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.
Three point-like charges are placed as shown in the attach image, where r1 = r2 = 44.0 cm. Find the magnitude of the electric force exerted on the charge q3. Let q1 = -1.90 uC, q2 = -2.60 uC, and q3 = +3.60 uC. Thank you.
The drawing attached shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.90 m², while Surface (2) has an area of 3.90 m². The electric field in magnitude of 215 N/C. Find the magnitude of the electric flux through surface (1 and 2 combined) if the angle theta made between the electric field with surface (2) is 30.0 degrees. Thank you.
A car driving at 27m/s veers to the left to avoid a deer in the road. The maneuver takes 2.0s and the direction of travel is altered by 20 degrees. What is the average acceleration during the constant speed maneuver? Do this in accordance with the example in the chapter.
Campbell Essential Biology with Physiology (5th Edition)
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