Question

1. When a dime is placed 3.5 cm in front of a lens, it forms an upright image that is 2.5 times as large as the dime. [Note: It is very important in these calculations to keep track of the (+/-) sign of every variable] What is the image distance from the lens? Is the image on the same side as the coin or on the other side of the lens? 5pts) a) b) What is the focal length of the lens? Is the lens a converging lens or a diverging lens, and tell how you know which it is? 5pts) Draw a ray tracing for the above situation. The coin is indicated by an arrow. Label the positions of the primary and secondary focal points, show all three principle rays, and draw an arrow indicating the position and orientation of the image. Draw the ray diagram to as precise of a scale as you can. You may use a straight edge (ruler or edge of a sheet of paper) to help draw straight lines. Ipts)

Answer 1

Here am conventin cartesian We Knous that me gni ficahim due to lens mot2.5 Here 4=-3-50m 2.5= V -3-S 2.SXT 3.5) (&nags the imege on thme same Cide of the coin. VE-8-75 cm So fhe imae alis tanee fro m tne lens of the 2-75 cm. Ond Same side coin uirg lens equahi f M-3.5(m Hert v-8-75 V -354815- S.25 2-75X3-5 +. -975 3.5 &75X3-5 87SX3.5 y f -> S.25 f S.83 cm lens it is eonverging Since fis positine so imge PPriany S condrng