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PHYS-2020: General Physics IICourse Lecture NotesSection XIDr. Donald G. LuttermoserEast Tennessee State UniversityEdition 3.3AbstractThese class notes are designed for use of the instructor and students of the course PHYS-2020:General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. Thesenotes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.XI. Reflection and Refraction of LightA. Huygens’ Principle.1. In 1678, Hu ygens proposed t he wave theory of light.a) At that same time, Newton maintained that light actedlike a particle.b) As we saw in t he last section, P lanck derived in 1900 thatlight has the characterists of both a wave and a p article=⇒ a wavicle , which he called a photon.c) In the last sections of the notes, we discussed the parti-cle characteristics of a photon. In this section, we willconcentrate on the wave-like characteristics.2. All points on a wavefront can be considered point sources forthe production of spherical secondary wavelets. After time t,the new position of the wavefront will be the surface of tangencyto these secondary wavelets. This geometrical argument is calledHuygens’ Principle.t = 0oldwavefrontr = ctnewwavefrontXI–1XI–2 PHYS-2020: General Physics I IB. Reflection of Light.1. When light travels from one mediu m to another, part of the lightcan be reflected at the media interface.a) Reflection off of a smooth surf ace is called specular re-flection (which we will assume from this point forward ).b) Reflection off of a rough surface is called diffuse reflec-tion.2. Law of Reflection: The angle of incidence with r espect to thenormal of the reflecting surface, θi, equals the angle of reflection,θr:θi= θr. (XI-1)ReflectingSurfacenormallineincidentrayreflectedrayθiθr3. The amou nt of energy that is reflected compared to the amountincident is called the reflectivity of the surface.a) This also is called albedo.b) The reflectivity of a mirror is about 96% (albed o = 0.96).Donald G. Luttermoser, ETSU XI–3C. Refraction of Light.1. When light travels from one mediu m to another, part of the lightcan be transmitted across t he media surface and refracted.a) Refraction means that th e light beam bends.b) This bending takes place because the light beam’s (i.e.,photon’s) velocity changes as it goes from one medium tothe next, following the relation:sin θrsin θi=vrvi= constant . (XI-2)i) vrand θrare the velocity and the angle of therefracted beam with respect to the normal line ofthe sur face.ii) viand θiare the velocity and the angle of theincident beam with respect to the normal line ofthe sur face.normallineincidentrayrefractedrayθiθrvivrni = n1nr = n22. The index of refraction of a material isn ≡speed of light in vacuumspeed of light in medium=cv. (XI-3)XI–4 PHYS-2020: General Physics I Ia) n for some common substances:i) Vacuum: 1.000000ii) Air (0◦C, 1 atm) : 1.000293iii) Ice (H2O, 0◦C): 1.309iv) Glass (crown): 1.52b) n also is a function of wavelength. For the index of re-fraction in air we havenair= 1+6.4328×10−5+2.949810 × 1061.46 × 1010− ν2+2.5540 × 1044.1 × 109− ν2,(XI-4)where ν = 1/λairis called the wavenumber and is mea-sured in cm−1in this equation (just include the value ofthe wavenumber without its units in t he equation above,remember, n is un itless).3. Eq. (XI-2) can be re-expressed as a function of n =⇒ Law ofRefraction better known as Snell’s Law:n1sin θ1= n2sin θ2, (XI-5)where the ‘1’ label indicates t he first medium the light is in andthe ‘2’ label indicates t he second med ium.4. Besides a change in light beam direct ion, the wavelength of lightalso changes when light goes from one medium to the next:λ1n1= λ2n2. (XI-6)a) For starlight coming in from outer space,λair=λvacnair. (XI-7)Donald G. Luttermoser, ETSU XI–5b) Spectral lines are shifted in wavelength as they pass throughthe Earth’s atmosphere which needs to be taken into ac-count when planetary, stellar, or galactic spectra are beinganalyzed.Example XI–1. Problem 22.18 (Page 785) from the Ser-way & Vuille textbook: A ray of light strikes a flat, 2.00-cm thickblock of glass (n = 1.50) at an angle of 30◦with respect to thenormal (see Fig. P22.18 in the text and the figure below). (a) Findthe angle of refracti on at the top surface. (b) Find the angle of i nci-dence at the bottom surface and the refracted angle at this surface.(c) Find the lateral distance d by which the light beam is shifted. (d)Calculate the speed of light in the glass and (e) the time requir edfor the light to pass through the glass block. (f) Is the travel timethrough the block affected by the angle of incidence? Explain.surface2surface1ab c= 2.00 cmhddθi1θi1θr1αnangnaab chdθi1θr1αSolution (a):At the top surface (labeled 1), the angle of incidence is given tous as θi1= 30.0◦. We will assume that th e beam is hitting th eglass from air, hence na= 1.00. Snell’s law gives the refractedXI–6 PHYS-2020: General Physics I Iangle in th e glass (with ng= 1.50) at surface 1 asnasin θi1= ngsin θr1sin θr1=nasin θi1ngθr1= sin−1nasin θi1ng= sin−1(1.00) sin 30.0◦1.50= 19.5◦.Solution (b):Since the second surface (labeled 2) is parallel to the first, theangle of incidence at the bottom surface is exactly the same asthe refracted angle at th e top surface following the theorem ofgeometry, so θi2= θr1= 19.5◦. The angle of refraction at thisbottom surface (back into the air) is thenngsin θi2= nasin θr2sin θr2=ngsin θi2naθr2= sin−1"ngsin θi2na#= sin−1(1.50) sin 19.5◦1.00= 30.0◦.Thus, the light emerges traveling parallel to t he incident beam.Solution (c):Let ` = 2.00 cm be the thick ness of the glass. The angle ofrefraction at the first surface from Part (a) is θr1= 19.5◦. Let hrepresent the d istan ce from point ‘a’ to ‘c’ (i.e., the hypotenuseof tr iangle abc), thenh =`cos θr1=2.00 cmcos 19.5◦= 2.12 cm .Donald G. Luttermoser, ETSU XI–7From the drawing above, n ote that angle α = θi1− θr1= 30.0◦−19.5◦= 10.5◦and also that d represents the opposite sid e ofthe right-angle triangle defined by angle α with h being the hy-potenuse of this triangle. Thend = h sin α = (2.12 cm) sin 10.5◦= 0.386 cm .Solution (d):The speed of light in the glass is given by Eq. (XI-3):v =cng=3.00 × 108m/s1.50= 2.00 × 108m/s .Solution (e):For th is question, we only need to use the definition of


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