R   h h’h’ s-Rs-R R-s’R-s’ s s’s’

Today... Overview : Nothing new here! –objects and images –convex mirrors Concave Spherical Mirrors –The Mirror Eqn, Magnification, Sign Conventions Planar & Convex Spherical Mirrors Text Reference: Chapter 34.1 examples: 34.1 and 2

Nothing New! For the next two lectures we will be studying geometric optics. You already know the fundamentals of what is going on!!! –Light propagates as rays in situations in which the length scales are >> than the light’s wavelength incident ray reflected ray 11 rr n1n1 refracted ray 22 n2n2 We will use these laws to understand the properties of mirrors (perfect reflection) and lenses (perfect refraction). We will also discover properties of combinations of lenses which will allow us to understand such applications as microscopes, telescopes, and eyeglasses. –Reflection: –Refraction:

Objects and Images Object: Light reflects off an object and rays go off in all directions Some rays travel to observer and are “processed” by her or him

Objects and Images Light reflects off an object and rays go off in all directions Some rays travel to lens and are bent to form “image” im ag e obj ect Rays from image are seen by observer This image formation is behind magnifying lens, microscopes and telescopes….. Real images and virtual images…….

Common misbeliefs on reflection –“One-way mirrors” or “one-way glass” These are often shown in “interrogation rooms” or police line-ups. A more familiar example is mirrored sunglasses. In all cases the transmission through the system has to be the same no matter which way the light is going. The sunglasses only appear (to the person looking at them) not to transmit any light because there is no light source behind them. –Mirrors reverse right-left, but not up-down Actually, mirrors don’t invert up-down or left-right! They invert the other axis -- distance from the mirror

Concave Spherical Mirrors We start by considering the reflections from a concave spherical mirror in the paraxial approximation (i.e., small angles of incidence close to a single axis): First draw a ray (light blue) from the tip of the arrow through the center of the sphere. This ray is reflected straight back since the angle of incidence = 0. Now draw a ray (white) from the tip of the arrow parallel to the axis. This ray is reflected with angle  as shown.   R

Concave Spherical Mirrors … continued Note that the blue and white rays intersect in a point, suggesting an inverted image. To check this, draw another ray (green) which is incident at angle , as shown. Note that this ray intersects the other two at the same point, as it must if an image of the arrow is to be formed there.   Note also that the green ray intersects the white ray at another point along the axis. We will call this point the focal point ( ).   R These 3 rays are the “principal rays”

Lecture 25, ACT 1 –Where do the rays which are reflected from the convex mirror shown intersect? (a) Inverted and in front of the mirror (b) Inverted and in back of the mirror (c) Upright and in back of the mirror What is the nature of the image of the arrow? 1B 1A (a) to left of (b) to right of (c) they don’t intersect

Lecture 25, ACT 1 (a) to left of (b) to right of (c) they don’t intersect The angle of incidence = angle of reflection. Blue ray has normal incidence and is reflected straight back. White ray is reflected at larger angle than blue ray. Therefore the reflected rays are diverging!! They do not intersect! R –Where do the rays which are reflected from the convex mirror shown intersect? 1A

Lecture 25, ACT 1 What is the nature of the image of the arrow? (a) Inverted and in front of the mirror (b) Inverted and in back of the mirror (c) Upright and in back of the mirror 1B Have you ever been to a 7-11, and looked at those mirrors in the corner? What about the right-side mirror on a car or Jeep (see Jurassic Park / T-Rex scene) To find the image of the arrow, we need to find the point where the reflected rays APPEAR to come from. The intersection of the extrapolations of the reflected rays gives the image position as shown. Well, you are looking at a convex mirror like this. So, what is the answer? and smaller ! R

The Mirror Equation We will now transform the geometric drawings into algebraic equations: !! we want to relate s, s ’, and R !! R     object  h image s s’s’ from triangles, eliminating , Plugging these back into the above equation relating the angles, we get: Defining the focal length f = R/2, This eqn is known as the mirror eqn. Note that there is no mention of  in this equation. Therefore, eqn works for all small , i.e., we have an image! For  a small angle we can approximate:

Magnification We have derived the mirror eqn which determines the image distance in terms of the object distance and the focal length: What about the size of the image? How is h ’ related to h ?? Use similar triangles …. R   h h’h’ s-Rs-R R-s’R-s’ s s’s’

Magnification … continued What about the size of the image? How is h ’ related to h ?? From similar triangles: Now, we can introduce a sign convention. We can indicate that this image is inverted if we define its magnification M as the negative number given by: R   h h’ s-Rs-R R-s’ s s’s’

More Sign Conventions Consider an object distance s which is less than the focal length: h’h’   s’s’ Ray Trace: Ray through the center of the sphere (light blue) is reflected straight back. R h s f We call this a “virtual image”, meaning that no light from the object passes through the image point. This situation is described by the same mirror equations as long as we take the convention that images behind the mirror have negative image distances s΄. I.e.: s ΄ < 0 → virtual image M>0 → not inverted Ray parallel to axis (white) passes through focal point f. These rays diverge! I.e., these rays look like they are coming from a point behind the mirror.

Concave-Planar-Convex What happens as we change the curvature of the mirror? –Plane mirror: »R =  IMAGE: virtual upright (non-inverted) h’h’  h  s s’s’ f IMAGE: virtual upright (non-inverted) –Convex mirror: »R < 0

Lecture 25, ACT 2 In order for a real object to create a real, inverted enlarged image, a) we must use a concave mirror. b) we must use a convex mirror. c) neither a concave nor a convex mirror can produce this image.

Lecture 25, ACT 2 In order for a real object to create a real, inverted enlarged image, a) we must use a concave mirror. b) we must use a convex mirror. c) neither a concave nor a convex mirror can produce this image. A convex mirror can only produce a virtual image since all reflected rays will diverge. Therefore, b) is false. To create a real image with a concave mirror, the object must be outside the focal point. The example we just did gave a real, inverted reduced image. Is it possible to choose the parameters such that the image is enlarged?? The easy (but clever) answer: h’ is a real image. Therefore consider the OBJECT to be h ’. The IMAGE will be h !!! Therefore it certainly IS POSSIBLE!! equations follow…

Lecture 25, ACT 2 In order for a real object to create a real, inverted enlarged image, a) we must use a concave mirror. b) we must use a convex mirror. c) neither a concave nor a convex mirror can produce this image. The example we just did gave a real, inverted reduced image. Is it possible to choose the parameters such that the image is enlarged?? Thus to get a real, inverted enlarged image we need: Thus to get a real image we need: Thus to get an inverted enlarged image we need:

Summary We have derived the equations for mirrors : when the following sign conventions are used: Variable f > 0 f < 0 s > 0 s < 0 s ’ > 0 s ’ < 0 Mirror concave convex real (front) virtual (back) real (front) virtual (back) Principal rays “connect” object and image one goes through the center of the mirror the other goes parallel to the optic axis and then is reflected through a focal point the third one is like the second one….

More reflections… We already know two ways to get a reflection –Off a conductor (e.g., aluminum or silver mirror) –Total internal reflection (e.g., binoculars) There are at least two other common methods –“Fresnel” reflection at an interface between two dielectrics with different indices of refraction, n 1 and n 2 : For light at normal incidence e.g., n 1 = 1, n glass =1.5  R = 4% –Interference between Fresnel reflections at interfaces separated by 1/4 or 1/2 the wavelength of light [Phys. 114]  this can lead to very high reflectivities (> %, used to couple photons to atoms) or very low reflectivities (<0.1%, used as “anti-reflection” coatings on glasses, camera lenses, etc.)

Selective reflection: color in metals “plasma edge” Al Cu Au Ag Silver (Ag): Gold (Au): Copper (Cu): In some systems, e.g., metals and the atmosphere, transparency and color influenced by the oscillation of free (mobile) charges

Next time Lenses –Lensmaker’s Formula –Lens Equation Reading Assignment: Chapter 34.2 Examples: 34.3,4,6,7 and 8