Question

A match is held a distance so = 14.7 cm to the left of a lens. The corresponding image for the match is located at si = 24.2 cm to the right of the lens (not pictured). Answer the two questions below, using three significant digits. Part A: What is the value of the focal length(f) of the lens? f = 9.144 cm Part B: What is the value of the magnification (M) of this configuration? (Remember: the magnification could be either a positive or a negative number.) -So- M = -1.646

Answer 1

f = 9.14498 cm 9. 14 cm (3 significant figure) yf S' Magnification, a So 24.2. 14.7 - 1.6462 . 1.65 (3 significant figure)

24.2 cm to Object distance, so = 14.7 com to left of lens. distance, Si Image lens. reight of Taking left of lens to be negative and right of to be positive. So = -14 7 cm and Si= + 24.2 cm. f be the focal length. lens formula, lens So, a) Let So by ㅗ So Si f 1) + Ź Ź 24 f 24.2 14.7 -1945 1 14 17787