23 1 Model Light rays travel in straight lines Solve a The time is t x c 1 0 m 8 3 10 m s 9 3 3 10 s 3 3 ns 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 x water v water t c n water t 8 3 10 m s 1 33 3 33 10 s 9 0 75 m Likewise the distances traveled in the glass and cubic zirconia are x 0 667 m and x glass cubic zirconia 0 458 m 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 23 2 Model Light rays travel in straight lines Solve Let tglass toil and tplastic be the times light takes to pass through the layers of glass oil and plastic The time for glass is t glass x v glass x c n glass x n glass c 2 1 0 10 m 1 50 3 0 10 m s 8 0 050 ns t 0 243 ns t and plastic Likewise oil 0 399 ns 0 40 ns Assess The small time is due to the high value for the speed of light 0 106 ns Thus ttotal tglass toil tplastic 0 050 ns 0 243 ns 0 106 ns 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 w 2 0 m 2 0 m 12 0 cm 1 0 m w 8 0 cm 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 w 1 m 3 m x 1 m x 2 w 2 1 0 m 2 0 m The red light illuminates the wall from x 0 50 m to x 4 50 m For the green light 4 w 1 m 3 m x 1 1 m x 1 1 0 m 4 w 3 1 m 3 m x 2 1 m 2 x 3 0 m 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 23 5 Visualize Note the similar triangles in this figure Solve 15 cm 5 cm 180 cm d d 15 cm 5 cm 180 cm 540 cm 5 4 m Assess This is a typical distance for photographs of people 23 6 Model Use the ray model of light Visualize According to the law of reflection Solve From the geometry of the diagram i r Using the law of reflection we get 90 i r 60 90 90 90 60 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 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 x1 5 0 cm From the geometry of the diagram tan i tan r x 1 x 2 10 cm x 1 10 cm x 2 15 cm Because i we have r x 1 x 2 10 cm x 1 10 cm 15 cm 15 cm 15 cm x 1 100 cm 10 2 x 1 x1 4 0 cm Thus the ray strikes a distance 9 0 cm below the top edge of the mirror 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 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 distance from the wall to the corner of the hexagon just as it enters the laser beam is 2 0 m d the base of a large right triangle whose side on the wall is x and whose angle opposite x is 60 Therefore the 0 20 m sin 60 This becomes d Solve Solve the large right triangle for x tan 60 x 2 0 m d x d tan 60 2 0 m tan 60 2 0 m 3 06 m 0 20 m sin 60 Now because of symmetry double x to get the total length of the streak of laser light 2 Assess The 50 cm distance from the laser to the center of the hexagon is irrelevant x 6 1 m 23 9 Model Light rays travel in straight lines and follow the law of reflection Visualize we must know the point P on the mirror where the ray is incident P is a distance x2 To determine the angle from the far wall and a horizontal distance x1 from the laser source The ray from the source must strike P so that the angle of incidence i is equal to the angle of reflection r Solve From the geometry of the diagram tan 1 5 m 3 m x 2 x 1 x1 x2 5 m 1 5 m 5 m x 1 3 m x 1 1 5 m x 1 15 m 2 3 m x 1 x 1 m 10 3 tan 0 90 42 3 m 9 x 10 1 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 …
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