lecture 12

Physics 123A,B Waves and Modern Physics Paul Boynton Professor of Physics Lecture 13 (T&M: 32.1-2) Lenses & Mirrors (ray optics) February 4, 2009 Plane Mirrors: reflection optics 1.? Any point P on an object acts as a point source of light, producing many rays that are reflected by the mirror in different directions. 2.? The reflection of each incident ray can be constructed using the law of reflection. 3.? The reflected rays can be extrapolated backward to the point P? from which they seem to emanate. Point P? is the virtual image of point P, when viewed by reflection. This construction shows that the object distance S and the image distance S? are equal. Example: How High the Mirror? If your height is h: a) What is the shortest mirror on the wall that will show your full image? b) Where must the top of the mirror be hung? l 1 =eye to top of head l 2 to floor h=l 1 + 2 height of miror= 1 2 l+ 1 2 l= 1 2 h top ofil 1 ?l 1 Lenses: Forming Images by Refraction n 1 ?= 2 (smal-ngle verion of Snel's Law) Thin Lenses When parallel rays pass through a converging lens, the rays converge at the focal point f of the lens. They form a real image of an object at infinity. When parallel rays pass through a diverging lens, the rays diverge away from the focal point f of the lens, and must be extrapolated back to find it. The image is virtual. (f>0) (f<0) Converging Lenses and Special Rays A thin lens is defined here as a lens with a thickness that is small relative to its focal length. We can approximate the lens behavior by assuming that the incident rays are bent as they pass through the mid-plane of the lens. 1.? Parallel incident rays are brought to a focus beyond the lens at the downstream focal point f. 2.? Incident rays that initially pass through the upstream focal point f become parallel beyond the lens. 3.? Incident rays that pass through the center of the lens are not deflected. 1.? Draw the lens. 2.? Draw the optical axis through the center of the lens, with the focal points f placed symmetrically on both sides. 3.? Represent the object with an upright arrow of height h at distance s. 4.? Draw the three ?special? rays from the tip of the object arrow: (a) A ray from the arrow tip parallel to the axis => right focus; (b) A ray from the arrow tip through the left focus => parallel; (c) A ray from the arrow tip through the lens center => straight. 5.? Extend these rays until they converge (at the image arrow tip). 6.? Measure the image height h? and the image distance s?. Converging Lens Ray Tracing Real Images M?? s' Q' s' = Thin les quation(s): 1 s + ? = 1 f Example: Finding the image of a flower A 4.0 cm diameter flower is 200 cm from the 50-cm-focal-length lens of a camera. How far should the film be placed behind the lens to record a well- focused image? What will be the diameter of the flower image on the film? s=200 cm f=50 c h=4.0 cm s'=66.7 cm (by construcion) or, Finaly, h' s' =;h' s' =(4.0 cm) (66.7 c) 200 =1.3 cm from 1 + ? 1 f ? f ? 50×200 ? c66.7 c Virtual Images The rays diverge from point P on the object (which is inside f ), are refracted by the lens, and still diverge after the lens. If the downstream rays are then extrapolated backwards, these extrapolated rays converge to a virtual image at P?. In the case shown, the virtual image is upright and has a magnification greater than 1. The rays reaching the eye appear to be coming from the virtual image. Magnification 1.? A positive value of M indicates that the image is upright relative to the object. A negative value of M indicates that the image is inverted relative to the object. 2.? The absolute value of M gives the size ratio of image to object: h?/h = |M|. Note: Even though M is called the ?magnification?, its magnitude can be less than 1, indicating that the image is smaller than the object, (i.e., it is demagnified). Example: Magnifying a flower To better view a flower, a naturalist holds a 6.0-cm-focal- length magnifying glass 4.0 cm from the flower. What is the magnification? s'=?12 cm (by construcion), or So, M=? s'(?12 cm) 4.0 =3.0 ?s fs6.0×4.0 4. c?12.0 cm Diverging Lenses: Special Rays A lens that is thicker at the edges than at the center is called a diverging lens. Special rays: 1.? Parallel incident rays, after passing through the lens, will diverge from a virtual focus point behind the lens, at the upstream focal point f. 2.? Incident rays converging toward the downstream focal point f become parallel after passing through the lens. 3.? Incident rays that pass through the center of the lens are ?undeviated.? Diverging Lens Ray Tracing 1.? Draw the lens. 2.? Draw the optical axis through the center of the lens, with the focal points f placed symmetrically on both sides. 3.? Represent the object with an upright arrow of height h at distance s. 4.? Draw the three ?special? rays from the tip of the object arrow: (a) A ray from the arrow tip parallel to the axis => from left focus; (b) A ray from the arrow tip toward the right focus => parallel; (c) A ray from the arrow tip through the lens center => straight. 5.? Extrapolate these rays backwards until they converge. 6.? Measure the image height h? and the image distance s?. Example: Demagnifying a flower A diverging lens with a 50-cm- focal-length is placed 100 cm from the flower. Where is the image? What is the magnification? s'=?33 cm (by construcion), or from 1 s + ? = 1 f ?s= fs ? 50×100 ?() cm=33.3 c And, M? ' s (33 cm) 100 0.33 saler, but upright. Spherical Mirrors If a mirror is only a small portion* of a spherical surface, so that the small-angle approximation is valid, incident parallel rays will focus at a distance that is ½ the mirror?s radius of curvature, r. The location of this convergence on the optic axis is called the focal point f. *Mirror diameter much smaller than radius of parent Sphere Ray Tracing For spherical mirrors, one can geometrically construct the image point using ?special? rays: 1.? A ray from the object that is parallel to the principal axis will be reflected through the focal point. 2.? A ray from the object through the focal point will be reflected parallel to the principal axis. 3.? A ray from the object along a radius will be reflected back along the radius. 4.? A ray (dashed) to the mirror center (A) will be reflected at an equal and opposite angle. 5.? An image is formed at the point where all these lines cross. 3 3 Object Outside the Focal Point When the object is farther from the mirror than the focal point f, a real, inverted image is formed. Here ?real? means that rays actually pass through the image point I? and ?inverted? means that the image is upside-down with respect to the object. Object Inside the Focal Point When the object is closer to the mirror than the focal point f, a virtual, upright image is formed. Here ?virtual? means that rays do not pass through the image point I? and ?upright? means that the image has the same up/down orientation as the object. This gives the magnifying effect of a shaving mirror. The Mirror Equations 1 s + ' = 1 f 2 r m? y' =? s' s s? y y' Convex Mirrors Convex mirrors have a virtual focus, in that reflected parallel rays diverge, but can be extrapolated backwards to the focal point F. The image of an object in front of a convex mirror is always virtual and upright with a magnification less than 1. s? s Paul Physics123A,B L12b.ppt