Question

Problem 12. TWO POSITIVE lenses each has a focal length of 20. Cm. The lenses are 100. Cm apart. An OBJECT is placed 30 cm from the left of the first lens # 1. The object has height of 4.Cm. A. What is the power of each lens? B. Find the final image: Location, size, and its nature Upside down or right side up and if it is real or virtual? C. Draw the rays for the final image? D. IF THE DISTANCE between the two lenses is 70. Cm ANSWER B in this case. Cased i g Fe iz Vaada loocm 30cm 30см Tolm

Answer 1

now ho -2 30 2x at -40cm magnification, m,2 = m =-60 A Leight boy images height obiect (inventall - height of inge inge = 2xy = 8 cmc linvented) so lens i form a real and inverted image a distance to cm and of height bem Now this image act of object for II'd ling So now = (100-60) f = 20cm tad & V = 40 cm 16 magnification yo - bi Tho height of image cremains same but again its inverted final Image charactuistist height of image abomi Location Yo cm from second lens (In Right) Image a Red and ito a upward. (Same orientation of as of Object) २० 2 m₂ = t so پیامط

tol Ray diagram: lens! lns 2 find image 1 s object Cariert fi fa fi 300m form. Toom ५०m. (cm. To 10 how distance blu lens reduled to 70cm. So image formed by first here will still act of a object but this tine a distance 6 70-60) = 10 cm. so for lins a f = Qocm. tott va -20 mamification ho so height of image = 2x 6 = 16 cm. final. characteristics - location 20cm som bes 2 on befan Size height of Image = 16 cm virtual Image formed is downwand directim. and it is in Conientation final image opposite to objet) ma v -20 -22h, u -10 So now away from len 1. how how de

2 th A Since power ft focal length food length of both lens = 20cm = 0.2m. 5 dioptre P= -1 0.2 So Power g both lins id s diopta (B) 4--2o Cm, have Since f. 20cm. * -t 1 2 1 4 to bo to u a Go cu cm. so the i first dens lens form distance of Go cm image towards bright