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Topic 5 - Image Formation Image formation • Based on 2 simple laws – Law of reflection – Snell’s Law • The two laws can be applied by simply tracing the geometrical paths of the light rays • Two types of images • Real images – the rays from the object after passing through the optical element really do pass through the image point. E.g. images formed on photographic film, on a projection screen and in the eye • Virtual images – the rays from the object after passing through the optical element do not pass through the image point, they only appear to come from the image, cannot be projected onto a screen, only view through the optical element Reflection - Plane Mirror • Nomenclature P = object P’ = image s = object distance from the mirror s’ = image distance from the mirror Where is the image located? the image point P’ is located exactly opposite the object point P, as far behind the mirror as the object point is in front of the Textbook – University Physics mirror Young and Freedman Figure 35-4 Page 1087 Extended Object • Object of finite size y = object height y’ = image height Lateral magnification m y′ = y Upright image m>0 Inverted image m<0 Plane mirror, m = 1 But the image is reversed Textbook – University Physics Young and Freedman Figure 35-5 and 35-6 Page 1087 and 1088 Finding Images • Lab experiment – how to find images in practice • Method of No Parallax Parallax occurs when you move your head from side to side, things near move faster than those further away but if they coincide they appear to move together – then there is no parallax Reflection at a Concave Mirror • Application: astronomical telescopes use large concave mirrors to photograph celestial objects • R = radius of curvature • C = centre of curvature • V = vertex of the mirror (i.e. centre of the mirror surface) • CV = the optic axis http://surendranath.tripod.com Textbook – University Physics Young and Freedman Figure 35-9 Page 1089 From geometry and using the paraxial approximation can obtain 1 1 2 + = s s' R Young and Freedman Section 35-3 Spherical Aberration • Hubble Telescope, 1990 • Image “smeared out” Textbook – University Physics Young and Freedman Figure 35-10 Page 1090 Concave Mirror 2 • When the object point P is very far from the mirror and the incoming rays are parallel, then the image is formed at the focal point • The distance from the vertex to the focal point is the focal length, f 1 1 2 + = s s' R Case (a) s = ∞, s ' = f 1 =0 s 1 2 = f R R f = 2 1 1 1 + = s s' f Sign Rules 1 1 1 + = s s' f 1) When the object is on the same side of the reflecting or refracting surface as the incoming light the object distance s is positive, otherwise it is negative 2) When the image is on the same side of the reflecting or refracting surface as the outgoing light the image distance s’ is positive (real), otherwise it is negative (virtual) 3) When the centre of curvature C is on the same side as the outgoing light the radius of curvature is positive otherwise it is negative so for the concave mirror R>0, f>0 Concave Mirror 3 Orientation of the extended image? From the geometry, similar triangles give the relationship m= y' s' =− y s Young and Freedman Page 1092 Minus because the object and image are on opposite sides of the optic axis • Can also look at plane mirror with the equation 1 1 2 + = s s' R 1 R = ∞, = 0 R s = − s' As expected from the sign rules And this also gives m=1 Question An object is placed 2.0. cm in front of a concave sherical mirror whose radius of curvature is 8.0 cm. Locate the position of the image and its size. Graphical Approach – Ray Diagrams • Principal Rays 1) A ray drawn parallel to the optic axis, after reflection passes through the focal point F of a concave mirror 2) A ray through the focal point F is reflected parallel to the optic axis 3) A ray through the centre of curvature C intersects the surface normally and is reflected back along the original path 4) A ray to the vertex V is reflected forming equal angles with the optic axis If the principal rays do not converge at a real image point, extend them back to locate a virtual image point as for convex mirror case Concave Mirror – ray diagrams a) s > R The 4 rays converge after reflection at point Q’ Image is real, inverted and smaller than the object, s’ > 0, 0 > m > -1 b) s < f Rays appear to diverge from Q’, Virtual image, upright and larger than object s’ < 0, m > 1 As in earlier numerical example Concave Mirror • What will be the effect on the image if half of the mirror surface is covered with a non-reflecting coating? a) b) c) d) No effect Can’t see any image Can only see half the object at the image The image is of the whole object but less bright Convex Mirror - Summary • See text book • Focus is virtual f < 0, R < 0 • Image is always virtual, lies on the opposite side of the mirror to the outgoing light, s’ < 0 • Image is upright and reduced in size, 0 < m < 1 Example: car side door mirrors – “objects are closer than they appear” • The same formulae can be used with the same sign convention Question A flea is located 3.0 cm from a convex spherical mirror of radius 10 cm. Where is the image of the flea? Thin Lenses - Converging • Converging lens – when a beam of rays parallel to the optic axis pass through the lens they converge to a point F2 and form a real image at that point (a) Rays passing through F1 emerge from the lens as a beam of parallel rays (b) The focal length is a positive quantity and the lens is also called a positive lens Thin lenses have 2 foci each the same distance f from the centre of the lens even when the two sides have different curvatures Thin Lenses - Diverging • Diverging lens – a beam of parallel rays incident on this lens diverges after refraction, the rays appear to come from the virtual image at F2 (a) Incident rays converging toward F1 emerge from the lens parallel to the optic axis (b) The focal length is a negative quantity and the lens is also called a negative lens NB positions of F1 and F2 have swapped compared with the converging lens Thin Lens Equation • From the geometry can derive the same equations as for the 1 1 1 spherical mirrors and use the same + = s s' f sign conventions y' s' m= =− y s For a converging lens see: s > f: the image is real and inverted, s’ > 0, on rhs of lens s < f: image is virtual, located on the same side of the lens as the object, it is upright and larger than the object Ray Diagrams for Lenses • Three principal rays for lenses 1) A ray parallel to the optic axis emerges from the lens in a direction that passes through F2 for a converging lens, or appears to come from this point for a diverging lens 2) A ray through the centre of the lens is not deviated appreciably so this ray emerges from the lens following the same line 3) A ray through (or proceeding toward) the F1 emerges parallel to the axis When the image is real, the position of the image point is determined by the intersection of any 2 rays When the image is virtual, we extend the diverging outgoing rays backward to their intersection point to find the image point Many cases examined for the converging lens in text book Converging Lens - ray diagram example • NB the 3 principal rays 1) A ray || to the axis emerges from the lens in a direction that passes through F2 2) Ray through the centre is undeviated 3) Ray through F1 emerges || to the axis Real inverted smaller image Diverging Lens – ray diagram example • NB the 3 principal rays 1) A ray || to the axis emerges from the lens in a direction that passes through F2 2) Ray through the centre is undeviated 3) Ray through F1 emerges || to the axis Virtual upright smaller image Lens Combination 1 • Image from the 1st lens is the object for the 2nd lens The Simple Magnifier • Simple imaging tool – magnifying glass • Angle subtended by the object at the eye = angular size • To examine something small in detail - increase the angular size (fill retina with image of the object) • Eye cannot focus sharply on objects closer to the eye than the near point 20-30 years old: 10 – 14 cm Standard used is 25 cm θ = h h = near point 25 cm Magnifying Glass Textbook – University Physics Young and Freedman Figure 35-10 Page 1126 A converging lens can be used to create an image that is larger and further from the eye than the object Virtual image most comfortable to view at infinity The angular size of the image can be substantially larger than the angular size of the object at the near point without a lens A MAGNIFYING GLASS y To find M: Angular Magnification θ' M= θ θ = 25 cm y θ '= f θ' M = = θ y f y 25 cm = 25 cm f Microscope s1 ' m1 = − ; s1 = f1 s1 s1 ' m1 = − f1 25 cm M2 = f2 M = m1M 2 = • • • • • 25 s1 ' f1 f 2 Lens combination – objective and eyepiece Objective – converging lens - forms real enlarged image Eyepiece – converging lens – magnifier of the real image Overall magnification M = m1 M2 Objective: small f1, increases size of image I, maximize overall magnification • Eyepiece: small f2 • Manufacturers specify m1 and M2 rather than f1,2 or s1’ Telescope θ′ M= θ ab y ′ θ= = f1 f1 cd y ′ = θ′ = f2 f2 f1 M =− f2 http://www.walter-fendt.de/ph11e/refractor.htm Textbook – University Physics Young and Freedman Figure 36-15 Page 1129