Question

The type of lens is not given.

(10%) Problem 6: Two lenses are mounted d= 29 cm apart on an optical bench. The focal length of the first lens is fı = 7.9 cm and that of the second lens is f2 = 5.6 cm. An object of height ho = 3.4 cm is placed at a distance of do = 29 cm in front of the first lens. 25% Part (a) Ignoring the second lens for now, at what distance, in centimeters, behind the first lens is the object's image formed by that lens? di = 10.86 Correct! 25% Part (b) Calculate the magnification of that image, including its sign. m= -0.3744 m=-0.3744 Correct! >> A 25% Part (c) Which choice best describes that image? Real and inverted. Virtual and upright. Real and upright. None of these. Virtual and inverted. Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 4 (25% per attempt) detailed view Submit Hint Feedback I give up! Hints: 1% deduction per hint. Hints remaining: 1 Feedback: 2% deduction per feedback. 25% Part (d) Now consider the two-lens system and the final image it forms, i.e., the image created by the second lens. What is the distance, in centimeters, between the object and its final image?

Answer 1

Both lenses are positive so both focal lengths are convex lens. • - for lens } lens formula? / - - 1 -29 t = = £ 1 V 7.9 = v= 10.86 cm. Image is formed on other side of object so rays are reblly meeting there & Real} m= ty = i he = thi = - 1-273 cm -0-3744 = so Inverted -ve port (6) Read and Inverted (A)

10.86cm a u au -> u=-/29 - 10.86) cm = -18.14 6m E for lens @ Last - I v -18.14 - 5.6 5.6 V = between 8.1 cm. object & distance Find Image - ut 29 tu' = 29 +29+ 8.1 = 66. 1 cm.