Chapter 23 - 23.1 Model Light rays travel in straight lines Solve(a The time is t= x 1.0 m = = 3.3 109 s = 3.3 ns c 3 108 m/s(b The refractive

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23.1. Model: Light rays travel in straight lines. Solve: (a) The time is 9 8 1.0 m 3.3 10 s 3.3 ns 31 0 m / s x t c Δ == = × (b) The refractive indices for water, glass, and cubic zirconia are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, light will travel the following distance in water: () 8 9 water water water / s 3.33 10 s 0.75 m 1.33 c xv t t n ⎛⎞ × Δ= = = × = ⎜⎟ ⎝⎠ Likewise, the distances traveled in the glass and cubic zirconia are glass 0.667 m x and cubic zirconia 0.458 m. x Assess: The higher the refractive index of a medium, the slower the speed of light and hence smaller the distance it travels in that medium in a given time.
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23.2. Model: Light rays travel in straight lines. Solve: Let t glass , t oil , and t plastic be the times light takes to pass through the layers of glass, oil, and plastic. The time for glass is ( ) () 2 glass glass 8 glass glass 1.0 10 m 1.50 0.050 ns 3.0 10 m/s xn xx t vc n c × Δ ΔΔ == = = = × Likewise, oil 0.243 ns t = and plastic 0.106 ns. t = Thus, t total = t glass + t oil + t plastic = 0.050 ns + 0.243 ns + 0.106 ns = 0.399 ns = 0.40 ns. Assess: The small time is due to the high value for the speed of light.
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23.3. Model: Light rays travel in straight lines. The light source is a point source. Visualize: Solve: Let w be the width of the aperture. Then from the geometry of the figure, 12.0 cm 2.0 m 2.0 m 1.0 m w = + w = 8.0 cm
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23.4. Model: Light rays travel in straight lines. Also, the red and green light bulbs are point sources. Visualize: Solve: The width of the aperture is w = 1 m . From the geometry of the figure for red light, 2 1 m 3 m 1 m wx = + ( ) 2 2 1.0 m 2.0 m xw == = The red light illuminates the wall from x = 0.50 m to x = 4.50 m. For the green light, 1 4 1 m 3 m 1 m = + 1 1.0 m x = 2 34 1 m 3 m 1 m = + 2 3.0 m x = Because the back wall exists only for 2.75 m to the left of the green light source, the green light has a range from x = 0 m to 3.75 m. x =
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23.5. Visualize: Note the similar triangles in this figure. Solve: 15 cm 5 cm 180 cm d = 15 cm (180 cm) 540 cm 5.4 m 5 cm d == = Assess: This is a typical distance for photographs of people.
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23.6. Model: Use the ray model of light. Visualize: According to the law of reflection, ri . θ = Solve: From the geometry of the diagram, i 90 θφ += ° ( ) r 60 90 + °− = ° Using the law of reflection, we get ( ) 90 90 60 φ °− = 30 = ° Assess: The above result leads to a general result for plane mirrors: If a plane mirror rotates by an angle relative to the horizontal, the reflected ray makes an angle of 2 with the horizontal.
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23.7. Model: Light rays travel in straight lines and follow the law of reflection. Visualize: Solve: We are asked to obtain the distance h = x 1 + 5.0 cm. From the geometry of the diagram, 1 i tan 10 cm x θ = 2 r tan 15 cm x = 12 10 cm xx += Because ri = , we have 1 10 cm 10 cm 15 cm 15 cm x == ⇒ ( ) 2 11 15 cm 100 cm 10 x x =− x 1 = 4.0 cm Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror.
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23.8. Model: Think of the view in the figure as a horizontal view of a vertical wall and the laser beam and hexagonal mirror in a vertical plane for ease of labeling. The laser beam will strike the highest spot on the wall when a new corner rotates into the laser beam and the angle the laser makes with the normal is greatest. We will
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This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 23 - 23.1 Model Light rays travel in straight lines Solve(a The time is t= x 1.0 m = = 3.3 109 s = 3.3 ns c 3 108 m/s(b The refractive

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