ࡱ> y Z'bjbj 4{{tfffzzzz&zkmmmmmm$ާf|͞͞͞Dfk͞k͞͞:ۣ,"Dc`QZ W0Rcfc͞7 : Convex Lenses Practice Worksheet Name _____________________ 1) An object is placed 8 cm in front of converging lens. A real image is produced at 12 cm. Find the focal distance of the lens. Diagram: (Given + Unknowns) f = ???? cm di = 12 cm do = 8 cmEquation: 1 = 1 + 1 f di doSubstitute: 1 = 1 + 1 1 = 0.08333 + .125 f di do f 1 = 1 + 1 1 = .20833 (DO A SWITCHAROO) f 12 8 f 1 = f .20833 Solve: f = 4.8 cm  2a) A 15.0 cm object is placed 60.0 cm from a convex lens, which has a focal length of 15.0 cm. Draw a ray diagram and use the information from the ray diagram to fill in the box.   2b) A 15.0 cm object is placed 60.0 cm from a convex lens, which has a focal length of 15.0 cm. Use the thin lens equation to find the distance of the image. Diagram: (Given + Unknowns) f = 15 cm di = ??? cm do = 60 cmEquation: 1 = 1 + 1 f di doSub 1 = 1 + 1 .0667 = 1 + .01667 (subtract .01667) f di do di (both sides) 1 = 1 + 1 .0500 = 1 (DO A SWITCHAROO) 15 di 60 di di = 1 .0500 Solve: di = 20 cm3a) A 15.0 cm object is placed 30.0 cm from a convex lens, which has a focal length of 15.0 cm. Draw a ray diagram and use the information from the ray diagram to fill in the box.  3b) A 15.0 cm object is placed 30.0 cm from a convex lens, which has a focal length of 15.0 cm. Use the thin lens equation to find the distance of the image. Diagram: (Given + Unknowns) f = 15 cm di = ??? cm do = 30 cmEquation: 1 = 1 + 1 f di doSub 1 = 1 + 1 .0667 = 1 + .0333 (subtract .0333) f di do di (both sides) 1 = 1 + 1 .0334 = 1 (DO A SWITCHAROO) 15 di 30 di di = 1 .0334 Solve: di = 30 cm  4a) A 15.0 cm object is placed 25.0 cm from a convex lens, which has a focal length of 15.0 cm. Draw a ray diagram and use the information from the ray diagram to fill in the box.   4b) A 15.0 cm object is placed 25.0 cm from a convex lens, which has a focal length of 15.0 cm. Use the thin lens equation to find the distance of the image. Diagram: (Given + Unknowns) f = 15 cm di = ??? cm do = 25 cmEquation: : 1 = 1 + 1 f di doSub 1 = 1 + 1 .0667 = 1 + .04 (subtract .04) f di do di (both sides) 1 = 1 + 1 .0267 = 1 (DO A SWITCHAROO) 15 di 25 di di = 1 .0267 Solve: di = 37.4 cm  5a) A 15.0 cm object is placed 15.0 cm from a convex lens, which has a focal length of 15.0 cm. Draw a ray diagram and use the information from the ray diagram to fill in the box.   5b) Explain why no image can be formed when the object is placed at the focal point. The light rays never intersect, so no image is formed.  6a) A 15.0 cm object is placed 10.0 cm from a convex lens, which has a focal length of 15.0 cm. Draw a ray diagram and use the information from the ray diagram to fill in the box.  6b) A 15.0 cm object is placed 10.0 cm from a convex lens, which has a focal length of 15.0 cm. Use the thin lens equation to find the distance of the image. Diagram: (Given + Unknowns) f = 15 cm di = ??? cm do = 10 cmEquation: 1 = 1 + 1 f di doSub 1 = 1 + 1 .0667 = 1 + .1 (subtract .1) f di do di (both sides) 1 = 1 + 1 - .033 = 1 (DO A SWITCHAROO) 15 di 10 di di = 1 -.033 This is a virtual image sincedi is a negative number.Solve: di = - 30.0 cm 7) A 2-meters-tall person is located 5 meters from a camera lens (camera lenses are convex lenses). The lens has a focal length of 35 millimeters. Do not forget to convert millimeters to meters before substituting into the equation. Find the distance where the image would appear. Diagram: (Given + Unknowns) f = 35 mm = .035 m di = ??? m do = 5 mEquation:Substitute: 1 = 1 + 1 28.57 = 1 + .2 (subtract .2) f di do di (both sides) 1 = 1 + 1 28.37 = 1 (DO A SWITCHAROO) .035 di 5 di di = 1 28.37 Solve: di = .03525 m 8) A 1.0 cm object is placed 30.0 cm from a convex lens, which has a focal length of 10.0 cm. (Notice, I did not label the F distance and 2F distance. Label these first and it help you determine where to put your object. Use an arrow for your object..) 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