Physics Solution Manual for 1100 and 2101

The intensity i of the light wave is the power p that

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Unformatted text preview: nce is 90.0° – 40.0° = 50.0°. The index of refraction for material c is nc = na sin θ c = (1.80 ) sin 50.0° = 1.38 Chapter 26 Problems 1367 As expected, the ranking of the indices of refraction, highest-to-lowest, is na = 1.80, nc = 1.38, nb = 1.16. ______________________________________________________________________________ 31. REASONING AND SOLUTION a. Using Equation 26.4 and the refractive index for crown glass given in Table 26.1, we find that the critical angle for a crown glass-air interface is 1.00 = 41.0° 1.523 θ c = sin −1 The light will be totally reflected at point A since the incident angle of 60.0° is greater than θc. The incident angle at point B, however, is 30.0° and smaller than θc. Thus, the light will exit first at point B . b. The critical angle for a crown glass-water interface is 1.333 = 61.1° 1.523 θ c = sin −1 The incident angle at point A is less than this, so the light will first exit at point A . ______________________________________________________________________________ 32. REASONING Total internal reflection can occur only when light is traveling from a higher index material toward a lower index material. Thus, total internal reflection is possible when the material above or below a layer has a smaller index of refraction than the layer itself. With this criterion in mind, the following table indicates the various possibilities: Is total internal reflection possible? Layer Top surface of layer Bottom surface of layer a Yes No b Yes Yes c No Yes 1368 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION The critical angle for each interface at which total internal reflection is possible is obtained from Equation 26.4: nair −1 1.00 = sin = 50.3° na 1.30 θ c = sin −1 Layer a, top surface ∆θ = 75.0° − 50.3° = 24.7° na −1 1.30 = sin = 60.1° 1.50 nb Layer b, top surface θ c = sin −1 ∆θ = 75.0° − 60.1° = 14.9° nc −1 1.40 = sin = 69.0° nb 1.50 Layer b, bottom surface θ c = sin −1 ∆θ = 75.0° − 69.0° = 6.0° nair −1 1.00 = sin = 45.6° nc 1.40 Layer c, bottom surface θ c = sin −1 ∆θ = 75.0° − 45.6° = 29.4° ______________________________________________________________________________ 33. REASONING The time it takes for the light to travel from A to B is equal to the distance divided by the speed of light in the substance. The distance is known, and the speed of light v in the substance is equal to the speed of light c in a vacuum divided by the index of refraction n1 (Equation 26.1). The index of refraction can be obtained by noting that the light is incident at the critical angle θc (which is known). According to Equation 26.4, the index of refraction n1 is related to the n2 = 1.63 n1 d θc B A critical angle and the index of refraction n2 by n1 = n2 / sin θc . SOLUTION The time t it takes for the light to travel from A to B is t= Distance d = Speed of light v in the substance (1) Chapter 26 Problems 1369 The speed of light v in the substance is related to the speed of light c in a vacuum and the index of refraction n1 of the substance by v = c/n1 (Equation 26.1). Substituting this expression into Equation (1) gives t= dn d d = =1 v c c n 1 (2) Since the light is incident at the critical angle θc, we know that n1 sin θ c = n2 (Equation 26.4). Solving this expression for n1 and substituting the result into Equation (2) yields n d 2 ( 4.60 m ) 1.63 sin θc dn sin 48.1° = 3.36 ×10−8 s t= 1= = c c 3.00 × 108 m/s ______________________________________________________________________________ 34. REASONING In the ratio nB/nC each refractive index can be related to a critical angle for total internal reflection according to Equation 26.4. By applying this expression to the A-B interface and again to the A-C interface, we will obtain expressions for nB and nC in terms of the given critical angles. By substituting these expressions into the ratio, we will be able to obtain a result from which the ratio can be calculated SOLUTION Applying Equation 26.4 to the A-B interface, we obtain sin θ c, AB = nB nA or nB = nA sin θ c, AB Applying Equation 26.4 to the A-C interface gives sin θ c, AC = nC nA or nC = nA sin θ c, AC With these two results, the desired ratio can now be calculated: nB nA sin θ c, AB sin 36.5° = = = 0.813 nC nA sin θ c, AC sin 47.0° ________________________________________________________________________ 1370 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 35. SSM REASONING Total internal reflection will occur at point P provided that the angle α in the drawing at the right exceeds the critical angle. This angle is determined by the angle θ2 at which the light rays enter the quartz slab. We can determine θ2 by using Snell’s law of refraction and the incident angle, which is given as θ1 = 34°. P α θ2 θ1 SOLUTION Using n for the refractive index of the fluid that surrounds the crystalline quartz slab and nq for the refractive index of quartz and applying Snell’s law give n sin θ1 = nq sin θ 2 sin θ 2 = or n sin θ1 nq (1) But when α eq...
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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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