F1 The ray diagram for a simple astronomical refracting telescope is shown in Figure 23. The focal points of the f o f e objective and the eyepiece F e, F are normally made to o θ o θ coincide so that an inverted, h I magnified image is viewed at infinity. This means that the observer s eye is fully relaxed and subject to less strain. The angular limit of objective lens virtual image resolution for a telescope at infinity eyepiece lens objective of diameter d is 1.22λ/d where λ is the Figure 233A simple astronomical telescope. The objective has a long focal length whilst wavelength of the that of the eyepiece is much shorter. radiation1 1the larger the objective diameter, the greater the ability to resolve fine detail on the object. You will find a discussion of
angular resolving power in Subsection 2.3.
F2 An achromatic doublet is formed by cementing together two lenses with equal and opposite chromatic aberrations (see Figure 10). One lens is converging and made from crown glass, with a low average index and a low dispersive power; the other is diverging and made from flint glass, with a high average index and a high dispersive power. The shapes of the lenses are chosen so that minimum spherical aberration is produced for zero coma. The cement used is Canada balsam, which is a resin with refractive index approximately equal to the average value of the two glasses. low dispersive power converging lens crown glass high dispersive power diverging lens flint glass Canada balsam Figure 103An achromatic doublet.
F3 The transverse magnification M tran is the ratio of transverse image size h to transverse object size h, i.e. M tran = h /h. The angular magnification M ang is the ratio of the angle θ0 subtended at the observer s eye by the image to the angle θ subtended there by the object, i.e.1m ang = θ0 /θ. The magnifying power M power is the ratio of the angle θ0 subtended by the image to the angle θ0 D subtended by the object when at the eye s near point (or least distance of distinct vision), i.e. M power = θ0 /θ D. (a) For a microscope we compare the angular sizes of the image seen through the microscope, with the best (i.e. largest) view we can obtain of the object when it is seen directly, which will be when it is at the near point. The magnifying power is therefore the term used to describe the performance of a microscope. (b) A film projector projects a greatly enlarged image of the film so as to fill the screen. The transverse dimensions are therefore important, so transverse magnification is the quantity used. (c) The object being viewed in a telescope is usually at a large distance which cannot be reduced. Normally, we adjust the eyepiece for the relaxed eye so that the image is at infinity and transverse dimension is a meaningless term; angular magnification therefore describes the telescope s performance.
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
R1 Three rays are considered: (i) the axial ray which travels along the optical axis from the bottom of the object and passes undeflected through the centre of the lens; (ii) a ray parallel to the optical axis from the top of the object which passes through the lens and is refracted so as to cross the optical axis at the focal point F (distance f from the lens); (iii) a second ray from the top of the object which passes undeviated through the centre of the thin lens. The top of the image is at the point where rays (ii) and (iii) meet. Using this method an inverted and enlarged real image is formed beyond 2f. (In fact the image is inverted, 41cm in height and 241cm from the lens.)
R2 Using the Cartesian sign convention: (1/v) (1/u) = 1/f which, for u = 41cm and f = +61cm, becomes 4(1/v) 1/( 41cm) = 1/(61cm) i.e. 1/v = (1/6 1/4)1cm 1 = 1/(121cm)4 so4v = 121cm So, the image is positioned 121cm from the lens. Such an image will be virtual and erect.
R3 The image is constructed using three rays as before but this time the paraxial ray from the top of the object diverges and so appears to come from the first focal point on the same side of the lens as the object1 1a dotted construction line is drawn back to this point where it meets the axial ray. The third ray is drawn from the top of the object passing undeviated through the centre of this thin lens. The eye sees a virtual image produced at the intersection of this third ray and the construction line since that is the point from which these rays appear to come. A careful diagram gives an image 1.41cm in height, near to the calculated distance from the lens of 5.51cm.
R4 In the dispersion of white light by a glass prism the violet end of the spectrum is refracted more than the red. The reason for the deviation is that the light is slowed down on entering the glass, since glass has a refractive index which is greater than that of air (refractive index = speed of light in vacuum/speed of light in the medium concerned). Consequently, a glass of large refractive index causes greater refraction (or deviation) than one of low refractive index. Since violet light is deviated more than red light, the refractive index of the glass must be larger for the shorter wavelengths of violet light than for the longer wavelengths of red light. Common mistakes here are to reverse the order of the spectrum or to show dispersion only taking place at the second face. Dispersion begins at the first air/glass interface and the different coloured rays are already separated when they reach the second face.
R5 In constructing the ray diagram to show image formation by a concave mirror, three rays are used. The first of these is the axial ray, (i.e. from the bottom of the object along the optical axis of the mirror), which is simply reflected back on itself, showing that the bottom of the image is on the axis. A second ray, from the top of the object, passes through the focal point, strikes the mirror and reflects back parallel to the optical axis. The third ray, from the top of the object, strikes the point where the optical axis meets the mirror, where it is reflected back at the same angle beneath the axis. The top of the image is situated where the second and third rays intersect. The remainder of the image will be perpendicular to the optical axis. In this problem, the initial object distance (201cm) exceeds 2f (151cm) and the above construction then shows that the image is real, diminished, inverted and situated between f and 2f. If the object is moved along the axis towards the mirror then the inverted image becomes the same size as the object, when it is at 2f, and is located directly under it. In its final position, (101cm) the inverted image is enlarged. (If the object were to reach f, 7.51cm from the mirror, then the inverted image would be located at infinity.) If you feel unsure of any of the terms referred to in Questions R1 to R5, consult the Glossary.
T1 The diameter of a five pence piece is about 11cm and the angular size is the actual diameter divided by the distance, so we have: Earth Moon distance 1m = Moon diameter 10 2 m The distance to the Moon is then: 3.5 10 6 10 2 1m= 3.5 10 8 1m.
T2 The larger the refractive index of glass, the larger the angle through which light of a given wavelength is deviated or refracted when crossing an air glass boundary (at non-normal incidence). As violet light is refracted through a larger angle than red light, the refractive index of the glass must be larger for the shorter wavelengths of violet light than for the longer wavelengths of red light. Since refractive index is the ratio of the speed of light in vacuo to the speed in the medium concerned, a large index means a low speed and therefore the violet light travels more slowly in the glass than the red light.
T3 (a) For the eye: θ = (1.22 5 10 7 1m)/(5 10 3 1m)2= 1.22 10 4 1rad (b) For the binoculars (taking a typical front lens diameter of 51cm) θ = (1.22 5 10 7 1m)/(5 10 2 1m) = 1.22 10 5 1rad (c) For the optical telescope θ = (1.22 5 10 7 1m)/(11m) = 6 10 7 1rad (d) For the radiotelescope θ = (1.22 0.221m)/(76.21m) = 3.5 10 3 1rad For best results we want θ small.
T4 The angle from the axis to the first minimum is given by Equation 5 1. 22λ θ = (Eqn 5) d as 1. 22λ θ = = (1.22 101cm/ 3001cm)1rad = 0.04071rad d The angular spread of the central maximum is therefore twice this angle, i.e. 0.08141rad. The diameter of this central beam when it reaches the Earth will then be 0.0814 2510001km = 20351km.
T5 The resolution limit of the normal eye occurs when the two sources are separated by about 1 of arc i.e. 0.071mm at 251cm. (a) If we take the individual letters to be about 11mm in size, separated by about 11mm then, by proportion, the furthest distance of distinguishability is about 3.51m. (b) The Earth Moon distance from Answer T1 is 3.5 10 8 1m and so the minimum distinguishable size is: 3.5 10 8 1m 0.071mm/(0.251m) = 981km.
T6 The ray diagram of Figure 15b is similar to that for this system, except that the final image is formed just beyond the near point. We substitute v = 301cm and f = +101cm in the thin lens equation (1/v) (1/u) = 1/f and obtain 1/u = 1/(301cm) 1/(101cm) = 4/(301cm), so that u = 7.51cm. The transverse magnification M tran = h /h = v/u = 301cm/( 7.51cm) = 4. (b) h' D h θ I u Figure 15b3Ray diagrams for a magnifying glass with the object brought closer (i.e. between F and the lens) and the image formed at the near point. It is assumed that the eye is placed very close to the lens so that D can be measured from the lens. θ I
T7 In order to maintain the same exposure, when the shutter time is halved, the aperture must be doubled in area which is achieved by changing to the next smallest f-number. So, if f/11 is the proper stop for (1/125)1s, then we require f/8 for (1/250)1s, f/5.6 for (1/500)1s and f/4 for (1/1000)1s. Since the f-number is the focal length f divided by the aperture diameter d, for f/11, d = 501mm/11 = 4.551mm; for f/4, d = 501mm/4 = 12.51mm.
T8 For the final image to be formed at infinity, the intermediate image must be formed at the (first) focal point of the eyepiece, i.e. 251mm from the eyepiece. This means that the intermediate image is (180 25)1mm i.e. 1551mm from the objective. From the thin lens equation we have 1/u = 1/v 1/f, so that applying this to the objective and substituting f = 21mm and v1=11551mm gives 1/u = 1/(1551mm) 1/(21mm) = (2 155)/(3101mm) 3333221 = 153/(3101mm) and so u = 2.031mm i.e. the object must be placed 2.031mm in front of the objective.
T9 For a telescope, for which both the object and final image are effectively at infinity, the length of the tube is f O + f e = 1001cm. The1angular magnification = f O /f e = 40, so f O = 40f e. Substituting for f O 40f e + f e = 1001cm and f e = 1001cm/41 = 2.441cm Hence f O = (100 2.44)1cm = 97.561cm. To calculate the exit pupil size we know that the angular magnification is also given by d O /d e = 40, so 121cm/d e = 40 and d e = 31mm.
T10 Looking at Figure 28 you will see how the introduction of a third converging lens between F O and F e produces an erect image at infinity which makes the telescope suitable for terrestrial use. erect virtual image at infinity F o F i F e objective lens intermediate (erecting) lens eyepiece lens