HW01solutions - 23.4. Model: Light rays travel in straight...

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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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23.6. 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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23.14. Model: Use the ray model of light. The sun is a point source of light. Visualize: A ray that arrives at the diver 50 ° above horizontal refracted into the water at θ water = 40 ° . Solve: Using Snell’s law at the water-air boundary air air water water sin sin nn θθ = water air water air 1.33 sin sin sin 40 1.0 n n ⎛⎞ = ⎜⎟ ⎝⎠
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HW01solutions - 23.4. Model: Light rays travel in straight...

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