23.3 (1) The first image in the left-hand mirror is 5.00 ft behind the mirror, or 10.0 ftfrom the person

Size: px

Start display at page:

Download "23.3 (1) The first image in the left-hand mirror is 5.00 ft behind the mirror, or 10.0 ftfrom the person"

Meredith Gibbs
5 years ago
Views:

1 irrors and Lenses Answers to Even Numbered Concetual Questions Chater (d). The entire image would aear because any ortion 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 Solutions 23.3 () The first image in the left-hand mirror is 5.00 ft behind the mirror, or 0.0 ftfrom the erson (2) The first image in the right-hand mirror serves as an object for the left-hand mirror. It is located 0.0 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 ftfrom the erson (3) The first 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 ftfrom the erson 23.5 Since the mirror is convex, R < 0. Thus, R = m. With a real object, > 0, so =+ 0.0 m. The mirror euation then gives the image distance as 2 2 = = R m 0.0 m, or = m Thus, the image is virtual and located m behind the m irror The magnification is m = = = 0.0 m Therefore, the image is uright ( since 0) > and dim inished in size ( since < ) 267

2 268 CHAPTER The radius of curvature of a concave mirror is ositive, so R = cm euation then gives. The mirror cm = = =, or R 0.0 cm 0.0 cm ( ) ( 0.0 cm ) = 0.0 cm (a) If = 40.0 cm, =+ 3.3 cm and 3.3 cm = = = cm The image is 3.3 cm in frontofthe m irror,real,and inverted (b) When = 20.0 cm, = cm and 20.0 cm = = = cm (c) If The image is 20.0 cm in frontofthe m irror,real,and inverted = 0.0 cm, ( 0.0 cm )( 0.0 cm ) = 0.0 cm 0.0 cm and no im age is form ed. Parallelrays leave the m irror 23.3 The image is uright, so > 0, and we have = =+ 2.0, or = 2.0= ( cm ) = 50 cm The radius of curvature is then found to be 2 2 = + = =, or R 25 cm 50 cm 50 cm 0.50 m R = 2 =.0 m A convex mirror ( R < 0) roduces uright, virtual images of real objects. Thus, > 0 giving = =+, or 3 = 3 Then, 2 + = R becomes 3 = 2, and yields =+ 0.0 cm 0.0 cm The object is 0.0 cm in frontofthe m irror

3 irrors and Lenses 269 n n2 n2 n 23.2 From + =, with R, the image osition is found to be R n n 2 = = =.00 ( 50.0 cm ) 38.2 cm.309 or the virtual image is 38.2 cm below the uer surface ofthe ice From the thin lens euation, + =, the image distance is found to be f ( 20.0 cm ) f = = f 20.0 cm (a) If = 40.0 cm, then = 40.0 cm and 40.0 cm = = = cm The image is real,inverted,and 40.0 cm beyond the lens (b) If = 20.0 cm, No im age form ed. Parallelrays leave the lens. (c) When = 0.0 cm, = 20.0 cm and ( 20.0 cm ) = = = cm The image is virtual,uright,and 20.0 cm in frontofthe lens

4 270 CHAPTER (a) The real image case is shown in the ray diagram. Notice that + = 2.9 cm, or = 2.9 cm. The thin lens euation, with f= 2.44 cm, then gives + = 2.9 cm 2.44 cm or ( ) 2.9 cm cm = Object f = 2.44 cm F F Image + = 2.9 cm Using the uadratic formula to solve gives = 9.63 cm or = 3.27 cm Both are valid solutions for the real image case. (b) The virtual image case is shown in the second = 2.9 cm +, diagram. Note that in this case, ( ) so the thin lens euation gives = 2.9 cm cm or ( ) cm 3.5 cm = Virtual Image f = 2.44 cm Object 2.9 cm The uadratic formula then gives = 2.0 cm or = 5.0 cm Since the object is real, the negative solution must be rejected leaving = 2.0 cm It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magnification then gives h.80 m = = =, or = 75.0 h m Also, we know that + = 3.00 m. Therefore, = 3.00 m giving (b) 3.00 m = = = m 39.5 mm

5 irrors and Lenses 27 (a) The thin lens euation then gives + = 76.0 = f = = 39.5 m m = 39.0 mm or f ( ) Since the light rays incident to the first lens are arallel, = and the thin lens euation gives = f= 0.0 cm. The virtual image formed by the first lens serves as the object for the second lens, so = 30.0 cm + = 40.0 cm. If the light rays leaving the second lens are arallel, then 2 = and the thin lens euation gives f2 = 2 = 40.0 cm Consider an object O at distance in front of the first lens. The thin lens euation gives the image osition for this lens as =. f O f f 2 2 = I 2 I = O 2 2 The image, I, formed by the first lens serves as the object, O 2, for the second lens. With the lenses in contact, this will be a virtual object if I is real and will be a real object if I is virtual. In either case, if the thicknesses of the lenses may be ignored, = and 2 = = + f 2 Alying the thin lens euation to the second lens, + = becomes f = or f f = + f f 2 2 Observe that this result is a thin lens tye euation relating the osition of the original object O and the osition of the final image I 2 formed by this two lens combination. Thus, we see that we may treat two thin lenses in contact as a single lens having a focal length, f, given by = + f f f 2

6 272 CHAPTER 23

( ) = + ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

( ) = + ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS Mirrors and Lenses 39 7. A concave mirror forms inverted, real images of real objects located outside the focal oint ( > f ), and uright, magnified, virtual images of real objects located inside the focal