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Speed of light (in vacuum) Foucault’s experiment Speed of light (in vacuum) Michelson’s 1878 Rotating Mirror Experiment • German American physicist A.A. Michelson realized, on putting together Foucault’s apparatus, that he could redesign it for much greater accuracy. • Instead of Foucault's 60 feet to the far mirror, Michelson used 2,000 feet.. • Using this method, Michelson was able to calculate c = 299,792 km/s • 20 times more accurate than Foucault • Accepted as the most accurate measurement of c for the next 40 years. Nature of light Waves, wave fronts, and rays • Wave front: The locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same. rays source wave fronts spherical wave plane wave Reflection and refraction Reflection and refraction • When a light wave strikes a smooth interface of two transparent media (such as air, glass, water etc.), the wave is in general partly reflected and partly refracted (transmitted). reflected rays incident rays r a a b b refracted rays b a Reflection and refraction Reflection • The incident, reflected, and refracted rays, and the normal to the surface all lie in the same plane. • The angle of reflection r is equal to the angle of incidence a for all wavelengths and for any pair of material. reflected rays incident rays r a a r a b b refracted rays Reflection and refraction Reflection (cont’d) Reflection and refraction Refraction • The index of refraction of an optical material (refractive index), denoted by n, is the ratio of the speed of light c in vacuum to the speed v in the material. wavelength in vacuum. Freq. stays the same. n c / v ; 0 / n f=v Reflection and refraction Refraction • The index of refraction of an optical material (refractive index), denoted by n, is the ratio of the speed of light c in vacuum to the speed v in the material. wavelength in vacuum. Freq. stays the same. reflected rays incident rays n c / v ; 0 / n • The ratio of the sines of the angles a and b , where both angles are measured from the normal to the surface, is equal to the inverse ratio of the two indices of refraction: sin a nb sin b na r a a b b Snell’s law refracted rays Example: depth of a swimming pool Pool depth s = 2m person looks straight down. 2 1 L the depth is judged by the apparent size of some object of length L at the bottom of the pool (tiles etc.) na sin 1 sin 2 2 tan 1 L s tan 2 L L s s s ' s tan 1 ( s s ) tan 2 for small angles: tan ->sin 1 s sin 1 ( s s ) sin 2 s sin 1 ( s s ) na sin 1 L s s na 1 1 (2 m) 50 cm. na 4 Example: Flat refracting surface • The image formed by a flat refracting surface is on the same side of the surface as the object – The image is virtual – The image forms between the object and the surface – The rays bend away from the normal since n1 > n2 L n1 n2 n2 s' s s s' n1 | s ' | tan 1 L, | s | tan 2 L s' tan 1 s tan 2 tan sin for 1 s' sin 1 s sin 2 n1 s' n2 s ( n1 sin 1 n2 sin 2 ) s’ s Total internal reflection Total internal reflection n2 sin 2 , sin 2 1 when n2 / n1 1& n2 n1 sin 1. Since sin 1 n1 When this happens, 2 is 90o and 1 is called critical angle. Furthermore when 1 crit , all the light is reflected (total internal reflection). Total internal reflection Optical fibers Prism example • Light is refracted twice – once entering and once leaving. • Since n decreases for increasing , a spectrum emerges... Analysis: (60 glass prism in air) sin 1 = n2 sin 2 n2 sin 3 = sin 4 n1 = 1 60 Example: 1 = 30 1 2 3 sin(30 ) o 19 .5 1.5 2 sin 1 4 3 (60 o 2 ) 40 .5o 4 sin 1 1.5 sin 3 76 .9 o n2 = 1.5 ++60o = 180o 3 = 90 - = 90 - 2 3 = 60 - 2 Prism Applications of prism • A prism and the total reflection can alter the direction of travel of a light beam. • All hot low-pressure gases emit their own characteristic spectra. A prism spectrometer is used to identify gases. Diversion Diversion • The index of refraction of a material depends on wavelength as shown on the right. This is called diversion. • It is also true that, although the speed of light in vacuum does not depends on wavelength, in a material, wave speed depends on wavelength. Diversion Examples Huygens’ principle Huygens’ principle Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the wave. Plane waves Huygens’ principle (cont’d) Huygens’ principle for plane wave • At t = 0, the wave front is indicated by the plane AA’ • The points are representative sources for the wavelets • After the wavelets have moved a distance st, a new plane BB’ can be drawn tangent to the wavefronts Huygens’ principle (cont’d) Huygens’ principle for spherical wave Huygens’ principle (cont’d) Huygens’ principle for spherical wave (cont’d) • The inner arc represents part of the spherical wave • The points are representative points where wavelets are propagated • The new wavefront is tangent at each point to the wavelet Huygens’ principle (cont’d) Huygens’ principle for law of reflection • The law of reflection can be derived from Huygen’s Principle • AA’ is a wave front of incident light • The reflected wave front is CD • Triangle ADC is congruent to triangle AA’C • Angles 1 = 1’ • This is the law of reflection Huygens’ principle (cont’d) Huygens’ principle for law of refraction • In time t, ray 1 moves from A to B and ray 2 moves from A’ to C • From triangles AA’C and ACB, all the ratios in the law of refraction can be found: n1 sin 1 = n2 sin 2 sin 1 v1t; sin 2 v2 t v1t v t c c 2 , v1 , v2 sin 1 sin 2 n1 n2 AC Atmospheric Refraction and Sunsets • Light rays from the sun are bent as they pass into the atmosphere • It is a gradual bend because the light passes through layers of the atmosphere – Each layer has a slightly different index of refraction • The Sun is seen to be above the horizon even after it has fallen below it Mirages • A mirage can be observed when the air above the ground is warmer than the air at higher elevations • The rays in path B are directed toward the ground and then bent by refraction • The observer sees both an upright and an inverted image Example Exercises The prism shown in the figure has a refractive index of 1.66, and the angles A are 25.00 . Two light rays m and n are parallel as they enter m the prism. What is the angle between them they emerge? n A A Solution na sin a 1 1.66 sin 25.0 na sin a nb sin b b sin ( ) sin ( ) 44.6. nb 1.00 Therefore the angle below the horizon is b 25.0 44.6 25.0 19.6, and thus the angle between the two emerging beams is 39.2. 1 Exercises Example Light is incident in air at an angle on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other. (a) t Prove that a a' . (b) Show that this is true for any number of different parallel plates. (c) Prove that the lateral displacement D of the emergent beam is given by the sin( a b' ) relation: d t , a n Q n’ n b' b P a' d cos b' where t is the thickness of the plate. (d) A ray of light is incident at an angle of 66.00 on one surface of a glass plate 2.40 cm thick with an index of refraction 1.80. The medium on either side of the plate is air. Find the lateral Displacement between the incident and emergent rays. Exercises Problem Solution a (a) For light in air incident on a parallel-faced plate, Snell’s law yields: n sin a n' sin b' n' sin b n sin a' sin a sin a' a a' . n t n’ b' Q (b) Adding more plates just adds extra steps in the middle of the above equation that P n b always cancel out. The requirement of L d a' parallel faces ensures that the angle n n' and the chain of equations can continue. (c) The lateral displacement of the beam can be calculated using geometry: t t sin( a b' ) d L sin( a ), L d . cos b' cos b' ' b (d) n sin a sin 66.0 ) sin 1 ( ) 30.5 n' 1.80 (2.40cm ) sin( 66.0 30.5) d 1.62 cm. cos 30.5 b' sin 1 (