Mirrors and Lenses 309 7. A concave mirror forms inverted, real images of real objects located outside the focal point ( )p f> , and upright, magnifi ed, virtual images of real objects located inside the focal point ( )p f< of the mirror. Virtual images, located behind the mirror, have negative image distances by the sign convention of Table 23.1. Choices (d) and (e) are true statements and all other choices are false. 8. With a real object in front of a convex mirror, the image is always upright, virtual, diminished in size, and located between the mirror and the focal point. Thus, of the available choices, only choice (d) is a true statement. 9. A convergent lens forms inverted, real images of real objects located outside the focal point ( ).p f> When pf> 2, the real image is diminished in size, and the image is enlarged if 2 f p f> >. For real objects located inside the focal point ( )p f< of the convergent lens, the image is upright, virtual, and enlarged. In the given case, p f> 2, so the image is real, inverted, and diminished in size. Choice (c) is the correct answer.10. For a real object ( )p > 0 and a diverging lens ( )f < 0, the image distance given by the thin lens equation is qp fp fp fp fp fp f=−=−()− −()= −+< 0 and the magnifi cation is Mqpqp= − = −−> 0 Thus, the image is always virtual and upright, meaning that choice (b) is a true statement while all other choices are false.ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Chromatic aberration is produced when light passes through a material, as it does when passing through the glass of a lens. A mirror, silvered on its front surface never has light passing through it, so this aberration cannot occur. This is only one of many reasons why large telescopes use mirrors rather than lenses for their primary optical elements. 4. All objects beneath the stream appear to be closer to the surface than they really are because of refraction. Thus, the pebbles on the bottom of the stream appear to be close to the surface of a shallow stream. 6. An effect similar to a mirage is produced except the “mirage” is seen hovering in the air. Ghost lighthouses in the sky have been seen over bodies of water by this effect. 8. Actually no physics is involved here. The design is chosen so your eyelashes will not brush against the glass as you blink. A reason involving a little physics is that with this design, when you direct your gaze near the outer circumference of the lens you receive a ray that has passed through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally distorts the direction to the object you are looking at.10. Both words are inverted. However, OXIDE looks the same right side up and upside down. LEAD does not.56157_23_ch23_p305-334.indd 30956157_23_ch23_p305-334.indd 309 3/20/08 1:27:54 AM3/20/08 1:27:54 AM310 Chapter 2312. (a) No. The screen is needed to refl ect the light toward your eye. (b) Yes. The light is traveling toward your eye and diverging away from the position of the image, the same as if the object was located at that position.14. (d). The entire image would appear because any portion of the lens can form the image. The image would be dimmer because the card reduces the light intensity on the screen by 50%.PROBLEM SOLUTIONS23.1 If you stand 40 cm in front of the mirror, the time required for light scattered from your face to travel to the mirror and back to your eye is ∆tdc= =()×= ×−22 0 403 0 102 7 1089... m m s s Thus, the image you observe shows you ~109− s younger than your current age.23.2 (a) With the palm located 1.0 m in front of the nearest mirror, that mirror forms an image, IP1,of the palm located 1.0 m behind the nearest mirror. (b) The farthest mirror forms an image, IB1, of the back of the hand located 2.0 m behind this mirror and 5.0 m in front of the nearest mirror. This image serves and an object for the nearest mirror, which then forms an image, IB2, of the back of the hand located 5.0 m behind the nearest mirror. (c) The image IP1 (see part a) serves as an object located 4.0 m in front of the farthest mirror, which forms an image IP2 of the palm, located 4.0 m behind that mirror and 7.0 m in front of the nearest mirror. This image then serves as an object for the nearest mirror, which forms an image IP 3 of the palm, located 7.0 m behind the nearest mirror. (d) Since all images are located behind the mirror, all are virtual images.23.3 (1) The fi rst image in the left-hand mirror is 5.00 ft behind the mirror,or 10.0 ft from the person. (2) The fi rst image in the right-hand mirror serves as an object for the left-hand mirror. It is located 10.0 ft behind the right-hand mirror, which is 25.0 ft from the left-hand mirror. Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror,or 30.0 ft from the person. (3) The fi rst image in the left-hand mirror serves as an object for the right-hand mirror. It is located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that mirror. This image then serves as an object for the left-hand mirror. The distance from this object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is 35.0 ft behind the mirror, or 40.0 ft from the person.56157_23_ch23_p305-334.indd 31056157_23_ch23_p305-334.indd 310 3/20/08 1:27:54 AM3/20/08 1:27:54 AMMirrors and Lenses 31123.4 The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror. The image of the choir is 0 800 5 30 6 10. . . m m m+ = from the organist. Using similar triangles gives ′=h0 6006 100 800... m m m or ′=()=h 0 6006 100 8004 58.... m m m m 23.5 In the fi gure at the right, ′=θ θ since they are vertical angles formed by two intersecting straight lines. Their complementary angles are also equal or ′=α α. The right trianglesPQRand ′P QR have the common side QR and are then congruent by the angle-side-angle theorem. Thus, the corresponding sides PQ and ′P Q are equal, or the image is as far behind the mirror as the object is in front of it.23.6 (a) Since the object is in front of the mirror, p > 0. With the image behind the mirror, q < 0. The mirror equation gives the radius of curvature as 2 1 1 11 00110 0 10 0R p q= + = − =−. . . cm cm10 1 cm or R == +210 02