Question

Question 4 (2.5 Points) This assignment is set up for sequential assessment. Complete each question and submit the answer before moving on to the next question. Correct answers for each question will be made available once the maximum number of attempts have been made for each submission. Lenses in combination 1 (case 1 and 2) Pay attention to the SIGN Convention. The sign of each quantity has meaning. Lenses in Combination 1 Cering lens Diverging lens As shown in the above figure, a converging thin lens with focal points indicated by F1's is placed to the left of a diverging thin lens with focal points indicated by F2's. The focal length of the converging lens is 20.0 cm. The focal length of the diverging lens is -28.0 cm. The distance between the two lenses is 56.0 cm. Case 1: An object is put 42.00 cm to the left of the converging lens (Lens 1). The object is 13.0 cm tall. Find the image distance di1 of the First image formed by the Converging lens (Lens 1). Include a proper sign. Keep 2 decimal places. Enter a number cm Submit (5 attempts remaining) The height of the object is ho = 13.0 cm, what is the height of the First image hil? Include a proper sign, keep 2 decimal places. Enter a number cm Submit (5 attempts remaining) On which side of the Converging lens (Lens 1) is the First image located? Select your answer from one of the following options. a. Left side b. Right side Submit (5 attempts remaining) CONVERGING Lens image formation 1 2 REAL - image F 2 2

Answer 1

Image distance =) 4 = -42cm hon height of image hi = Object height for lens hog 4, d=56cm F = 20 F2 = -28 According to sign convention Ruce u is ove beacuse it is left side of a lans Lens formula L-1=1 F vadi, (-42) (20) 38.18cm Image form Converging 4 + F va U, F Lens -42+ 20 B ma -0.90 magnification ma Image distance from Cent object distance forom Cent 38.18 -42 he me height of Image height of object (h1), (hi); m, hoa - 0.90 x 13=-11.81cm, © The sign of image distance is tue meane left side af lenta image for va lens-2 vis object for Consa Ug=-(2-42) =-|56-38:18) ka=diz Ug=-17.82(m) Robject diftance from Cen-a ug få -17.82x1-28) Ug+fa -17.82-28 Va is negative means left side of Diverging Cent - 10.899 O. Gll -17.82 -0.611 x 11.81=-7.216 cm mg = 1h 1/2 ) /h) Da mihaela (hola -10.889 cm mg

9 F, = 20 F22-28 cm 4,2-28 ho, = +13cm I d=56cm P 4 (-28) 20 u, f, 4,+ fi 70 cm -28 +20 m² dis 70 = - 2.5 -28 hil ho, b -mho, = -2.5x 13 = -32.5 cm 1 du is the means Right side Left side © dog = -(d-d() = -31.5cm f dog fa = -37.5+ (-28)=-16.03 cm -37.5-28 dogtfa 9 Final image is right side forom diverging Cent © virtual image beacuse diverging lens always forma virtual image -16.03 0.427 -31.5 (1) mga dig dog ma mixha -2.sto.4970 -1.068 hiz hol hia =mho, -1.068+13= -13.893 cm