946_PartUniversity Physics Solution

946_PartUniversity Physics Solution - 33-4 Chapter 33...

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33-4 Chapter 33 33.14. IDENTIFY: Apply the law of reflection. SET UP: The mirror in its original position and after being rotated by an angle θ are shown in Figure 33.14. α is the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled φ are equal and the two angles labeled φ are equal because of the law of reflection. The two angles labeled are equal because the lines forming one angle are perpendicular to the lines forming the other angle. EXECUTE: From the diagram, 222 ( ) φφ =−= and = . 2 = , as was to be shown. EVALUATE: This result is independent of the initial angle of incidence. Figure 33.14 33.15. IDENTIFY: Apply sin sin aa bb nn = . SET UP: The light refracts from the liquid into the glass, so 1.70 a n = , 62.0 a = ° . 1.58 b n = . EXECUTE: 1.70 sin sin sin62.0 0.950 1.58 a ba b n n θθ ⎛⎞ ⎛ ⎞ == = ⎜⎟ ⎝⎠ ° and 71.8 b = ° . EVALUATE: The ray refracts into a material of smaller n , so it is bent away from the normal. 33.16. IDENTIFY: Apply Snell’s law. SET UP: a and b are measured relative to the normal to the surface of the interface. 60.0 15.0 45.0 a °° ° . EXECUTE: 1.33 arcsin sin arcsin sin45.0 38.2 1.52 a b n n ° = ° . But this is the angle from the normal to the surface, so the angle from the vertical is an additional 15& because of the tilt of the surface. Therefore, the angle is 53.2&. EVALUATE: Compared to Example 33.1, a is shifted by 15 ° but the shift in b is only 53.2 49.3 3.9 −= ° . 33.17. IDENTIFY: The critical angle for total internal reflection is a that gives 90 b = ° in Snell’s law. SET UP: In Figure 33.17 the angle of incidence a is related to angle by 90 a + = ° . EXECUTE: (a) Calculate a that gives 90 b = ° . 1.60 a n = , 1.00 b n = so sin sin = gives (1.60)sin (1.00)sin90 a = ° . 1.00 sin 1.60 a = and 38.7 a = ° . 90 51.3 a = = °− ° . (b) 1.60 a n = , 1.333 b n = . (1.60)sin (1.333)sin90 a = ° . 1.333 sin 1.60 a = and 56.4 a = ° . 90 33.6 a ° . EVALUATE: The critical angle increases when the ratio a b n n increases. Figure 33.17
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The Nature and Propagation of Light 33-5 33.18. IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will occur at the cube surface if the angle of incidence is greater than or equal to the critical angle. SET UP: At the critical angle θ c , Snell&s law gives n glass sin c = n air sin 90± and likewise for water. EXECUTE: (a) At the critical angle c , n glass sin c = n air sin 90±. 1.53 sin c = (1.00)(1) and c = 40.8±. (b) Using the same procedure as in part (a), we have 1.53 sin c = 1.333 sin 90± and c = 60.6±. EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air. 33.19. IDENTIFY: Use the critical angle to find the index of refraction of the liquid. SET UP: Total internal reflection requires that the light be incident on the material with the larger n , in this case the liquid. Apply sin sin aa bb nn = with a = liquid and b = air, so liq a = and 1.0.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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946_PartUniversity Physics Solution - 33-4 Chapter 33...

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