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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-26 34-23 An insect 3.00 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location of the image this lens forms of the insect. (b) Calculate the size of the image. (c) Is the image real or virtual? Erect or inverted? (a) Using the formula for the focal length of a lens with two curved surfaces (the Lensmakers Equation), with n = 1 . 7, R 1 = , and R 2 =- 13 cm, we have 1 f = ( n- 1) ( 1 R 1- 1 R 2 ) = 0 . 7 ( 1 - 1- 13 ) = 1 18 . 57 cm The value of f determined can be substituted into the lens equation: 1 f = 1 d o + 1 d i The result is: 1 18 . 57 = 1 25 + 1 d i = d i = 1 1 18 . 57- 1 25 = 72 . 2 cm b) To calculate the size of the image, use the magnification m =- d i d o =- 72 . 2 25 . =- 2 . 89 The image height h i is h i = mh o =- 2 . 89(3 . 00 mm) =- 8 . 67mm (c) The image is real because the image distance is positive. The image is inverted because the magnification is negative. (d) If the lens is reversed, then R 1 = 13 cm and R 2 = . The focal length is therefore: 1 f = (0 . 7) ( 1 13- 1 ) = 1 18 . 57 cm The value of 1 /f is exactly the same as we had before. All the results are the same. 34-28 A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is theinverted?...
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This note was uploaded on 12/18/2011 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .
- Spring '08