Physics Solution Manual for 1100 and 2101

8 refractive power e 1 fe 1 to obtain the focal length

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Unformatted text preview: is a virtual image located to the left of the diverging lens. This first image is also the object for the converging lens and is located within its focal point. From the drawing, we can see that the corresponding object distance is do = 30.0 cm − 5.00 cm = 25.0 cm. To determine the final image distance, we again use the thin-lens equation: 111 1 1 =− = − = −0.0067 cm-1 di f do 30.0 cm 25.0 cm or di = −150 cm The minus sign means that the final image is virtual and located to the left of the converging lens. Furthermore, the size of the image distance indicates that the final image is located to the left of both lenses. b. Using the magnification equation we can determine the size of the first image: hi d =− i ho do or d ( −5.00 cm ) hi = ho − i = ( 3.00 cm ) − = 1.50 cm d 10.0 cm o The fact that the image height (which is also the object height for the converging lens) is positive means that the image is upright with respect to the original object. Using the magnification equation again, we find that the height of the final image is d ( −150 cm ) hi = ho − i = (1.50 cm ) − = +9.0 cm d 25.0 cm o Since the final image height is positive, we conclude that the final image is upright with respect to the original object. 71. SSM REASONING We begin by using the thin-lens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ] to locate the image produced by the lens. This image is then treated as the object for the mirror. Chapter 26 Problems 1395 SOLUTION a. The image distance from the diverging lens can be determined as follows: 111 1 1 =− = − di f do −8.00 cm 20.0 cm or di = −5.71 cm The image produced by the lens is 5.71 cm to the left of the lens. The distance between this image and the concave mirror is 5.71 cm + 30.0 cm = 35.7 cm. The mirror equation [Equation 25.3: (1 / do ) + (1 / di ) = (1 / f ) ] gives the image distance from the mirror: 111 1 1 =− = − di f do 12.0 cm 35.7 cm or di = 18.1 cm b. The image is real , because di is a positive number, indicating that the final image lies to the left of the concave mirror. c. The image is inverted , because a diverging lens always produces an upright image, and the concave mirror produces an inverted image when the object distance is greater than the focal length of the mirror. ______________________________________________________________________________ 72. REASONING 1 1 1 + = (Equation 26.6), where do1 = 2.00 m is do1 di1 f1 the object distance and f1 = −3.00 m is the focal length of the diverging lens, to determine the image distance di1 when the friend is viewed through the diverging lens only. Because this image is the image of a real object formed by a diverging lens, the image is virtual, appearing to the left of the diverging lens. The magnification m1 of the first image is given d by the magnification equation m1 = − i1 (Equation 26.7) d o1 a. We will use the thin-lens equation, b. When the converging lens is placed a distance L = 1.60 m to the right of the diverging lens, the image of the friend formed by the diverging lens becomes the object for the converging lens (see the drawing). The object distance do2 is, therefore, the distance between the converging lens and the image formed by the diverging lens. This object is to the left of the converging lens, so the object distance is positive, and equal to the sum of the distance L and the magnitude |di1| of the first image distance: do2 Friend 1st image di1 do1 L 1396 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS do2 = L + di1 (1) We will use the thin-lens equation (Equation 26.6) to determine the second image distance di2, and then the magnification equation (Equation 26.7) to determine the magnification m2 of the second image relative to the first image. The overall magnification m, that of the final image relative to the friend, is equal to the product of the two magnifications: m = m1m2 (2) For example, if the first image is upright and one-half the size of the friend, then m1 = +0.50. If the second image is inverted and is three times the size of the first image, then m2 = −3.0. The overall magnification provided by the two-lens system is, then, equal to m = m1m2 = (+0.50)(−3.0) = −1.5. This means that the second image is inverted, and oneand-a-half times as tall as the visitor’s friend. SOLUTION a. Solving the thin-lens equation (Equation 26.6) for 1/di1 and taking the reciprocal of both sides yields d −f 1 1 1 =− = o1 1 di1 f1 do1 d o1 f1 or di1 = do1 f1 do1 − f1 = ( 2.00 m ) ( −3.00 m ) = −1.20 m 2.00 m − ( −3.00 m ) (3) Substituting this result into Equation 26.7, we find that m1 = − di1 ( −1.20 m ) = +0.600 =− d o1 2.00 m b. Adapting Equation (3) to the formation of the second image produced by the converging lens, we see that di2 = d o2 f 2 (4) do2 − f 2 Substituting Equation (4) into the magnification equation (Equation 26.7), we obtain d o2 f 2 m2 = − di2 d o2 =− do2 − f 2 d o2 =− Substituting Equati...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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