B using the thin lens formula 1 1 1 s s f 1 1 1 15 cm

Info icon This preview shows pages 63–69. Sign up to view the full content.

View Full Document Right Arrow Icon
(b) Using the thin-lens formula, 1 1 1 s s f + = 1 1 1 15 cm 20 cm s + = 1 1 60 cm s = − 60 cm s ′ = − The image height is obtained from 60 cm 4 15 cm s M s = − = − = + Thus, the image is 4 times larger than the object or ( ) 4 4 2.0 cm 8.0 cm. h Mh h ′ = = = = The image is upright. These values agree with those obtained in part (a).
Image of page 63

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.64. Model: Use ray tracing to locate the image. Assume the converging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. After refraction, the three special rays converge and give an image 50 cm away from the converging lens. Thus, s = + 50 cm. The image is inverted and its height is 0.65 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 75 cm 30 cm s s f s + = + = 1 1 50 cm s = 50 cm s ′ = The image height is obtained from 50 cm 2 75 cm 3 s M s = − = − = − The image height is ( )( ) 2/3 1 cm 0.67 cm. h Mh ′ = = − = − Because of the negative sign, the image is inverted. These results agree with those obtained in part (a).
Image of page 64
23.65. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction, the three special rays do not converge. The rays, on the other hand, appear to meet at a point that is 8.5 cm on the same side of the lens as the object. So 8.5 cm. s ′ = − The image is upright and has a height of 1.1 cm. (b) Using the thin-lens formula, 1 1 1 s s f + = 1 1 1 15 cm 20 cm s + = 1 7 60 cm 8.6 cm 60 cm 7 s s = − = − = − The image height is obtained from ( ) 60/7 cm 4 0.57 15 cm 7 s M s = − = − = + = Thus, the image is 0.57 times larger than the object, or (0.57)(2.0 cm) 1.14 cm. h Mh ′ = = = The image is upright because M is positive. These values agree, within measurement accuracy, with those obtained in part (a).
Image of page 65

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.66. Model: Use ray tracing to locate the image. Assume the diverging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. After refraction from the diverging lens, the three special rays do not converge. However, the rays appear to meet at a point that is 20 cm on the same side as the object. So 20 cm. s ′ = − The image is upright and has a height of 0.3 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 30 cm 60 cm 20 cm s f s = = = − 20 cm s ′ = − The image height is obtained from 20 cm 1 0.33 60 cm 3 s M s = − = − = = Thus, ( )( ) 0.33 1.0 cm 0.33 cm, h Mh ′ = = = and the image is upright because M is positive. These values for s and h agree with those obtained in part (a).
Image of page 66
23.67. Visualize: Refer to Figure 23.57. 1 1 1 fs s s s f s f + = = We are given 60 cm, 20 cm, f s = = and 1.0 cm. h = Solve: (60 cm)(20 cm) 30 cm 20 cm 60 cm fs s s f ′ = = = − The negative sign means the image is behind the mirror; it is a virtual image. The magnification is 30 cm 20 cm 1.5. m s s = − = = This means the image is upright and has a height of (1.5)(1.0 cm) 1.5 cm. h mh ′ = = = Assess: Ray tracing will confirm these results.
Image of page 67

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
23.68.
Image of page 68
Image of page 69
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern