normal angle of incidence increases special angle no light is reflected

Transcription

1 Reflection from transparent materials (Chapt. 33 last part) When unpolarized light strikes a transparent surface like glass there is both transmission and reflection, obeying Snell s law and the law of reflection, respectively, θ i θ r θ t For glass (n ~ 1.5) at normal incidence (θ i = 0) the reflected ray is ~ 4% of the incident intensity. As the angle of incidence increases more light is reflected (and less transmitted) for one polarization component, but the other component exhibits a minimum at a special angle at which no light is reflected.

2 We consider the origin of this phenomena. When radiation is incident on a material, the oscillating electric field from the EM wave induces the electrons associated with the atoms near the surface of the material to oscillate about their nuclei at the same frequency. For polarized incident radiation, the induced oscillations in the electronic distribution around the nuclei induce the atom to emit in a so called dipole radiation pattern. Colors here indicate zero weak source of light. Out going radiation has in same f (color) as incoming. + Strong slice through vertical plane including atom zero weak in top view

3 in zero weak cylindrical symmetry + Strong slice through vertical plane including atom zero weak in top view Note that no radiation appears in the direction along the line of the induced oscillation. The oscillating charge has no component of acceleration (generally what causes radiation) perpendicular to that direction to launch EM waves in that direction. In a transparent material part of this radiation is propagated forward into the material as the transmitted wave and a fraction is sent out from the surface as the reflected wave.

4 The directional nature of the transmitted and reflected rays, that makes them consistent with Snell s law and the law of reflection occurs because of interference between the waves from nearby atoms causing reinforcement and cancellation of field amplitudes in the various directions (HRW chapt. 35). Consider what happens when the incident radiation is polarized in the plane of incidence ( the plane that includes the surface normal and incident ray, called p-polarization) along with the condition that θ t + θ r = 90 o. p-polarized θ i θ r Requires θ t θ r + θ t

5 In that case the oscillating dipoles are to θ t and along the line of θ r, a direction in which they can not radiate, so all the incident beam is transmitted and none reflected. θ t + θ r = 90 o θ i θ r + θ t This special angle is called Brewster s angle, designated θ B. Since the law of reflection requires that θ r = θ i we also have that, θ i = θ B. Now, θ B + θ t = 90 o Snell s law then gives (with θ i = θ B ), θ t = 90 o θ B n sinθ = n sinθ = n sin(90 θ ) = n cosθ o 1 B 2 t 2 B 2 B

6 nsinθ = ncosθ 1 B 2 B sinθ = cosθ n n B 2 B 1 tanθ = B θ = B n n 2 1 tan n n If medium 1 is air (n 1 ~ 1) θ = tan n B 2 If medium 2 is glass of index n 2 = 1.5, θ B = 56.3 o. Applications θ B vacuum 1 When high power lasers must be coupled into vacuum systems through windows (to maintain the vacuum), Brewster angle windows are used to avoid the loss of power to reflection.

7 Note that if the angle of incidence for the p-polarized light goes steeper than Brewster s angle the reflected intensity increases again, so Brewster s angle constitutes a minimum (= 0) in the reflection. Light polarized perpendicular to the plane of incidence (called s-polarized light) induces electronic oscillation parallel to the surface and doesn t produce such a minimum. s-polarized

8 Unpolarized incident light can be resolved into s and p polarized components. At Brewster s angle only the s-polarized component is reflected so the surface polarizes the light on reflection. Sunlight reflecting off the surface of water near Brewster s angle is horizontally polarized. This is why polarizing sunglasses have their polarizing axis vertical. To block the polarized s-component that remains in the refection. p s θ B polarizing direction vertical (allowed to pass) blocking horizontal polarization.

9 Image formation in mirrors (chapt. 34) Plane mirrors object p i virtual image Rays of scattered light diverge from each point of the object. same orientation and size as object. image is erect (as opposed to inverted) The rays that reflect from the mirror, in the direction of the pupil, extended back, appear to converge from a virtual image at a distance behind the mirror that equals the object distance in front of the mirror, i = p. Distances for virtual images are by convention negative so, i = p

10 Spherical mirrors Concave (Innie) Two general types: Convex (Outtie)

11 The concave mirror Has radius of curvature r. Light in and out Parallel rays of light (plane waves) coming in along the central axis reflect to a single point, a distance f from c, called the real focus, F. (CA) This focal distance is completely determined by radius of curvature r. Like a plane mirror the concave mirror also forms images, but the location, size and type is distinct from the plane mirror.

12 Image location for the concave mirror by ray tracing. From tip of object draw two rays. One parallel to CA (reflection goes through focus) and one through focus (reflection goes parallel to CA). p F CA Where those rays converge locates the tip of the image. i In contrast to the plane mirror, the image here is a real image (a replica of the object will exist on a screen placed in the image plane). Note also that the image is inverted and a different size than the object. As the object moves toward the mirror, the image moves away from the mirror and grows (magnifies). i p F f CA

13 If the object is closer to the concave mirror than the mirror focal length we do not get a real image. p Again consider 2 rays coming from the object tip. F CA The ray along the direction from the focus to the mirror (green ray), must reflect back parallel to the CA. The other ray parallel to the CA (black ray), must, on reflection, go through the focus. f i Note that these two reflected rays diverge so there is no real image, however, their projections behind the mirror surface converge to form a magnified, erect, virtual image, that an observer sees.

14 The convex mirror Also has radius of curvature r. Light in and out Parallel rays of light (plane waves) coming in along the central axis diverge as if coming from a single point, a distance f from c, called the virtual focus, F. (CA) This focal distance is completely determined by radius of curvature r. Also forms (virtual) images.

15 Image location for the convex mirror by ray tracing. Again draw two rays from the tip of the object. p f Draw one ray parallel to the CA (green ray), which must reflect along the line to the virtual focus. F CA Draw the other ray (black) toward the virtual focus, which must reflect parallel to the CA. Since these reflected rays diverge they can not form a real image. Their projections back into the mirror, however converge, so we again get a virtual image, which in this case is erect and smaller. Note that as the object approaches the mirror, the black ray from the tip going toward the focus steepens raising the reflection (parallel to CA) making the image approach the surface and grow. You ve all probably experienced your nose growing as you approach a convex mirror. i

16 These geometrical constructs are useful for understanding the image formation characteristics of these mirrors, and very carefully made drawings and measurements can give quantitative information but an algebraic formulation is preferred. Two simple relations (also useful for image formation with thin lenses) and some conventions (rules) provide this (HRW sec for proof) = p i f Where, as above (all measured along the central axis), p = object-mirror distance i = image-mirror distance f = focal length (real or virtual) And, m = i p Where m is the lateral magnification (i.e. image height/object height).

17 =, p i f m = i p The conventions are as follow: The object distance p is positive. The focal length f is positive for a real focus (concave mirror) and negative for a virtual focus (convex mirror). The image distance i is positive for a real image (the image in front of a mirror) and negative for a virtual image (the image behind the mirror) A positive magnification, m, means that the object and image have the same orientation while negative means they have opposite orientation.

18 Example A 4 cm high toy car lies 25 cm (along the CA) from a concave mirror that has a 50 cm focal length. Where is the image? = p i f = 25 i 50 positive because concave real focus = = = i i = 50cm Negative so virtual image, 50 cm behind mirror. What s the magnification and orientation? i 50 m= = = 2 Image is twice as large as object and erect p 25 (positive sign means erect or same orientation).

19 Example A 4 cm high toy car lies 25 cm (along the CA) from a convex mirror that has a 50 cm focal length. Where is the image? = p i f = 25 i 50 negative because convex virtual focus = = = i i = 16.7cm Negative so virtual image, 17 cm behind mirror. What s the magnification and orientation? i 16.7 m= = = 0.67 Image is 2/3 as large as object and erect. p 25