Physics Solution Manual for 1100 and 2101

# 0 103 rad 064 rad the magnitude of this angular size

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lies inside the focal point of the converging (#1) lens. According to Figure 26.28, such an object produces a virtual image that lies to the left of the lens. This image act as the object for the diverging (#2) lens. Since a diverging lens always produces a virtual image that lies to the left of the lens, the final image lies to the left of the diverging lens. In part (b), the diverging (#1) lens produces a virtual image that lies to the left of the lens. This image act as the object for the converging (#2) lens. Since the object lies outside the focal point of the converging lens, the converging lens produces a real image that lies to the right of the lens (see Figure 26.26). Thus, the final image lies to the right of the converging lens. SOLUTION a. The focal length of lens #1 is f1 = 15.00 cm, and the object distance is do1 = 10.0 cm. The image distance di1 produced by the first lens can be obtained from the thin-lens equation, Equation 26.6: 1392 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 1 1 1 1 1 =− = − = − 3.33 × 10−2 cm−1 or di1 = − 30.0 cm di1 f1 do1 15.00 cm 10.00 cm This image is located to the left of lens #1 and serves as the object for lens #2. Thus, the object distance for lens #2 is di2 = 30.0 cm + 50.0 cm = 80.0 cm. The image distance produced by lens #2 is 1 1 1 1 1 −2 −1 = − = − = − 6.25 × 10 cm or di2 = −16.0 cm di2 f 2 do2 −20.00 cm 80.0 cm The negative value for di2 indicates that, as expected, the final image is to the left of lens #2. b. The focal length of the lens #1 is f1 = –20.0 cm, and the object distance is do1 = 10.00 cm. The image distance di1 produced by the first lens can be obtained from the thin-lens equation, Equation 26.6. 1 1 1 1 1 −1 −1 =− = − = − 1.50 × 10 cm or di1 = − 6.67 cm di1 f1 do1 −20.0 cm 10.00 cm This image is located to the left of lens #1 and serves as the object for lens #2. Thus the object distance for lens #2 is di2 = 6.67 cm + 50.0 cm = 56.7 cm. The image distance produced by lens #2 is 1 1 1 1 1 −2 −1 = − = − = 4.90 × 10 cm or di2 = 20.4 cm di2 f 2 do2 15.00 cm 56.7 cm The positive value for di2 indicates that, as expected, the final image is to the right of lens #2. ______________________________________________________________________________ 69. SSM REASONING The problem can be solved using the thin-lens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ] twice in succession. We begin by using the thin lens-equation to find the location of the image produced by the converging lens; this image becomes the object for the diverging lens. SOLUTION a. The image distance for the converging lens is determined as follows: 1 1 1 1 1 =– = – d i1 f d o1 12.0 cm 36.0 cm or d i1 = 18.0 cm This image acts as the object for the diverging lens. Therefore, Chapter 26 Problems 1 1 1 1 1 =– = – d i2 f d o2 –6.00 cm (30.0 cm –18.0 cm ) or 1393 d i2 = –4.00 cm Thus, the final image is located 4.00 cm to the left of the diverging lens . b. The magnification equation (Equation 26.7: hi / ho = – di / do ) gives d i1 18.0 cm mc = – =– = –0.500 d o1 36.0 cm 144444 2444444 4 3 d i2 –4.00 cm md = – =– = 0.333 d 12.0 cm 1444o2 4 2444444 44 3 Converging lens Diverging lens Therefore, the overall magnification is given by the product m c md = –0.167 . c. Since the final image distance is negative, we can conclude that the image is virtual . d. Since the overall magnification of the image is negative, the image is inverted . e. The magnitude of the overall magnification is less than one; therefore, the final image is smaller . ______________________________________________________________________________ 70. REASONING The following drawing shows the arrangement of the lenses and the location of the object. The thin-lens equation can be used to locate the final image produced by the converging lens. We know the focal length of this lens, but to determine the final image distance from the thin-lens equation, we also need to know the object distance, which is not given. To obtain this distance, we recall that the image produced by one lens (the diverging lens) is the object for the next lens (the converging lens). We can use the thin-lens equation to locate the image produced by the diverging lens, since the focal length and the object distance for this lens are also given. The location of this first image relative to the converging lens will tell us the object distance that we need. To find the height of the final image, we will use the magnification equation twice, once for the diverging lens and once for the converging lens. 20.0 cm 10.0 cm 30.0 cm 1394 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION a. Using the thin-lens equation to obtain the distance di of the first image from the diverging lens, which has a focal length f and an object distance do, we find 111 1 1 =− = − = −0.200 cm -1 di f do −10.0 cm 10.0 cm or di = −5.00 cm The minus sign indicates that this first image...
View Full Document

## 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.

Ask a homework question - tutors are online