The equation 1/ p 1/ i = 2/ r for spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle α with that axis. First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y = ( x + p – r ) tan α and x 2 + y 2 + r 2 , where p is the object distance and r is the mirror’s radius of curvature. Next, use tan β = y / x to find the angle β at the point of intersection, and then use α + γ = 2 β to find the value of γ . Finally, use the relation tan γ = y /( x + i – r ) to find the distance i of the image. (a) Suppose r = 12 cm and p = 20 cm. For each of the following values of α , find the position of the image — that is, the position of the point where the reflected ray crosses the central axis: 0.500, 0.100, 0.0100 rad. Compare the results with those obtained with the equation 1/ p + 1/ i = 2/ r . (b) Repeat the calculations for p = 4.00 cm.
The equation 1/ p 1/ i = 2/ r for spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle α with that axis. First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y = ( x + p – r ) tan α and x 2 + y 2 + r 2 , where p is the object distance and r is the mirror’s radius of curvature. Next, use tan β = y / x to find the angle β at the point of intersection, and then use α + γ = 2 β to find the value of γ . Finally, use the relation tan γ = y /( x + i – r ) to find the distance i of the image. (a) Suppose r = 12 cm and p = 20 cm. For each of the following values of α , find the position of the image — that is, the position of the point where the reflected ray crosses the central axis: 0.500, 0.100, 0.0100 rad. Compare the results with those obtained with the equation 1/ p + 1/ i = 2/ r . (b) Repeat the calculations for p = 4.00 cm.
The equation 1/p 1/i = 2/r for spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle α with that axis.
First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y = (x + p – r) tan α and x2 + y2 + r2, where p is the object distance and r is the mirror’s radius of curvature. Next, use tan β = y/x to find the angle β at the point of intersection, and then use α + γ = 2 β to find the value of γ. Finally, use the relation tan γ = y/(x + i – r) to find the distance i of the image.
(a) Suppose r = 12 cm and p = 20 cm. For each of the following values of α, find the position of the image — that is, the position of the point where the reflected ray crosses the central axis: 0.500, 0.100, 0.0100 rad. Compare the results with those obtained with the equation 1/p + 1/i = 2/r. (b) Repeat the calculations for p = 4.00 cm.
air is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cm
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