Kapittel 23
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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,
Because , we have
x1 = 4.0 cm
Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror.
23.12. Model: Use the ray model of light and Snell’s law.
Visualize:
Solve: According to Snell’s law for the air-water and water-glass boundaries,
From these two equations, we have
23.13. Visualize: Use Snell’s law We are given We look up in Table 23.1 and
Solve: Solve the equation for
Assess: Since the ray goes into a material with higher index of refraction we know it bends toward the normal, so we expect this is the case.
23.15. Model: Represent the laser beam with a single ray and use the ray model of light.
Solve: Using Snell’s law at the air-water boundary,
Assess: As expected, nliquid is larger than nair.
23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection.
Visualize:
Solve: The critical angle of incidence is given by Equation 23.9:
Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90° - 67.7° = 23.3°.
Assess: We can have total internal reflection because ncore ncladding.
23.17. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection.
Visualize:
Solve: The critical angle of incidence is given by Equation 23.9:
Assess: The critical angle exists because noil nglass.
23.48. Model: Use the ray model of light and the law of refraction.
Visualize:
Solve: (a) The ray of light strikes the meter stick at which is a distance L from the zero mark of the meter stick. So,
(b) The ray of light refracts at Phalf and strikes the meter stick a distance from the zero of the meter stick. We can find x1 from the triangle PfullPhalfO¢:
We also have Using Snell’s law,
(c) The ray of light experiences refraction at Pfull and the angle of refraction is the same as in part (b). We get
23.50. Model: Use the ray model of light and the law of refraction. Assume the sun is a point source of light.
Visualize:
When the bottom of the pool becomes completely shaded, a ray of light that is incident at the top edge of the swimming pool does not reach the bottom of the pool after refraction.
Solve: The depth of the swimming pool is We will find the angle by using Snell’s law. We have
23.58. Model: Use the ray model of light.
Solve: (a) Using Snell’s law at the air-glass boundary, with f being the angle of refraction inside the prism,
Considering the triangle made by the apex angle and the refracted ray,
Thus
(b) Using the above expression, we obtain
23.61. Model: Assume that the converging lens is a thin lens. Use ray tracing to locate the image.
Solve: (a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three rays after refraction converge to give an image at s¢ = 40 cm. The height of the image is h¢ = 2 cm.
(b) Using the thin-lens formula,
The image height is obtained from
The image is inverted and as tall as the object, that is, h¢ = 2.0 cm. The values for h¢ and s¢ obtained in parts (a) and (b) agree.
23.62. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens.
Solve: (a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays that experience refraction do not converge at a point. Instead they appear to come from a point that is 15 cm on the same side as the object itself. Thus s¢ = -15 cm. The image is upright and has a height of h¢ = 1.5 cm.
(b) Using the thin-lens formula,
The image height is obtained from
The image is upright and 1.5 times the object, that is, 1.5 cm high. These values agree with those obtained in part (a).
23.63. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens.
Solve: (a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays after refracting do not converge. Instead the rays appear to come from a point that is 60 cm on the same side of the lens as the object, so The image is upright and has a height of 8.0 cm.
(b) Using the thin-lens formula,
The image height is obtained from
Thus, the image is 4 times larger than the object or The image is upright. These values agree with those obtained in part (a).
23.65. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens.
Solve: (a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction, the three special rays do not converge. The rays, on the other hand, appear to meet at a point that is 8.5 cm on the same side of the lens as the object. So The image is upright and has a height of 1.1 cm.
(b) Using the thin-lens formula,
The image height is obtained from
Thus, the image is 0.57 times larger than the object, or The image is upright because M is positive. These values agree, within measurement accuracy, with those obtained in part (a).
23.66. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens.
Solve: (a)
The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction from the diverging lens, the three special rays do not converge. However, the rays appear to meet at a point that is 20 cm on the same side as the object. So The image is upright and has a height of 0.3 cm.
(b) Using the thin-lens formula,
The image height is obtained from
Thus, and the image is upright because M is positive. These values for s¢ and h¢ agree with those obtained in part (a).