Physics Solution Manual for 1100 and 2101

We begin by using the thin lens equation to find the

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Unformatted text preview: n2 sin θ 2 ) at each face of the prism. At the first interface where the ray enters the prism, n1 = 1.000 for air and n2 = ng for glass. Thus, Snell's law gives (1) sin 60.0° = ng sin θ 2 or sin θ 2 = sin 60.0° ng (1) We will represent the angles of incidence and refraction at the second interface as θ1′ and θ 2′ , respectively. Since the triangle is an equilateral triangle, the angle of incidence at the ′ second interface, where the ray emerges back into air, is θ1 = 60.0° – θ 2 . Therefore, at the second interface, where n1 = ng and n2 = 1.000, Snell’s law becomes ′ ng sin (60.0° – θ 2 ) = (1) sin θ 2 (2) ′ We can now use Equations (1) and (2) to determine the angles of refraction θ 2 at which the red and violet rays emerge into the air from the prism. Chapter 26 Problems 1377 SOLUTION The index of refraction of flint glass at the wavelength of red light is Red Ray ng = 1.662. Therefore, using Equation (1), we can find the angle of refraction for the red ray as it enters the prism: sin θ 2 = sin 60.0° = 0.521 1.662 θ 2 = sin –1 0.521 = 31.4° or Substituting this value for θ2 into Equation (2), we can find the angle of refraction at which the red ray emerges from the prism: ′ sin θ 2 = 1.662 sin ( 60.0° – 31.4° ) = 0.796 Violet Ray or ′ θ 2 = sin –1 0.796 = 52.7° For violet light, the index of refraction for glass is ng = 1.698. Again using Equation (1), we find sin θ 2 = sin 60.0° = 0.510 1.698 θ 2 = sin –1 0.510 = 30.7° or Using Equation (2), we find ′ sin θ 2 = 1.698 sin ( 60.0° – 30.7° ) = 0.831 or ′ θ 2 = sin –1 0.831 = 56.2° ______________________________________________________________________________ 48. REASONING At the slanted face, the refracted ray(s) emerge into the surrounding material, which has an index of refraction n2. Together, the index of refraction n2 of the 45.00° surrounding material and the index of refraction n1 of θ2 crown glass (which varies with the color of the light; see θ1 Table 26.2) determine the critical angle θc for total internal n reflection, as we see from sin θc = 2 (Equation 26.4). To n1 guarantee that the colors that are to emerge from the prism at the slanted face do not undergo total internal reflection there, we must choose n2 such that the angle of incidence θ1 is less than the critical angle θc for each such color. The incident ray and the totally-internally-reflected ray are perpendicular to one another, and the 90.00° angle between them is bisected by the normal to the slanted surface (see the drawing). Therefore, the angle of incidence is θ1 = 45.00°. Thus, all colors of light for which the critical angle is greater than 45.00° will not undergo total internal reflection and will emerge from the slanted face of the prism. 1378 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION a. Solving sin θc = n2 (Equation 26.4) for n2, we obtain n1 n2 = n1 sin θc (1) Since only red light is to emerge from the slanted face of the prism, we will use n1 = 1.520, the refractive index for red light in crown glass (see Table 26.2). The indices of refraction for the remaining colors are all larger than 1.520, so all other colors of light will have critical angles less than 45.00° and will undergo total internal reflection at the slanted face of the prism. Thus, we have n2 = (1.520 ) sin 45.00o = 1.075 b. All colors except violet are to emerge from the slanted face. Therefore, we will use n1 = 1.531, the index of refraction of blue light in crown glass (see Table 26.2), in Equation (1): n2 = (1.531) sin 45.00o = 1.083 ______________________________________________________________________________ 49. SSM REASONING The ray diagram is constructed by drawing the paths of two rays from a point on the object. For convenience, we will choose the top of the object. The ray that is parallel to the principal axis will be refracted by the lens so that it passes through the focal point on the right of the lens. The ray that passes through the center of the lens passes through undeflected. The image is formed at the intersection of these two rays. In this case, the rays do not intersect on the right of the lens. However, if they are extended backwards they intersect on the left of the lens, locating a virtual, upright, and enlarged image. SOLUTION a. The ray-diagram, drawn to scale, is shown below. Scale: 20 cm 20 cm F Image F Object Chapter 26 Problems 1379 From the diagram, we see that the image distance is d i = –75 cm and the magnification is +2.5 . The negative image distance indicates that the image is virtual. The positive magnification indicates that the image is larger than the object. b. From the thin-lens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ], we obtain 111 1 1 =– = – di f d o 50.0 m 30.0 cm or d i = –75.0 cm The magnification equation (Equation 26.7) gives the magnification to be d –75.0 cm m=– i =– = +2.50 do 30.0 cm _______________________________________________...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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