Question

please provide explaination because i have hard time understand this problem

The same object (height = y) is placed at several different distances s to the left of the same lens (focal length = f). For each case, draw the 3 principal rays to locate the image. Then use the "thin lens" equations to calculate: • the image height y' (in terms of y) • the image distance from the lens s' (in terms of s and f) • whether the image is real or virtual • whether the image is upright or inverted = 3f s = 2f s=f s= tf

Answer 1

According to Chegg guidelines, i can only answer first four subpart of a question. If you want answer to the rest, post them separately
magnification, m=t & - 1 ū P U = -3f i v=uf = -3fxf . ütf -347€ = 20.756 magnification=m= -0.757 -35 [m=0.25 a) image height = 9/ 1.

b) image distance, = 0.75 f = $' <- 0 - 35 s'= 45 c) sine mot V=-34 f, it is kithual inge, d) sime m= + 4 image is upright.