Physics Solution Manual for 1100 and 2101

00 sin 503 na 130 c sin 1 layer a top

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Unformatted text preview: equations by eliminating the speed v yields n = c/(f λ). SOLUTION The index of refraction of the substance is n= 2.998 × 108 m/s c = = 1.632 fλ 5.403 × 1014 Hz 340.0 × 10 −9 m ( )( ) An examination of Table 26.1 shows that the substance is carbon disulfide . ______________________________________________________________________________ 7. REASONING The refractive index n is defined by Equation 26.1 as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in a material medium. The speed in a vacuum or in the liquid is the distance traveled divided by the time of travel. Thus, in the definition of the refractive index, we can express the speeds c and v in terms of the distances and the time. This will allow us to calculate the refractive index. SOLUTION According to Equation 26.1, the refractive index is n= c v Using dvacuum and dliquid to represent the distances traveled in a time t, we find the speeds to be dliquid d c = vacuum and v = t t Substituting these expressions into the definition of the refractive index shows that n= 8. c d vacuum / t d vacuum 6.20 km = = = = 1.82 v dliquid / t dliquid 3.40 km REASONING Distance traveled is the speed times the travel time. Assuming that t is the time it takes for the light to travel through the two sheets, it would travel a distance of ct in a vacuum, where its speed is c. Thus, to find the desired distance, we need to determine the travel time t. This time is the sum of the travel times in each sheet. The travel time in each sheet is determined by the thickness of the sheet and the speed of the light in the material. The speed in the material is less than the speed in a vacuum and depends on the refractive index of the material. 1354 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION In the ice of thickness di, the speed of light is vi, and the travel time is ti = di/vi. Similarly, the travel time in the quartz sheet is tq = dq/vq. Therefore, the desired distance ct is d dq c c ct = c ti + tq = c i + = di + d q vi vq vi vq ( ) Since Equation 26.1 gives the refractive index as n = c/v and since Table 26.1 gives the indices of refraction for ice and quartz as ni = 1.309 and nq = 1.544, the result just obtained can be written as follows: ct = di c c + dq = di ni + d q nq = ( 2.0 cm )(1.309 ) + (1.1 cm )(1.544 ) = 4.3 cm vi vq ______________________________________________________________________________ 9. SSM REASONING AND SOLUTION a. We know from the law of reflection (Section 25.2), that the angle of reflection is equal to the angle of incidence, so the reflected ray is reflected at 43° . b. Snell’s law of refraction (Equation 26.2: n1 sin θ1 = n2 sin θ 2 can be used to find the angle of refraction. Table 26.1 indicates that the index of refraction of water is 1.333. Solving for θ2 and substituting values, we find that n1 sin θ1 (1.000) (sin 43°) = = 0.51 or θ 2 = sin –1 0.51 = 31° n2 1.333 ______________________________________________________________________________ sin θ 2 = 10. REASONING a. The refracted ray is shown correctly. When light goes from a medium of lower index of refraction (n = 1.4) to one of higher index of refraction (n = 1.6), the refracted ray is bent toward the normal, as it does in part (a). b. The refracted ray is shown incorrectly. When light goes from a medium of lower index of refraction (n = 1.5) to one of higher index of refraction (n = 1.6), the refracted ray must bend toward the normal, not away from it, as part (b) of the drawing shows. c. The refracted ray is shown correctly. When light goes from a medium of higher index of refraction (n = 1.6) to one of lower index of refraction (n = 1.4), the refracted ray bends away from the normal, as it does part (c) of the drawing. d. The refracted ray is shown incorrectly. When the angle of incidence is 0°, the angle of refraction is also 0°, regardless of the indices of refraction. Chapter 26 Problems 1355 SOLUTION a. The angle of refraction θ2 is given by Snell’s law, Equation 26.2, as n1 sin θ1 −1 (1.4 ) sin 55° = sin = 46° 1.6 n2 θ 2 = sin −1 b. The actual angle of refraction is n1 sin θ1 −1 (1.5 ) sin 55° = sin = 50° 1.6 n2 θ 2 = sin −1 c. The angle of refraction is n1 sin θ1 −1 (1.6 ) sin 55° = sin = 69° n2 1.4 θ 2 = sin −1 d. The actual angle of refraction is n1 sin θ1 −1 (1.6 ) sin 0° = sin = 0° n2 1.4 ______________________________________________________________________________ θ 2 = sin −1 11. REASONING The angle of refraction θ2 is related to the angle of incidence θ1 by Snell’s law, n1 sin θ1 = n2 sin θ 2 (Equation 26.2), where n1 and n2 are, respectively the indices of refraction of the incident and refracting media. For each case (ice and water), the variables θ1, n1 and n2, are known, so the angles of refraction can be determined. SOLUTION The ray of light impinges from air (n1 = 1.000) onto either the ice or water at an angle of incidence of θ1 = 60.0°. Using n2 = 1.309 for ice and n2 =...
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