Hwksolns7_Publishers - 23.60 Model Use the ray model of...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
23.60. Model: Use the ray model of light. The surface is a spherically refracting surface. Visualize: Solve: Because the rays are parallel, s = . The rays come to focus on the rear surface of the sphere, so s = 2 R , where R is the radius of curvature of the sphere. Using Equation 23.21, 1 2 2 1 1 1 2.00 2 n n n n n n n s s R R R + = + = =
Image of page 1

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

View Full Document Right Arrow Icon
23.69. Model: Assume the lens is a thin lens and the thin-lens formula applies. Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s and s are positive. This also implies that the spider’s image is inverted, so 1 2 M s s = − = − . Using the thin-lens formula with 1 2 , s s ′ = 1 2 1 1 1 1 1 1 s s f s s f + = + = 3 1 3 s f s f = = We also know that the spider is 2.0 m from the wall, so s + s = 2.0 m = s + 1 2 s ( ) 1 3 4.0 m 133.3 cm s = = Thus, 1 3 44 cm f s = = and 2.0 m 1.33 m 0.67 m 67 cm. s ′ = = = We need a 44 cm focal length lens placed 67 cm from the wall.
Image of page 2
23.73. Visualize: The lens must be a converging lens for this scenario to happen, so we expect f to be positive. In the first case the upright image is virtual 2 ( 0) s ′ < and the object must be closer to the lens than the focal point. The lens is then moved backward past the focal point and the image becomes real 2 ( 0). s ′ > 1 1 1 ss f s s f s s + = = + We are given 1 10cm s = and 1 2. m = Solve: Since the first image is virtual, 0. s ′ < We are told the first magnification is 1 1 1 1 2 20 cm. m s s s = = − = − We can now find the focal length of the lens. 1 1 1 1 (10 cm)( 20 cm) 20 cm 10 cm 20 cm s s f s s = = = + After the lens is moved, 2 2 2 2 . m s s = − = − Start with the thin lens equation again. 2 2 1 1 1 s s f + = Replace 2 s with 2 2 . m s 2 2 2 1 1 1 s m s f + = Now solve for 2 . s 2 2 2 2 2 2 1 ( ) m s s s m s f + = 2 2 2 2 2 2 1 m s s m s f + = Cancel one 2 . s 2 2 2 1 1 m m s f = Multiply both sides by 2 . fs 2 2 2 1 2 1 (20 cm) 30 cm 2 m s f m = = = The distance the lens moved is 2 1 30 cm 10 cm 20 cm. s s = = Assess: We knew 2 s needed to be bigger than f ; it is, and is in a reasonable range. The final answer for the distance the lens moved also seems reasonable.
Image of page 3

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

View Full Document Right Arrow Icon
24.1. The picture will be underexposed if the shutter speed remains the same. The problem stated that D is not changed. 2 2 . D I f If f is increased then I is decreased so the picture is underexposed.
Image of page 4
24.2. The answer is C. Rays from all parts of the object hit all parts of the lens, so the whole image still appears, but because the area of the aperture is smaller then the image will be less bright. Contrary to choice a, the image actually gets sharper because the off-axis rays contribute to blurriness and a smaller aperture eliminates many of those rays.
Image of page 5

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

View Full Document Right Arrow Icon
24.3.
Image of page 6
Image of page 7
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