Image formation Lenses and Mirrors

Object:p Image: q Focal length : f Magnification :M Real : light at image Virtual : no light Mirrors Lenses +p +q +R +f 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 Refraction at curved surface +p +q +R +f Convex -f Concave

Simple optical components Imagine a spherical mirror A horizontal ray gets reflected according to… But if they area close to the axis they almost meet The rays don’t really meet Horizontal rays close to the axis focus

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 f Focal point Rays form a point close to the axis are also focused Image

A pinhole camera oatmeal box small hole

A goldfish in a bowl

Mirrors p q object image

Flat Mirrors Light rays appear to diverge from point behind mirror Law of Reflection: p =|q| independent of viewing angle Image is Virtual; behind mirror, can’t project on screen, light rays don’t actually pass through image point h h’ Principle axes p q q Magnification: M = h’/h = 1 Image is Upright p q p q

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?

q1q1 q q ’ q2q2 q2q2 x x’ p q p-R R-q R u Just geometry… Reflection by spherical mirrors

R p q Look at the mirror formula: q=R/2 when p is very large All rays converge at R/2, the focal point

For concave mirror the image is in front of the mirror The ray is reflected straight back because it move in the radial direction and so it is normal to the mirror The image is inverted h h’ q p

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 The image is not inverted h h’ qp The mirror formula still holds, but with q<0 and R<0 The image is virtual The mirror is convex

Provided we use the following sign conventions The mirror formula holds for convex or concave mirrors for real or virtual images +p +q +R +f

Provided we use the following sign conventions The mirror formula holds for convex or concave mirrors for real or virtual images +p +q +R +f

Provided we use the following sign conventions The mirror formula holds for convex or concave mirrors for real or virtual images +p +q +R +f

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

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 f’ f p q h -h’

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 0.8m organist 5.3m choir 0.6m L l

Exercise 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 corrected apr 25 + side - side

So, to make an image just do some ray-tracing A ray trough a focal point is refracted to a horizontal ray q’q’ q f’ f A horizontal ray is refracted towards a focal point p q h -h’ The optical component may not be back-front symmetric so in general f≠f’ 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 n 1 (left) and refraction index n 2 (right) Thin lens of curvature radii R left, R right and of material with refraction index n

An object is: REAL if light diverges from it VIRTUAL if light converges toward it Definitions and sign conventions An image is: VIRTUAL if light diverges from it REAL if light converges toward it f f p q h h’ +p +q +R +f Convex -f Concave +p +q +R +f mirror lens

For example: image An image is: VIRTUAL if light diverges from it REAL if light converges toward it object An object is: REAL if light diverges from it VIRTUAL if light converges toward it +p +q +R +f Convex -f Concave

Sign conventions for mirrors: +p +q +R +f

Exercise 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 +p +q +R +f Convex -f Concave

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

two lens system Fig 36-32a, p.1151 first lens q=30 second lens p=40-30=10cm q=-20 cm M 1 =-30/15=-2 15 cm 40 M=M 1 M 2 =-8 M 2 =-(-20)/(10)=4 real inverted virtual, upright Total inverted +p +q +R

Microscope Fig 36-44a, p.1161

f=2cm f=20 cm f=2cm f=20 cm virtual object? first lens q=20 cm second lens p=5-20=-15cm VIRTUAL OBJECT! q=1.8 cm double lens q=20 cm +p +q +R

Fig 36-40, p.1157

Fig 36-41, p

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.

Rope with total mass m = 2 kg, L = 80 m and mass M = 20 kg at end L = 80 m M = 20 kg Tension in rope F and wave speed v If end of rope driven in SHM with f=0.056 Hz then wavelength of traveling wave up rope is If end of rope driven in SHM with f=0.056 Hz then wavelength of traveling wave up rope is ?

damped SHO driven and dampled SHO traveling waves ^ ^ p=E/c I=I 0 cos 2  positive direction is when one is moving toward the other

Object:p Image: q Focal length : f Magnification :M Real : light at image Virtual : no light Mirrors Lenses +p +q +R +f 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 Refraction at curved surface +p +q +R +f Convex -f Concave

virtual object? Fig 36-32a, p.1151 first lens q=30 cm second lens q=20-30=-10cm VIRTUAL OBJECT! q=6.7 cm 15 cm distance 20.0 cm