(20%) Problem 4: Consider the compound optical system shown in the diagram, wher
ID: 1770683 • Letter: #
Question
(20%) Problem 4: Consider the compound optical system shown in the diagram, where two thin lenses of focal lengths 5.5 cm (left lens) and 32 cm (right lens) are separated by a distance 23 cm 0 ©theexpertta.com 13% Part (a) If an object is placed a distance do real or virtual, and inverted or upright? 1 to the left of the first lens (the left lens), will the resulting image from the first lens be Grade Summary Deductions Potential OReal, UprightOVirtual, Inverted OReal, Inverted O Virtual, Upright 0% 100% Submissions Attempts remaining: 4 % per attempt) detailed view Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: Feedback: 0% deduction per feedback - 13% Part (b) If a 2.6 cm tall object is placed as indicated in part (a), and the image formed is 0.76 cm tall what is the magnification of the first lens? - 13% Part (c) Using the information from part (b), calculate the image distance, in centimeters, from the first lens 13% Part (d) Does the image formed by the first lens serve as a real or a virtual object for the second lens? 13% Part (f) What is the magnification of the second lens? 13% Part (h) Is the image created by the second lens real or virtual? Is it upright or inverted? D 13% Part (e) What is the image distance, in centimeters, for the second lens? D 13% Part (g) What is the total magnification of this compound optical system?Explanation / Answer
(a) if object distance > focal length
then image will be real.
(b) m = hi / ho = - 0.76/2.6 ( m is negative for real image)
m = - 0.292
(C) m = - di / do = - 0.292
di = 0.292 do
putting in 1/f = 1/do + 1/di
1/5.5 = 1/do + 1/0.292 d0
do = 24.3 cm
(d) now this image will work as object for second lens.
do' = 23 - 24.3 = -1.31 cm
so this is virtual object.
(e)
f = 32 cm
1/32 = 1/-1.31 + 1/di'
di' = 1.26 cm
(e) 1.26 cm
(f) m = - di'/do' = 0.965
(g) m = - 0.282
(h) image is real. and inverted.
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