RCW8%20solutions

RCW8%20solutions - Recitation ClassWork 8 solutions Physics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Recitation ClassWork 8 solutions Physics 204, morning summer 2008 Chapter 25 Conceptual Questions #7 Concepts & Calculations #47 Chapter 26 Conceptual Questions #7,8,14 Concepts & Calculations #119 7. REASONING AND SOLUTION a. The back side of a shiny teaspoon acts like a convex mirror. When an object is placed in front of a convex mirror, a virtual image is produced that is reduced in size and upright, as shown in Figure 25.22. Thus, when you look at the back side of a shiny teaspoon held at arms length, you will see yourself upright. b. The concave side of a shiny teaspoon acts like a concave mirror. If the teaspoon is held at arm's length, then you (the object) are farther from the reflecting surface than the center of curvature of the surface. The situation is similar to that of an object located beyond the center of curvature of a concave mirror. A real image is formed that is reduced in size and inverted with respect to the object, as shown in Figure 25.19b. Therefore, when you look at the concave side of a shiny teaspoon, you will see yourself upside down. 47. CONCEPT QUESTIONS a. You will need a concave mirror. Since you must measure the image distance and image height of the tree, the image must be a real image. Only a concave mirror can produce a real image. b. The image distance of the sun is equal to the focal length of the mirror. The sun is extremely far away, so the light rays from it are nearly parallel when they reach the mirror. According to the discussion in Section 25.5, light rays near and parallel to the principal axis are reflected from a concave mirror and converge at the focal point. c. With numerical values for the focal length f of the mirror and the image distance di, the mirror equation (Equation 25.3) can be used to find the object distance do: 1 1 1 = − do f di The height ho of the tree is related to the height hi of its image by the magnification m; ho = hi/m. However, the magnification is given by the magnification equation (Equation 25.4) as m = −di/do, where di is the image distance and do is the object distance, both of which we know. SOLUTION a. The distance to the tree is given by the mirror equation as 1 1 1 1 1 = − = − d o f di 0.9000 m 0.9100 m b. Since ho = hi/m and m = −di/do, we have that so d o = 82 m ho = hi m = hi ⎛ di ⎜− d ⎝ o ⎛ d = hi ⎜ − o ⎞ ⎝ di ⎟ ⎠ ⎞ ⎟ ⎠ Now hi = −0.12 m, where the minus sign has been used since the image is inverted relative to the tree. Thus, the height of the tree is ⎛ d ho = hi ⎜ − o ⎝ di ⎞ 82 m ⎞ ⎛ ⎟ = 11 m ⎟ = ( −0.12 m ) ⎜ − ⎝ 0.9100 m ⎠ ⎠ 7. REASONING AND SOLUTION a. The man is using a bow and arrow to shoot a fish. The light from the fish is refracted away from the normal when it enters the air; therefore, the apparent depth of the image of the fish is less than the actual depth of the fish. When the arrow enters the water, it will continue along the same straight line path from the bow. Therefore, in order to strike the fish, the man must aim below the image of the fish. The situation is similar to that shown in Figure 26.5a; we can imagine replacing the boat by a dock and the chest by a fish. b. Now the man is using a laser gun to shoot the fish. When the laser beam entersthe water it will be refracted. From the principle of reversibility, we know that if the laser beam travels along one of the rays of light emerging from the water that originates on the fish, it will follow exactly the same path in the water as that of the ray that originates on the fish. Therefore, in order to hit the fish, the man must aim directly at the image of the fish. 8. REASONING AND SOLUTION Two rays of light converge to a point on a screen, as shown below. point of convergence screen A plane‐parallel plate of glass is placed in the path of this converging light, and the glass plate is parallel to the screen, as shown below. As discussed in the text, when a ray of light passes through a pane of glass that has parallel surfaces, and is surrounded by air, the emergent ray is parallel to the incident ray, but is laterally displaced from it. The extent of the displacement depends on the angle of incidence, on the thickness, and on the refractive index of the glass. glass plate point of convergence screen As shown in the scale drawing above, the point of convergence does not remain on the screen. It will move away from the glass as shown. 14. SSM REASONING AND SOLUTION A person is floating on an air mattress in the middle of a swimming pool. His friend is sitting on the side of the pool. The person on the air mattress claims that there is a light shining up from the bottom of the pool directly beneath him. His friend insists, however, that she cannot see any light from where she sits on the side. Rays from a light source on the bottom of the pool will radiate outward from the source in all directions. However, only rays for which the angle of incidence is less than the critical angle will emerge from the water. Rays with an angle of incidence equal to, or greater than, the critical angle will undergo total internal reflection back into the water, as shown in the following figure. illuminated circle on water's surface air mattress θc pool-side observer θc light source Because of the geometry, the rays that leave the water lie within a cone whose apex lies at the light source. Thus, rays of light that leave the water emerge from within an illuminated circle just above the source. If the mattress is just over the source, it could cover the area through which the light would emerge. A person sitting on the side of the pool would not see any light emerging. Therefore, the statements made by both individuals are correct. 119. CONCEPT QUESTION When the light ray passes from a into b, it is bent toward the normal. According to the discussion in Section 26.2, this happens when the index of refraction of b is greater than that of a, or nb > na . When the light passes from b into c, it is bent away from the normal. This means that the index of refraction of c is less than that of b, or nc < nb . The smaller the value of nc, the greater is the angle of refraction. As can be seen from the drawing, the angle of refraction in material c is greater than the angle of incidence at the a‐b interface. Applying Snell’s law to the a‐b and b‐c interfaces gives na sin a = nb sin b = nc sin c. Since c is greater than a, the equation na sin a = nc sin c shows that the index of refraction of a must be greater than that of c, na > nc . Thus, theordering of the indices of refraction, highest to lowest, is nb, na, nc. SOLUTION The index of refraction for each medium can be evaluated from Snell’s law, Equation 26.2: a‐b interface nb = b‐c interface nc = na sin θ a sin θ b nb sin θ b sin θ c = (1.20 ) sin 50.0° = 1.30 = (1.30 ) sin 45.0° = 1.10 sin 45.0° sin 56.7° As expected, the ranking of the indices of refraction, highest to lowest, is nb = 1.30, na = 1.20, nc = 1.10 ...
View Full Document

Ask a homework question - tutors are online