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MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 16: OPTICS MIRRORS AND LENSES CLASS NOTES PLANE MIRRORS The image formed by a plane mirror has the following properties: 1. The image is as far behind the mirror as the object is in front. 2. The image is unmagnified, virtual, and erect. 3. The image has right-left reversal. IMAGES FORMED BY SPHERICAL MIRRORS A spherical mirror has the shape of a segment of a sphere. Real images are formed at a point where reflected light actually passes through the point. Virtual images are formed at a point where light rays appear to diverge from the point. RAY DIAGRAMS FOR MIRRORS The point of intersection of any two of the following rays for mirrors locates the image. There are three steps to drawing the rays. RAY DIAGRAMS FOR MIRRORS The following three rays form the ray diagram for a curved mirror. 1. The first ray is drawn from the top of the object (the object is drawn as an arrow). The ray is parallel to the optical axis and is reflected back through the focal point, F. 2. The second ray is drawn from the top of the object to the vertex (center) of the mirror and is reflected with the angle of incidence equal to the angle of reflection. 3. The third ray is drawn from the top of the object through the center of curvature, C, which is reflected back on itself. MIRROR DIAGRAM EXAMPLE RULES FOR MIRROR DIAGRAMS 1. If O > 2F, the image is inverted, smaller, and located between F and 2F. 2. If O = 2F (at C), the image is inverted, the same size as the object, i.e. the distances of both the object and image to the mirror are equal. 3. If 2F < O < F, the image is inverted, larger than the object, and located > 2F. PHYSICS PAGE 1 MR. SURRETTE VAN NUYS HIGH SCHOOL REAL IMAGES Real images form at the intersection of light rays. They are often projected on screens. Any image formed on a screen is a real inverted image. VIRTUAL IMAGES A virtual image is formed because the reflected rays diverge from the surface of the mirror. The virtual image is upright, enlarged, and behind the mirror. Virtual images are always upright. VIRTUAL IMAGE EXAMPLE THE MIRROR EQUATION The mirror equation determines the location of an image: 1/si + 1/so = 1/f si = distance of image to mirror so = distance of object to mirror f = focal length of mirror Important note: si is negative in the case of virtual images. MIRROR FOCAL POINTS The focal point of a spherical mirror is located midway between the center of curvature and the vertex of the mirror: f = R/2 Example 1. An object is placed at a distance of 40.0 cm from a convex mirror along an axis. If a virtual image forms at a distance of 50.0 cm from the mirror, on the same side of the object, what is the focal length of the mirror? 1A. (1) 1/si + 1/so = 1/f (2) 1 /(- 50 cm) + 1/ 40 cm = 1/f (3) 1/40cm – 1/50cm = 1/f (4) 1/f = 5/200cm – 4/200cm = 1/200cm (5) f = 200 cm MAGNIFICATION For spherical mirrors, the magnification can be expressed as the ratio of image distance to object distance. Magnification can also be expressed as the ratio between image and object height: M = hi / ho = - si / so Note: The negative sign in front of si means that the image is upside down. PHYSICS PAGE 2 MR. SURRETTE VAN NUYS HIGH SCHOOL Example 2. If a man’s face is 30.0 cm in front of a concave shaving mirror creating an erect image 1.50 times as large as the object, what is the mirror’s focal length? 2A. (1) M = - si / so (2) - si = M(so) (3) si = - (1.50)(30.0 cm) (4) si = - 45.0 cm (5) 1/so - 1/si = 1/f (6) 1/f = 1/30cm – 1/45cm (7) 1/f = 3/90cm – 2/90cm = 1/90cm (8) f = 90 cm Example 3. A concave spherical mirror has a radius of curvature of 40 cm. An object is placed 60 cm from the mirror’s vertex. 3a. What is the focal length of this mirror? A. (1) f = R / 2 (2) f = 40 cm / 2 (3) f = 20 cm 3b. Sketch a ray diagram. A. 3c. A. (1) (2) (3) (4) (5) (6) Calculate the image distance. 1/si + 1/so = 1/f 1/si = 1/f – 1/so 1/si = 1/20cm – 1/60cm 1/si = 3/60cm – 1/60cm 1/si = 2/60cm si = 30 cm PHYSICS PAGE 3 MR. SURRETTE VAN NUYS HIGH SCHOOL RAY DIAGRAMS FOR THIN LENSES The following three rays form the ray diagram for a thin lens. 1. The first ray is drawn parallel to the optic axis. After being refracted by the lens, this ray passes through (or appears to come from) one of the focal points. 2. The second ray is drawn through the center of the lens. This ray continues in a straight line. 3. The third ray is drawn through the focal point F, and emerges from the lens parallel to the optic axis. CONVERGING LENS EXAMPLES DIVERGING LENS EXAMPLE LENSES PLACED IN SERIES When two or more thin lenses are placed together, their combined focal point is: 1/ fs = (1/fi) PHYSICS PAGE 4 MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 16: OPTICS WAVE OPTICS CLASS NOTES WAVE THEORY OF LIGHT The wave theory of light analyzes phenomena that cannot be explained with ray optics. INTERFERENCE In order to observe sustained interference in light waves, the following conditions must be met: 1. The sources must maintain a constant phase (coherent). 2. The sources must be a single wavelength (monochromatic). 3. The superposition principle must apply. YOUNG’S DOUBLE-SLIT APPARATUS The two slits S1 and S2 serve as coherent monochromatic sources. The path difference: (r2 – r1) = d sin YOUNG’S DOUBLE-SLIT APPARATUS YOUNG’S DOUBLE-SLIT APPARATUS The maximum bright fringe m occurs in the center of the screen where m = 0. The bright fringes on either side of m= 0 are m = +1 and – 1. The equation to find the central bright fringe (bright central line on the screen) is: xm =mL / d YOUNG’S DOUBLE-SLIT APPARATUS To determine the position of all bright fringes, use the equation: dsin = m (m = 0, + 1, + 2, etc. The fringes get fainter as the value of m increases.) PHYSICS PAGE 5 MR. SURRETTE VAN NUYS HIGH SCHOOL Example 1. A Young’s double slit apparatus has a slit separation of 4.00 x 10-5 m on which a monochromatic light beam is directed. The resultant bright fringes on a screen 1.20 m away are separated by 2.15 x 10-2 m. What is the wavelength of the beam? 1A. (1) xm =mL / d (2) = dxm / mL (3) = [(4.00 x 10-5 m)(2.15 x 10-2 m)] / [(1)(1.20 m)] (4) = 7.17 x 10-7 m (5) = 717 nm Example 2. Light in the form of plane waves of a single wavelength are incident on two parallel slits. A viewing screen is a distance D from the slits. A point P on the screen is a distance r1 from one slit and r2 from the other. Example 2. Diagram 2a. If the difference in the distances (r2 – r1) is 1.5 wavelengths (3/2) what would be observed at P? A. A minimum. Destructive interference occurs when the extra distance is an odd integer multiple of the wavelength divided by 2. 2b. If the difference in the distances (r2 – r1) is 2.0 wavelengths (2) what would be observed at P? A. A maximum. Constructive interference occurs when the extra distance is a multiple of wavelength. CHANGE OF PHASE DUE TO REFLECTION An electromagnetic wave undergoes a phase change of 180o upon reflection from a medium that is optically more dense than the one it was traveling (n2 > n1). INTERFERENCE IN THIN FILMS Interference in thin films (like soap bubbles) will be constructive if the waves are out of phase by a multiple of ½ . Destructive interference will occur when the phase difference is ¼ , ¾ , etc. This is expressed: 2nt = (m + ½) PHYSICS PAGE 6 MR. SURRETTE VAN NUYS HIGH SCHOOL THIN FILM INTERFERENCE Example 3. Constructive interference occurs when light of wavelength 500 nm shines on soap bubble film (n = 1.46). What is the minimum thickness of the film? 3A. (1) 2nt = (m + ½) (2) 2nt = (m + ½) (3) 2nt = ½ (4) 4nt = (5) t = / 4n (6) t = (500 x 10-9 m) / (4)(1.46) (7) t = 8.56 x 10-8 m Example 4. A possible means for making an airplane radar-invisible is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would the coating be? 4A. (1) 2nt = (m + ½) (2) 2nt = (m + ½) (3) 2nt = ½ (4) 4nt = (5) t = / 4n (6) t = (3.00 x 10-2 m) / (4)(1.50) (7) t = 5.00 mm POLARIZATION Light can be polarized because light waves are transverse (the electric and magnetic fields that make up light are perpendicular to each other). Compressed longitudinal waves like sound cannot be polarized. Polarized light has vibrations confined to a single plane that is perpendicular to the direction of motion. POLARIZED LIGHT PHYSICS PAGE 7 MR. SURRETTE VAN NUYS HIGH SCHOOL BREWSTER’S LAW Brewster’s law gives the value of the polarizing angle for a surface of index of refraction n: n = tan p BREWSTER’S ANGLE Example 5. A beam of unpolarized light in air strikes a flat piece of glass at an angle of incidence of 57.33o. If the reflected beam is completely polarized, what is the index of refraction of the glass? 5A. (1) n = tan p (2) n = tan 57.33o (3) n = 1.56 PHYSICS PAGE 8