Question

A match is held a distance ?o=15.7 cmso=15.7 cm to the left of a lens. The corresponding image for the match is located at ?i=24.2 cmsi=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=

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.)

?=

Answer 1

Page No. so we assume that (A Guiven the distance of Object ( match) ferom. the lens is , u = 15.7 cm The distame of image from lens, se = 24.2 cm Also, it is given that the image is formed to the right of the lens the image formed is real. Therefore, the lens used will be convex lens. for a conver lens, the object distance u is taken aus negatwie and the iniage distance is taken as positive. from the lens maker formula, we have where f = focal length け ㅗ 그 u f of the lens I= I I to 24.2 cm 1517 cm 1 + L ㅗ 15.7 cm to 24.2 cm صم см 15. y Control + 24.2 mg LX 24:2 x 15.4 39.9 / 379.94 cm of

TE PAGE No. 2. 29....... L IN at Is 39.9 379.94 cm = 319.94 cm 39.9 9.52 cm Af Le m = 1 (8) Magnification is defined as the ratio of the height of the image formed to the cheight of the object . height of image height of object But magnifications can also be defined in terms of the distance of the image formed and the distance of the object from the lens Magnificatiei distance of image from Lens of object from lens 으 distane lele m = 브 lu 24.2 cm 1 5* 7 cm =-1.54 Negatie magnification means the lens used is Couver lens