Question

Converging- coovex F) Group B 1. An object is placed in front of a converging lens with a focal length of magnitude 10.0 cm such that the image is four times the size of the object. (a) At what distances must the object be placed from the lens such that this would occur? (b) For each object distance determined in part (a), what is the image distance? (c) Specify whether the image is upright or inverted and real or virtual for each image distance? (e) Draw ray diagrams to confirm your answers. ength 10. O cm

Answer 1

The image will be larger than the object occurs in two cases is for real image II) for virtual image Again magnification m = 4 therefore v = 4 n for real image v = -4 n for virtual image According to the lens law 1/v - 1/n = 1/f for real image 1/4 u + 1/u = 1/f [u is always - ve] = > 5/4 u = 1/10 or u = + 12.5 c m fro virtual image - 1/4 u + 1/4 = 1/f = > - 3/4 u = 1/f = > u = -30/4 = -7.5 c m so two position of object u = 12.5 c m, 7.5 c m \r\n for u = 12.5 c m 1/v = 1/f + 1/u = 1/10 - 1/12.5 = 1/50 or v = 50 c m i.e in front of lens for u = 7.5 c m