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 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 ( 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 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 . 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 ? . 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 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( ) . angular magnifying power (MP)  a angle subtended by object at eye(u ) . In principle, image erection can be provided by a

h single lens   tan   , in the paraxial approximation a a f e the eye relief has got larger h   tan    u u the erector limits the field of view f o f  MP   o , the  ve sign indicating an inverted image a larger eyepiece diameter is needed fe Stanley “spyglasses”

handlens object Rheita’s erector; prism erector eye a Simple yo

Primary eyepiece f image from Erect objective image . The closest distance that we can see objects clearly is fer 2fer fer called the nearest distance of distinct vision, do (~0.25 m) . The simple handlens forms a virtual image at infinity, . In practice, erection with is achieved using 2 which a relaxed eye can focus on  . The arrangement is really a telescope of . Magnifying power: MP  a(aided) u(unaided) magnification -1 within a telescope yo yo . Since   ; and  u  , a f d . Even more common is the use of a dual o prism erector d Simmons MP  o  0.25D spotter scope f

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Example of handlens magnification Compound . What is the power of a "10" eyepiece? eyepiece . Use: d The diagram above, from the first year lab manual, MP  o  0.25D . f introduces the field lens bends round pencils of image-forming light so that they therefore: 10 = 0.25 D pass through a smaller following lens hence D = 40 dioptres improves image quality allows control of the eye relief clearly defines the field of view (“acts as a field stop”) . Real eyepieces contain 2 or more lenses Courtesy: wsu.edu

Telescope summary

. Telescopic systems are designed to have the object at  and, for an observer, the final image at  . The minimum telescope will have an achromatic objective and a multi-element eyepiece . A terrestrial telescope will have a 2-lens or a prismatic erector . The magnification is given by the ratio of focal lengths of objective to eyepiece . Reflecting telescopes replace the objective lens by a mirror, which leads to a complete redesign of the system

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