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The Lens Equations

The Lens Equations. It is possible to calculate characteristics of images using formulae. You can find: d o = The distance of the object to the lens d i = The distance of the image from the lens h o = The height of the object h i = The height of the image

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The Lens Equations

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  1. The Lens Equations It is possible to calculate characteristics of images using formulae. You can find: do= The distance of the object to the lens di= The distance of the image from the lens ho = The height of the object hi = The height of the image M = The magnification of the lens

  2. The Thin Lens Equation • An explanation of the derivation of the lens equation is possible but not necessary. It has to do with "equal triangle" see page 563 in your text if you are interested. (See next slide) You must follow some conventions however. 1. Object distances are always positive. (do is a positive value) 2. Image distances (di) are: Positive for real images (image is on opposite side of lens) Negative for virtual images (image is on the same side of the lens as the object) 3. The focal length (f) is : positive for converging lenses. negative for diverging lenses.

  3. Similar Triangles Let’s say you wanted to calculate the height of the sun. 8 = x 6 60 000 8(60 000) = 6x X = 80 000 li

  4. Sample problems page 563 • A converging lens has a focal length of 17 cm . A candle is located 48 cm from the lens. What type of image will be formed and where will it be located? • Given: f= 17 cm • do = 48 cm • Required : di = ? • Analysis: 1/do + 1/di = 1/f • Solution 1/di = 1/f – 1/do • 1/di = 1/17 cm - 1/48 cm • 1/di = 0.058 cm – 0.021 cm = 0.038 cm • di = 1/ 0.038 cm = 26.3 cm = 26 cm • The image is real and located 26 cm behind the lens

  5. Lens problem for a Diverging lens • A diverging lens has a focal length of 29 cm. A virtual image of a marble is located 13 cm in front of the lens. Where is the marble located? • f = - 29 cm • di = - 13 cm • do = ? • 1/do + 1/di = 1/f • 1/do = 1/f – 1/di • 1/do = 1/ -29 cm – 1/-13 cm = ( - 0.034 cm) –( - 0.077 cm) = 0.043 cm • do = 1/0.043 cm = 23.2 cm = 23 cmin front of the lens.

  6. Magnification Equation • The two shaded triangles are called similar triangles because they share common angles. (See slide on height of sun.) • Because of this there is a relationship between the height and the distance of the object and the image.

  7. What are the Words to the song the seven dwarfs sang as they went off to solve lens equation problems? Di Do it’s off to work on lens equation problems we go!

  8. Conventions (to be remembered) • Object height (ho ) and image height (hi ) are: • Positive when measured upward from the p.a. • Negative when measured downward from the p.a. • Magnification (M) is: • Positive for and upright object. • Negative for an inverted object. • M does not have a unit because the units cancel out

  9. Sample problems page 565 • M of a converging lens • A 8.4 cm toy is placed in front of a converging lens. An inverted, real image with a height of 23 cm is formed on the other side of the lens. What is the magnification of the lens? • G ho = 8.4 cm • hi = -23 cm ( real image for converging lens is below p.a.) • R M = ? • A M = hi / ho • S = -23cm/ 8.4 cm • S = -2.7

  10. Locating Image • A small toy building block is placed 7.2 cm in front of a lens. An upright virtual image of magnification 3.2 is observed. Where is the image located? • G do = 7.2 cm • M = 3.2 ( upright image  positive value) • di = ? • A M = - di / do • S Mdo = - di • di = -Mdo • di = - 3.2 ( 7.2cm) = -23 cm in front of the lens.

  11. M of a Diverging Lens • A coin of height 2.4 cm is placed in front of a diverging lens. An upright, virtual image of height 1.7 cm is noticed on the same side of the lens as the coin. What is the Magnification? • G ho = 2.4 cm (both object and virtual image of a diverging lens • hi = 1.7 cm will be above the p.a.  positive h ) • R M = ? • A M = hi/ ho • S = 1.7 cm/ 2.4 cm • S = 0.71 • The lens has a magnification of 0.71

  12. Homework • Page 566 • # 1 - 8

  13. Homework pg 566 #1-4

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