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Optics We have developed a formalism which we can now apply to electromagnetic waves – light Electromagnetic waves are oscillations of the electric (E) and magnetic (B) fields
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Wave equation for electromagetic waves are two constants which describe how well waves propagate through electric and magnetic media the o subscript tells about the propagation in free space – the vacuum (c=299792458 ms -1 ). For materials with values of relative permittivity ( r ) and relative permeability ( r ) the velocity of the light is where n is the refractive index
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There are two guiding principles that we shall employ extensively: 1. Huygen’s principle Each point on a wavefront serves as the source of spherical secondary wavelets that advance with a speed and frequency equal to those of the primary wave.
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Fermat’s Principle (Pierre de Fermat) The actual path between two points taken by a beam of light is the one which is traversed in the least time (dt/dl=0). Stricter definition the optical path length must be extremal, which means that it can be either minimal, maximal or a point of inflection (a saddle point). Minima occur most often, for instance the angle of refraction a wave takes when passing into a different medium or the path light has when reflected off of a planar mirror. Maxima occur in gravitational lensing. A point of inflection describes the path light takes when it is reflected off of an elliptical mirrored surface.optical path lengthpoint of inflectionsaddle pointangle of refractiongravitational lensingpoint of inflectionelliptical
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Relationship between Huygen’s and Fermat’s Principles speed of light in a medium is less than vacuum. Speed is characterised by index of refraction (n) n=c/v For water n=1.333 air n=1.0003 http://ephysics.physics.ucla.edu/ntnujava/propagation/ereflection_and_refraction.htm
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When light strikes the boundary surface, there is a transmitted and reflected component (just as with waves on a string). reflected refracted n1n1 n2n2 n 1 <n 2 ii rr i’i’ i = i ’ (angle of incidence = angle of reflection) n 1 sin( i )= n 2 sin( r ) (Snell’s law)
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From Fermat to Snellius n1n1 n2n2 n 1 <n 2 11 22 d1d1 d2d2 l1l1 l2l2 L x
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Total internal reflection n1n1 n2n2 n 1 >n 2 11 22 n2n2 n1n1
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Dispersion the refractive index is slightly different for different wavelengths n1n1 n2n2 n 1 >n 2 11 22
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René Descartes, French philosopher
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Mirrors! image is virtual s s’ y y’ P P’ ray diagram
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Angled Mirrors Get multiple images
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Spherical Mirrors: image is real – light rays pass through it and image could be projected on a screen – not true for virtual images. Spherical aberrations -non axial rays (paraxial) come to a different focus, and thus image is blurred. The non-paraxial rays are usually removed using an aperture.
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C P P’ A s r s’ l
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when the object distance, s, is large compared with the radius of curvature, r
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Concave mirror Convex mirror - parallel rays strike the mirror and are focused at F at a distance r/2 -Incoming plane waves become spherical waves converging at F -the outgoing wavefronts appear to emanate from F behind the mirror. Rays are perpendicular to the wavefronts and appear to diverge from F
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Ray diagrams Draw 3 rays 1.parallel 2.focal 3.radial Note. if s < F image is behind mirror and virtual and need a different construction Convention for s is +ve/-ve if object is in front/behind mirror s’ is +ve/-ve if image is in front/behind mirror F/r +ve/-ve if mirror is concave/convex
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Ray diagram for convex mirror Magnification: for similar triangles: negative magnification hence image is inverted positive magnification: image is said to be erect
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Better do an example or two! Concave mirror, 40 cm radius of curvature, object 1cm high, placed 100 cm from the mirror – where is the image and what is the magnification r C 1 cm 100 cm F 20 cm Thus image is 0.25 cm high and inverted
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Concave mirror, 40 cm radius of curvature, object 1cm high, placed 10 cm from the mirror – where is the image and what is the magnification r C 1 cm 10 cm F 20 cm Thus image is 2 cm high and erect and virtual
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An object is 2 cm high and 10 cm from a convex mirror with a radius of curvature 10 cm. (a) locate the image (b) find the image height 10 2 c f
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Lenses! Believed light rays enter the eye – theory of perspective
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Apparent depth
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Images formed by refraction at a single surface Consider a spherical surface separating two media of different refractive indices P P’ n2n2 n1n1 s s’ Snell’s law 22 11 C l r A
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A
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Magnification s s’ 11 22 y y’
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Sign conventions incident ray n 1 refracted ray n 2 s+ (real object) s’ - (virtual image) r,F - if radius of curvature on incident side s- (virtual object) s’ + (real image) r,F + if radius of curvature on refracted side
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2121 Object is at the focus of 2 Image is at the focus of 1 Thin lenses Image is formed by refraction at each surface separately. Consider a lens of refractive index n l with the refractive index of the medium n m.
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v1v1 v2v2 A B C2C2 C1C1 P P1’P1’ C 1 and C 2 are the centres of curvature of the surfaces Av 1 B and Av 2 B respectively. Applying usual equation for a surface….to the first surface ss1’ss1’ In this case the image distance s 1 ’ is negative (virtual image to the left) So rays at second surface behave as if they came from P 1 ’, i.e. object at P 1 ’ (image of first surface becomes object of the second). If the lens has thickness d: s 2 =s 1 ’+d
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At second surface, medium on incident wave side has refractive index n l and refracted side of n m. s 2 is to the left and is hence positive, but s 1 ’ is -ve (s 2 =-s 1 ’-d) Add A+B For a thin lens d 0 (i.e. last two terms vanish)
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When s s’ f (the focal length) This is called the lens makers equation! The thin lens equation
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Sign convention Incident light Refracted light Object (s): +ve Image (s’): -ve r: -ve if C is on incident side Object (s): -ve Image (s’): +ve r: +ve if C is on refracted side Incident: r 2 <0 Refracted: r 1 >0 n=n l /n m
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Converging lens (positive lens) Diverging lens (negative lens)
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Maurits Cornelis Escher (Dutch, 1898-1972):Convex and Concave
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Diverging lens Incident: r 1 <0 Refracted: r 2 >0 First focal point Second focal point P=1/f is called the power of the lens measured in units of dioptres (1/m)
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Ray tracing F F’ Parallel ray Central Ray Focal Ray y y’ Note: central ray is undeflected as faces of lens are parallel – just like looking though a window (get slight displacement) F F’ Parallel ray Central ray Focal ray y y’
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Example Plane – convex lens of refractive index of 1.5 and convex radius of curvature of 15 cm. What is the focal length r 2 <0 r 1 >>0
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Example An object 1.2 cm high is placed 4 cm from a double convex lens (radii of curvature 10 and 15 cm refractive index 1.5). What is the focal length. Locate the image, and perform the ray tracing. Is the image real or virtual, and what is its height? r 2 =-15 r 1 =10 C1C1 10 cm 15 cm C2C2
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F F’ Focal Parallel Central So image is to left of lens and thus virtual Image is magnified
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Fresnel Lenses
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Systems with more than one lens F F’ Focal Parallel Central Imagine a second lens (f=+6 cm) is placed 12 cm to the right of the previous lens F2F2 F2’F2’ Changed image from virtual to real
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When two lenses are added together (no spacing) the focal length is The eye Far sighted Near sighted
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Apparent size of an object The near point is the closest distance for which the eye can form a sharp image (usually about 25 cm) – this changes with age!
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Optical correction A persons near point is 75 cm – what kind of lens is required to bring it to 25 cm? Hubble – celestial eye Eskimo nebula
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The Telescope 8 inch refractor at Chabot Space and Science Centre in Oakland, California. Job of telescope is to make objects which are far away appear close - and magnify them! Magnifying power
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Telescopes need to collect light -large lenses suffer from gravitational sagging Solution: reflector (Newtonian telescope) – using lighter mirrors Telescopes classified by F-number (ratio of focal length of objective/diameter)
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