# The laser beam will strike the highest spot on the

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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 compute how high on the wall this highest spot is from the center spot behind the laser; then we will multiply by two because symmetry says the reflected beam will hit the lowest spot just as the face rotates out of the laser beam (and the beam makes the largest angle with the normal in the downward direction), and then a new corner enters the beam with the reflection at the top again. Visualize: From the small right triangle inside the hexagon we deduce 0.20 m/sin60 . d = ° Therefore, the distance from the wall to the corner of the hexagon just as it enters the laser beam is 2.0 m . d This becomes the base of a large right triangle whose side on the wall is x and whose angle opposite x is 60 . ° Solve: Solve the large right triangle for x . tan60 2.0 m x d ° = 0.20 m (tan60 )(2.0 m ) (tan60 ) 2.0 m 3.06 m sin60 x d = ° = ° = ° Now, because of symmetry, double x to get the total length of the streak of laser light: 2 6.1 m. x = Assess: The 50 cm distance from the laser to the center of the hexagon is irrelevant.

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23.9. Model: Light rays travel in straight lines and follow the law of reflection. Visualize: To determine the angle , φ we must know the point P on the mirror where the ray is incident. P is a distance x 2 from the far wall and a horizontal distance x 1 from the laser source. The ray from the source must strike P so that the angle of incidence θ i is equal to the angle of reflection θ r . Solve: From the geometry of the diagram, 2 1 1.5 m 3 m tan x x φ = = x 1 + x 2 = 5 m ( ) ( ) 2 1 1 1 1 1.5 m 3 m 1.5 m 15 m 3 m 5 m x x x x = = 1 10 m 3 x = 1 3 m 9 tan 0.90 42 10 x φ φ = = = = °
23.10. Model: Use the ray model of light. Visualize: The arrow is at a distance s from the mirror, so its image is at a distance s behind the mirror. When you are at x = 0 m, a ray from the arrow’s head, after reflection from the mirror, is able to enter your eye. Similarly, a ray from the arrow’s tail, after normal incidence, is reflected into the eye. That is, the eye is able to see the arrow’s head and tail. While walking toward the right, a ray from the arrow’s head will reflect from the mirror’s right edge and enter your eye at P. A ray starting from the arrow’s tail will also enter your eye when you are at P. That is, while at P you will be able to see the entire image of the arrow. However, the light from the arrow’s head can never reach beyond point P. Solve: Point P is a distance x from the origin. From the geometry of the diagram, 1 m 1 m 2 m 2 m 1 2 m tan 3.5 m 2 m 1 m 3 m 2 3 m x x x x s s φ = = = = = = + Thus, the range of x over which you can see the entire arrow in the mirror is 0 m to 3.5 m.

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23.11.
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