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Physics Solution Manual for 1100 and 2101

# 520 2 sin 1 diamond the angle between the blue and

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Unformatted text preview: the glass and the (unknown) angle θ2 of refraction for the light entering the slab: (1.00 ) sin 35.0° = n2 sin θ 2 . When the incident light is in the liquid, we can again use Snell’s law to express the relation between the index of refraction n1 of the liquid, the angle of incidence (20.3°), the index of refraction n2 of the glass, and the (unknown) angle of refraction θ2: n1 sin 20.3° = n2 sin θ 2 . By equating these two equations, we can determine the index of refraction of the liquid. SOLUTION Setting the two equations above equal to each other and solving for the index of refraction of the liquid gives n1 sin 20.3° = (1.00 ) sin 35.0° n1 = (1.00 ) sin 35.0° = 1.65 sin 20.3° ______________________________________________________________________________ and 18. REASONING According to Equation 2.1, when an object moves with a constant speed v d and travels a distance d in an elapsed time t, these three quantities are related by v = . In t this situation, the stone sinks a distance d from the surface to the bottom, and the elapsed time is t. When viewed from above, the stone still sinks to the bottom in an elapsed time t, but the apparent depth d ′ to which it sinks is not equal to the actual depth d. Therefore, the apparent speed v′ of the stone is not the same as its actual speed v, and we have that Chapter 26 Problems v= d t and v′ = d′ t 1359 (1) n The apparent depth d ′ is given by d ′ = d 2 (Equation 26.3), where n2 = 1.000 is the n 1 index of refraction of air and n1 = 1.333 (See Table 26.1) is the index of refraction of water. SOLUTION Solving the first of Equations (1) for t yields t = d . Substituting this v expression into the second of Equations (1), we obtain v′ = d′ d′ d′ = = v t d d v (2) n Substituting d ′ = d 2 (Equation 26.3) into Equation (2), we find that n 1 n d 2 n n d′ 1.000 v′ = v = v 1 = v 2 = ( 0.48 m/s ) = 0.36 m/s n d d 1.333 1 ______________________________________________________________________________ 19. REASONING AND SOLUTION The horizontal distance of the chest from the normal is found from Figure 26.4b to be x = d tan θ1 and x = d ' tan θ2, where θ1 is the angle from the dashed normal to the solid rays and θ2 is the angle from the dashed normal to the dashed rays. Hence, d ' = d (tan θ1/tan θ2) Snell's law applied at the interface gives n1 sin θ1 = n2 sin θ2 For small angles, sin θ1 ≈ tan θ1 and sin θ2 ≈ tan θ2, so tan θ1/ tan θ2 ≈ sin θ1/sin θ2 = (n2/n1) Now d ' = d (tan θ1/tan θ2). Therefore, n d′ ≈ d 2 n 1 1360 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________ 20. REASONING Snell’s law will allow us to calculate the angle of refraction θ2, B with which the ray leaves the glass at point B, provided that we have a value for the angle of incidence θ1, B at this point (see the drawing). This angle of incidence is not given, but we can obtain it by considering what happens to the incident ray at point A. This ray is incident at an angle θ1, A and refracted at an angle θ2, A. Snell’s law can be used to obtain θ2, A, the value for which can be combined with the geometry at points A and B to provide the needed value for θ1, B. Since the light ray travels from a material (carbon disulfide) with a higher refractive index toward a material (glass) with a lower refractive index, it is bent away from the normal at point A, as the drawing shows. θ2, A B θ2, B Glass θ1, A = 30.0º A θ1, B θ2, A Carbon disulfide SOLUTION Using Snell’s law at point B, we have (1.52 ) sin θ1, B = (1.63) sin θ 2, B 14 244 4 3 14 244 4 3 Glass or 1.52 sin θ 2, B = sin θ1, B 1.63 (1) or θ1, B = 90.0° − θ 2, A (2) Carbon disulfide To find θ1, B we note from the drawing that θ1, B + θ 2, A = 90.0° We can find θ2, A, which is the angle of refraction at point A, by again using Snell’s law: (1.63) sin θ1, A = (1.52 ) sin θ 2, A 14 244 4 3 14 244 4 3 Carbon disulfide Glass or 1.63 sin θ 2, A = sin θ1, A 1.52 Chapter 26 Problems 1361 Thus, we have 1.63 1.63 sin θ 2, A = sin θ1, A = sin 30.0° = 0.536 1.52 1.52 or θ 2, A = sin −1 ( 0.536 ) = 32.4° Using Equation (2), we find that θ1, B = 90.0° − θ 2, A = 90.0° − 32.4° = 57.6° With this value for θ1, B in Equation (1) we obtain 1.52 1.52 sin θ 2, B = sin θ1, B = sin 57.6° = 0.787 1.63 1.63 21. SSM WWW The drawing at the right shows the geometry of the situation using the same notation as that in Figure 26.6. In addition to the text's notation, let t represent the thickness of the pane, let L represent the length of the ray in the pane, let x (shown twice in the figure) equal the displacement of the ray, and let the difference in angles θ1 – θ2 be given by φ. We wish to find the amount x by which the emergent ray is displaced relative to the incident ray. This can be done by applying Snell's law at each interface, and then making use of the geometric and trigonometric relations in the drawing. θ 2, B = sin −1 ( 0.787 ) = 51.9° or x θ3 Air ( n 3 =...
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