Question

(13%) Problem 5: An object of height 2.9 cm is placed 29 cm in front of a diverging lens of focal length 17 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. - A 50% Part (a) Find the location of the final image, in centimeters beyond the converging lens. Grade Summary Deductions 0% Potential 100% 1 E sin( cos() tan() cotan() asin() acos atan acotan sinh cosh() | tanh) | cotanh0 Degrees Radians ( 7 8 9 HOME 4 5 6 - * 1 2 3 - 0 . END VO BACKSPACE DEL CLEAR Submissions Attempts remaining: 20 (4% per attempt) detailed view Submit Hint Feedback I give up! Feedback: 0% deduction per feedback. Hints: 0% deduction per hint. Hints remaining: 2 50% Part (b) What is the magnification of the final image? Include its sign to indicate its orientation with respect to the object.

Answer 1

Part I final image 2,=-29cm Using lens equation for first lens. 1-1-I = I = 1 + 1 Here I am wring cartesian sign convention f = -17cm, u=-29 cm => #has 1x29 17%20 = -29-17 1729 - 46 17429 1729 Using v=17X22 2 V = -10.72cm I - show imny lens equation for and lens. L.1 ! 1 Here 5 м, ч, , , ч f2= nem =-(11+14) 2 = -21.72cm I= 1 1 - 21.72-17 = 4.72 k 17 21.12 17821.72 1782172 - 17X21,72 = 78.23cm 4.72 [ 12 = 50 = 78.23 cm] Part B magnification m = mim2 = 4 2 12- m= -10.72 29 x 18-23 = -1.33 21.72