus Onewouldliketohavebothnearandfarobjectsin focus. As shown in figure 3, relocating the object relocates the image resulting in an out-of-focus blurr of diameter on the emulsion. Since + = then = 2 Thus the width of the blurr on the emulsion 2 is given by: That is: = = 2 2 = 2 2 (Depth of focus) Considera35mmcamerawith =5, =25 and let the blurr width equal the emulsion resolution =0 5. These give a depth of focus for an object at =, of =!. That is, the depth of focus is only one centimeter, which is half the depth of your nose. To achieve good depth of focus one needs to use a very large F number, such as a pinhole camera. Figure 4 shows an example of deph of focus. Field of view An f=50mm focal length lens, on a 35mm camera, has about the field of view of the eye. A smaller focal length is required to obtain a wider field of view for a Figure 5 The human eye given size sensor. However, such a smaller focal length implies a smaller image for which film graininess will be more important. The wider field of view makes lens aberrations more noticeable. Barrel distortion is a typical problem using very wide-angle lenses. It is not possible to simultaneously satisfy the conflicting requirements of lens speed, shutter speed, resolution, depth of focus and field of view. The camera parameters are adjusted by each photographer to achieve his or her balance of these conflicting requirements for each picture. HUMAN EYE The optics of the human eye is effectively the same as that of the camera, and thus we can use the equations developed for the camera. The components of the eye are shown in figure 5. The focal length of the eye is 20, the iris ranges from =5 to 6, and thusthefnumberrangesfrom3to3whichisquite modest compared with a typical camera. The cones on the retina are clustered around the central region of view and have a spatial resolution of 45

6 Figure 7 Figure 6 Wavelength sensitivity of rods and cones of the human eye. 0 6 The cones are surrounded by the rods which have a resolution of Thus the cones have an angular resolution of 02 minutes of arc while the rods have an angular resolution of 06 minutes of arc. Diffraction at the pupil of the eye limits the resolution to about which is a good balance with the resolution of the rods and cones in the retina. Note that the moon and sun both subtend about 33 minutes of arc at the earth, thus the best angular resolution achievable by the eye is the order of 06% of the angular size of the moon. The F number of the eye only changes by a factor of 4, so the speed of the lens only changes by a factor of 6 due to opening of the iris. However, in the dark the eye sensitivity increases by a factor of 5000 over a period of 5 hours. This behavior is observed when you walk into a darkened room from bright daylight, initially you cannot see objects that are clearly visible by the dark-adapted eye. The explanation is that the cones, which have an excellent spatial resolution and color sensitivity, have low light sensitivity. However, the rods have high light sensitivity, lower spatial resolution and are insensitive to red light. Figure 6 compares the wavelength response of the cones and rods. The rods are saturated in bright light and their light sensitivity gradually recovers in the dark. Note the the peripheral vision of the eye, which focusses on the rods, is the most sensitive at low light levels, but it does not see red. Thus highest sensitivity is achieved by looking slightly to the side of an object at night rather than staring directly at it. Red light is used in aircraft cockpits to ensure that the eye remain dark adapted at night. The power of the eyeball lens decreases with age resulting in a steady increase in the near point as shown in figure 7. As a result it is necessary to use reading glasses which comprise converging lenses which produce a virtual image at the near point of the eye allowing one to focus as shown in figure 8. Myopia is the opposite effect where the focal length of the eye is too small resulting in an image in front of Figure 8 The use of a converging lens produces an enlarged virtual image that can be seen by the hyperopic eye. the retina. A negative power lens is required to weaken the power of the eye lens as shown in figure 9 COMPOUND MICROSCOPE The compound microscope is shown in figure 20. The objective forms an enlarged, inverted, real image of the object. This image is at a distance L, called the tube length, from the near focal point. Figure 9 The myopic eye has too short a focal length lens. A negative power corrective lens is required to focus the image on the retina. 46

7 Figure 20 Magnification The compound microscope. The lateral magnification by the objective is = (Objective magnification) The normal eye can focus objects at object distance from 25 to The image on the retina of the eye is largest when the object is closest, that is at the near point. A magnifying lens produces an image at the near point of the eye, of an object that is very close to the eye. Since the object of a magnifying glass is close to the eye it subtends a large angle leading to a large image on the retina of the eye. Since the focal lens of a microscope eyepiece is much smaller than 25, the object distance for the eyepiece almost equals the focal length of the eyepiece, producing a virtual noninverted image at 25. Thus the magnification of the eyepiece is (near point) = (Eyepiece magnification) The net magnification is the product of the objective and eyepiece magnifications. As an example consider a typical microscope with = 5, tube length = 20 eyepiece focal length = 2 and taking the near point at 25 cm. Then the objective has a magnification of = = 40 and the eyepiece magnification is = 25 2 =+25 Thus the microscope has a net transverse magnification of = 500. Resolution The discussion of diffraction pointed out that for a circular aperture the angle of the first minimum of a diffraction pattern occurs at sin min =22 where is the diameter of the circular object being imaged. Note that is not the diameter of the objective, it is the diameter of the circular object being magnified. The information about the fine structure of the Figure 2 Light collection of diffraction pattern by objective lens. object is carried by the diffraction pattern. To obtain a perfect image one must collect all the diffracted light. If the lens has a maximum acceptance angle then even the first minimum will not be accepted by 22 sin the lens for aperture widths less than = That is, there is a minimum spacing for which the lens will accept enough of the diffraction pattern to carry the information as to the spatial dimension. That is, the smallest spatial resolution min is given by: min = 22 sin (Spatial resolution of dry objective lens.) Assuming that the objective collects light to 65 to the normal, that is a maximum value of the numerical aperture sin =09 then diffraction sets a minimum resolved structure, for = 500 to be = 680 Ifthevolumebetweentheobjectandobjectivelens is filled with an oil of refractive index n, then the wavelength in the oil is = Since the refractive index exceeds unity, then the wavelength in oil is less than is vacuum. The above discussion about resolution is modified in that the wavelength of interest is the one in oil. Thus we get for an oil immersion objective, that 22 min = sin (Spatial resolution for an oil immersion objective lens.) Good oil immersion objective have a numerical aperture = sin =4. Such a lens will give a spatial resolution of 480nm. It is not possible to attain better spatial resolution using = 500 light. Shorter wavelength light is required. As will be discussed later, much shorter wavelengths, such as X rays or electron matter waves, are needed to attain spatial resolutions of 0 0 which istheorderofthesizeoftheatom. Modernintegrated circuit manufacture uses UV light to be able to make the 0.80 micron size traces on silican wafers. 47

8 Figure 22 The phase-contrast microscope restricts the angular range of the incident light through the condenser lens. After the light passes through the objective of the microscope the phase of the unscattered light, shown red, is retarded by 90 by a phase plate. This results in thesmallphasedifference in the scattered light shown in yellow, which is not easily visible by eye, being converted to differences in amplitude of the combined light which greatly enhances the visibility of thin biological samples. Figure 23 Comparison of the observed light scattered by a thin biological sample using a conventional microscope, shownontheleft,withthatfromaphase-contrast microscope, shown on the right. SUMMARY The performance of optical instruments is a compromise between several contradictory requirements due to geometrical optics and wave properties. Reading assignment: Giancoli, Chapter 34. Phase-contrast microscopy. One problem with biological samples is that thin slices tend to be transparent and thus little contrast is seen between different parts of the sample. Phase contrast microscopy uses a trick to change the phase between the primary and secondary diffraction maxima resulting in phase differences shown as differing image brightness. As shown in figure 22, an annular slit restricts the angular range of the light focussed by the condenser lens onto the object, shown in green. Most of the light is unscattered by the specimen, shown in red. The diffracted light, shown in yellow, contains the required spatial information regarding the specimen. Unfortunately the weak diffracted light is shifted in phase by 90 andthusasshownthenetvector,shown in blue, is essentially the same as the unscattered direct light, shown in red. That is, the length of the red and blue vectors are essentially the same. The eye is only sensitive to the intensity of the light, not the phase. The trick is to retard the phase of the central (red) unscattered light by 90 by use of a phase shift ring behind the objective. The sum of the in-phase red and yellow vectors resulting in a change in the intensity shown in blue. A further trick is to attenuate the strong unscattered light by use of a grey filter ring which enhances the ratio of the yellow and attenuated red intensities. Thus, variations in refractive index across a sample, which corresponds to phase differences, are displayed as differences in contrast. Figure 23 illustrates that the phase differences, shown on the left, become much more clearly defined by the phaseshift microscope as seen by the eye which is sensitive to differences in contrast, not phase differences. 48