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Imaging Forming an Image Vergence Effect of a Lens Graphic Examples Of Forming an image Stanley telescope Imaging . Each point of an object sends out light over a range of directions . This section covers . Seeing an object point requires intercepting a how images are formed Nikon waterproof binoculars pencil of light with our eye imperfect and perfect imaging the point is located by us at the apex of the pencil spherical image-forming surfaces pencils are characterised by their vergence locating images some imaging devices Nikkor 50 mm camera lens Olympus 2100 digital camera Vergence Effect of a lens object image Refractive index n1 n2 so si object image . A lens alters the vergence of a pencil by an amount equal to the power D of the lens so si n n n n 2 1 D or 2 1 D s s s s . Vergence measures the convergence (+ve) or i o i o divergence (-ve) of a pencil of rays or, equivalently, a . This is the fundamental imaging equation of a small section of travelling wavefront. At the aperture: lens (relating object and image distances) vergence of the wavefront forming the image = n /s 2 i remembering the relationship between rays and vergence of the wavefront from the object = -n /s 1 o wavefronts, this is essentially a wave view of imaging vergence is measured in dioptres ( m-1) Graphic examples of imaging Summary . Imaging requires the divergence or convergence of a . Imaging by a object bundle of neighbouring rays from each object point diverging lens eye . The apex of the pencil is the position of the image the vergence of the image incident pencils is . The image of an extended object is the composite image made more –ve by lens of all its points the lens . Each imaging surface takes for its object the image from . Imaging by a plane mirror eye the previous surface there is no alteration of object . Simple image forming surfaces alter pencils of light in vergence, only a change in direction of the axis of the two ways: pencil they alter the vergence of the pencil image they may bend the axis of the pencil 1 Imperfect imaging Perfect imaging - 1 reflector source . A perfect image forms when every part of the image-forming surface produces an image at the same place extended image of centre point . Mathematically, a perfect image creates a one- one mapping from Imaging . Different parts of an imaging forming surface object space to surface may form images in different places image space Perfect image the result is a blurred image . Image point and object Object the blur reduces when a smaller amount of image point are conjugate point forming surface used Perfect imaging - 2 The paraxial approximations . A perfect image mapping must be linear . Angles of incidence, i , and refraction, t , are small n1 n2 i t + + R V Object with . C Object with equi-spaced Image with non equi-spaced lines equi-spaced Image with object distance s image distance si lines equi-spaced lines o lines Non-linear mapping Linear mapping . Off-axis distances are small compared with . Perfect imaging only happens with a plane the curvature of the refracting surface and mirror with object and image distances (R, so, si) Spherical surfaces Sign convention n n1 2 . For an algebraic equation to give the right answer R C when images (and objects) can be on either side of a V . lens you need a sign convention to determine which image distance s object distance so i quantities are +ve and which are –ve real objects are those in front . Under the paraxial approximations of the image forming surface, in Direction of incident light spherical surfaces form ideal images front being taken with respect to the direction of incident light +ve objects -ve objects . All rays from a single object point pass virtual objects are those after O through a single image point after refraction the image forming surface -ve images +ve images n n n n the reverse is true for images 2 1 2 1 D , the power of the surface s s R i o . The sign convention is known as real is positive 2 Worked example Focal points and lengths . An object is placed on the axis of a refracting surface of radius of curvature 200 mm; the object is 600 mm from the surface in air . An imaging surface has 2 focal points (refractive index 1.00) and the surface has refractive index 1.64. Nikon Fi in image space and Fo in object space underwater lens Where is the image? Imaging surface Imaging surface F n1 = 1.00 n = 1.64 Fi o 2 V V V fi fo object distance so = 0.6 image distance si . Fi is the image point of an axial object at . Given: n1 = 1.000; n2 = 1.64; R = 200 mm; so = 600 mm 0.6 m . Fo is the axial object point whose image is at . Surface power D = (n2 –n1)/R = (1.64 – 1.00)/0.2 = 3.2 dioptres n n 1.64 1.00 2 1 . f = VF , the image focal length . Find si from D . Therefore 3 . 2 si = 1.07 m i i si so si 0.6 . fo = VFo, the object focal length Where are the focal points? Focal point example surface . The imaging equation tells us where the focal points are F o n = 1.00 n = 1.50 Fi n n n n 1 2 for Fi, so = ; 2 1 2 1 D , the power of the surface s s R i o R = 0.2 m n n n n n n n fo fi 2 1 2 1 D hence 2 2 1 D f i R f i R . An imaging surface between media of refractive n1 n2 n1 likewise D surface indices 1.00 and 1.50 has radius of curvature 200 f o R F o n1 n2 Fi mm. What is the power of the surface? What are . Fo and Fi are on either its focal lengths? f side of the surface o fi D = (1.50 – 1.00)/0.2 = 2.5 dioptres . For a given power D, f depends on the refractive fi = 1.50/D = 0.6 m; fo = 1.00/D = 0.4 m index of the medium The usefulness of focal points Imaging equation for a thin lens . If you know the focal lengths for a surface, and one other fact about it, e.g. its radius of curvature, . Application in succession of the you can calculate all its imaging single surface equation to each . Focal points are 2 out of 6 cardinal points that surface of a thin lens gives characterise an imaging component 1 1 1 1 1 . A lens is simply two imaging surfaces, one after n 1 D so si R1 R2 f the other lens Fo Fi n R1 R2 . This is known as the ‘thin a lens is specified by its rear (image) focal length lensmaker’s equation’ Olympus tele- focal points are usually equi-spaced from the centre converters 3 Lensmaker’s equation example Example of an enlarger screen lens . An enlarger lens has a focal length of slide 50 mm and is positioned 60 mm beneath Fi the negative being enlarged. How far 4 m 0.25 m below the lens must the printing paper be placed so that the image is in focus? . An image of a slide is to be formed 4 m in front . f = 50 mm; s = 60 mm; s ? of a lens of focal length 250 mm. How far from o i the lens must the object be placed? 1 1 1 1 1 1 , therefore , giving si 300 mm so si f 60 si 50 1 1 1 1 1 1 4 , therefore , giving so m 0.267 m so si f so 4 0.25 15 . Notice that so long as the same units are used for all the known distances, the result is in these units Transverse magnification Examples . The fact that a light ray passes through the centre of a Referring back to the previous questions thin lens undeviated, makes the magnification simply . What magnification did the slide the ratio of image distance to object distance projector give the projected picture? remembering the sign convention, that gives image and object heights different signs in the diagram si = 4 m; so = 4/15 m, hence, Lens Magnification = -si/so = -15 Object Image . What magnification did the enlarger produce of the negative? Object distance so Image dist si si = 300 mm; so = 60 mm, hence Image height s Magnification = -si/so = -300/60 = -5 Magnification i Object height so the –ve signs show that the images are upside down Recap of sign convention . Measurements are in object space and image space Olympus E-10 object . Real object +ve so . Real image +ve si image image . Real object +ve so object . Virtual image -ve si object . Virtual object -ve so image . Real image +ve si Courtesy: imaging-resource.com 4 The basic telescope - Kepler Telescopes objective eyelens Bushnell object at Primary eye image, a Leica height h u . Telescopes are the prime instruments of astronomy Observer eye sees final fo fe . In the form of binoculars, they are used widely by relief image at naturalists, sailors and many other groups . Telescopes are found widely, embedded in other . The objective forms an image of instruments such as theodolites and spectrometers an object at infinity . The eyelens forms a virtual image at infinity of the primary image TV102 Griffith Observatory refractor Telescope magnification Erecting telescope objective Observer eyelens objective eyelens sees final object at Primary eye image at image, a erector object at Primary eye height h image u u a Observer sees final f f eye o e image at relief f eye o 4fer fe relief angle subtended by image at eye( ) .
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