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Image formation Lenses and Mirrors. +p. +p. +q. +q. +f. +R. +R. +f Convex -f Concave. Refraction at curved surface. Object:p Image: q Focal length : f Magnification :M Real : light at image Virtual : no light. Mirrors. Lenses. Ray1- || to axis, then thru f Ray2-thru f then ||
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Image formation Lenses and Mirrors
+p +p +q +q +f +R +R +f Convex -f Concave Refraction at curved surface Object:p Image: q Focal length : f Magnification :M Real : light at image Virtual : no light Mirrors Lenses Ray1- || to axis, then thru f Ray2-thru f then || Ray3-thru R and back (Ray-4- to center of mirror then reflect at same angle) Ray1- || to axis, then thru f on back Ray2-thru f on front then || Ray3-thru center of lens
Simple optical components The rays don’t really meet Imagine a spherical mirror A horizontal ray gets reflected according to… But if they area close to the axis they almost meet Horizontal rays close to the axis focus
f • So, for • Optical components that Symmetric about an axis • With rays close to the axis • And not very thick when transparent Horizontal rays are focused to a point Axis of symmetry or principal axis Image Focal point Rays form a point close to the axis are also focused
A pinhole camera oatmeal box small hole
Mirrors image q p object
Flat Mirrors p q Light rays appear to diverge from point behind mirror Principle axes Law of Reflection: p =|q| independent of viewing angle q q p p q h h’ Magnification: M = h’/h = 1 Image is Upright Image is Virtual; behind mirror, can’t project on screen, light rays don’t actually pass through image point
In the overhead view of Figure 26.4, the image of the stone seen by observer 1 is at C. At which of the five points A, B, C, D, or E does observer 2 see the image?
Reflection by spherical mirrors R-q q u q ’ q1 q2 q2 p-R p q Just geometry… R x x’
p q Look at the mirror formula: R q=R/2 when p is very large All rays converge at R/2, the focal point
h p q For concave mirror the image is in front of the mirror h’ The image is inverted The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror
h p q For convex mirror the image appears to be behind the mirror, it is virtual The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror h’ The image is not inverted The mirror formula still holds, but with q<0 and R<0 The image is virtual The mirror is convex
You wish to reflect sunlight from a mirror onto some paper under a pile of wood to start a fire. Which would be the best choice for the type of mirror? (a) flat (b) concave (c) convex
h -h’ f f’ p q • Ray tracing for lenses (and mirrors) • ray parallel to the axis refracts (reflects) through the focal point • ray through the focal point emerges parallel to the axis • ray through the center of the lens (mirror) • does not bend (lenses) • goes out at an equal angle (mirror) Consequences of Snell’s law and the laws of reflection PLUS geometry
26.2 • In a church choir loft, two parallel walls are 5.3 m apart. The singers stand against the north wall. The organist faces the south wall sitting 0.8 m away from it. To enable her to see the choir, a flat mirror 0.6 m wide is mounted on the south wall, straight in front of her. What is the width of the north wall she can see.
Exercise 26.2 Sounds like ray tracing Will need the formula for reflected rays form mirrors 5.3m l organist 0.6m L choir 0.8m
Exercise 26.11 Sounds like an application of the formula A spherical convex mirror has radius of curvature with magnitude 40 cm. Find q and M when p=30 cm and p=60 cm - side + side corrected apr 25 http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html
A horizontal ray is refracted towards a focal point q h -h’ q’ f f’ p q The optical component may not be back-front symmetric so in general f≠f’ So, to make an image just do some ray-tracing A ray trough a focal point is refracted to a horizontal ray f=f’
Specific cases: Spherical concave mirror of radius R Spherical convex mirror of radius R Spherical boundary of radius R between a medium of refraction index n1 (left) and refraction index n2 (right) Thin lens of curvature radii Rleft, Rright and of material with refraction index n
Definitions and sign conventions +p +p +q +q +f +R +R +f Convex -f Concave p q h f h’ f mirror • An object is: • REAL if light diverges from it • VIRTUAL if light converges toward it • An image is: • VIRTUAL if light diverges from it • REAL if light converges toward it lens
+p +q +R +f Convex -f Concave An object is: REAL if light diverges from it VIRTUAL if light converges toward it An image is: VIRTUAL if light diverges from it REAL if light converges toward it For example:
+p +q +f +R Sign conventions for mirrors:
Exercise +p +q +R +f Convex -f Concave • Sounds like another application of the formula • The tricky part are the signs: • M=+2 since the image is not inverted • The object and the image are on same sides of the lens so p and q have the different signs Magnifying looks at a map. The image is virtual, of twice the size and |q|=2.84 cm. Find f
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the transparent cornea. Assume that this outer surface has a 6mm surface of curvature and assume that the eyeball contains just one fluid with index of refraction 1.4. Where is the image? Does it fall on the retina? Describe it. real, inverted
+p +q +R 15 cm two lens system first lens 40 q=30 M1=-30/15=-2 real inverted second lens p=40-30=10cm q=-20 cm M2=-(-20)/(10)=4 virtual, upright Total M=M1M2=-8 http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html inverted Fig 36-32a, p.1151
Microscope Fig 36-44a, p.1161
f=2cm f=2cm f=20 cm f=20 cm +p virtual object? +q q=20 cm +R double lens first lens q=20 cm second lens p=5-20=-15cm VIRTUAL OBJECT! q=1.8 cm
http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.htmlhttp://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html Fig 36-41, p.1158
6. If you cover the top half of the lens in Active Figure 26.24a with a piece of paper, which of the following happens to the appearance of the image of the object? (a) The bottom half disappears. (b) The top half disappears. (c) The entire image is visible but dimmer. (d) There is no change. (e) The entire image disappears.
http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html http://www.hazelwood.k12.mo.us/~grichert/optics/intro.html http://webphysics.davidson.edu/physlet_resources/dav_optics/Examples/eye_demo.html
Rope with total mass m = 2 kg, L = 80 m and mass M = 20 kg at end If end of rope driven in SHM with f=0.056 Hz then wavelength of traveling wave up rope is ? L = 80 m Tension in rope F and wave speed v M = 20 kg If end of rope driven in SHM with f=0.056 Hz then wavelength of traveling wave up rope is
traveling waves ^ ^ positive direction is when one is moving toward the other damped SHO p=E/c driven and dampled SHO I=I0 cos2
+p +p +q +q +f +R +R +f Convex -f Concave Refraction at curved surface Object:p Image: q Focal length : f Magnification :M Real : light at image Virtual : no light Mirrors Lenses Ray1- || to axis, then thru f Ray2-thru f then || Ray3-thru R and back (Ray-4- to center of mirror then reflect at same angle) Ray1- || to axis, then thru f on back Ray2-thru f on front then || Ray3-thru center of lens
virtual object? first lens 15 cmdistance 20.0 cm q=30 cm second lens q=20-30=-10cm VIRTUAL OBJECT! q=6.7 cm Fig 36-32a, p.1151