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.
Statistical thermodynamics. The number of imaginary replicas of a system of N particlesa) cannot be greater than Avogadro's numberb) must always be greater than Avogadro's number.c) has no relation to Avogadro's number.
Lab-Based Section
Use the following information to answer the lab based scenario.
A student performed an experiment in an attempt to determine the index of refraction of glass.
The student used a laser and a protractor to measure a variety of angles of incidence and
refraction through a semi-circular glass prism. The design of the experiment and the student's
results are shown below.
Angle of
Incidence (°)
Angle of
Refraction (º)
20
11
30
19
40
26
50
31
60
36
70
38
2a) By hand (i.e., without using computer software), create a linear graph on graph paper
using the student's data. Note: You will have to manipulate the data in order to achieve a
linear function.
2b) Graphically determine the index of refraction of the semi-circular glass prism, rounding your
answer to the nearest hundredth.
Use the following information to answer the next two questions.
A laser is directed at a prism made of zircon (n = 1.92) at an incident angle of 35.0°, as shown in
the diagram.
3a) Determine the critical angle of zircon.
35.0°
70°
55
55°
3b) Determine the angle of refraction when the laser beam leaves the prism.
Campbell Essential Biology with Physiology (5th Edition)
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