Question

lens! 1 f = 100cm lens 2 f2 = 50cm 30cm 40cm Two converging lenses are positioned 40 cm apart, as shown in the figure. The vertical lines represent the lenses, as we usually draw them. The first lens (left) has a focal length of 100 cm, and the second lends (right) has a focal length of 50 cm. An object is placed 30 cm to the left of the first (left) lens, as shown. a) How far is the final image from the vertex of the second (right) lens (enter a positive number)? (8 points) 159 cm b) What is the overall magnification for the final image formed by the two lens combination (don't forget to include the algebraic sign)? (8 points) 7.14 X c) Where is the final image? (3 points) O To the left of the first lens O Between the two lenses To the right of the second lens d) The final image is (2 points) real O virtual

Could you explain parts c through f as well please? Thank you!

Answer 1

6 = 100cm F₂-room u, = 300m f = 100 cm 1-30cm luoom I F₂ = room daroom or tatt ta 30cm BE100 am so tattu fois too so mita mrava V; = -42.91) . 42.91 - 1:43 30 m = 1.4 L image is left from first. Noru + + 2 = 742 tt 42 = you(-48.91 ) [u, =) 82.91 cm tos = 5e-82.91 ₂ = 126. 10 cm/ m₂ = 1/2 - 12 = -12 6.10 8 2.91 = 1.52 -2 -174 mr mix m₂ = - 1.52 x 1.43 In 2-20174 )

V2 = 126. 10 cm from second leng Right ) m= 2.174 to the right of second (4 126.10) lens. The final image is ustural Real ( The image (m= - 2.174] (is the right side) mel 8o The ge Than is laager object The final inverted . i age ma-ive . obsesves obferver e invested in is at the observer's side. the image This is Real 80