(8%) Problem 12: Consider the compound optical system shown in the diagram, wher
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(8%) Problem 12: Consider the compound optical system shown in the diagram, where two thin lenses of focal lengths 6-5 cm (left lens) and 42 cm (right lens are separated by a distance 29 cm. Otheexpertta.com 13% Part (a) If an object is placed a distance do fi to the left of the first lens (the left lens), will the resulting image from the first lens be real or virtual, and inverted or upright? 13% Part (b) If a 2.9 cm tall object is placed as indicated in part (a), and the image formed is 0.88 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? 32 13% Part (e) What is the image distance, in centimeters, for the second lens? 32 13% Part (f What is the magnification of the second lens? 13% Part (g) What is the total magnification of this compound optical system? Grade Summary M 1.6 Deductions Potential 100% sino cos0 tan0 7 8 9 HOME Submissions Attempts remaining: 14 4 5 6 cotano asin0 acos0 0% per attempt) detailed view atan0 acotan0 si cosh0 tanh0 cotanh END 0 BACKSPACE DEL CLEAR Degrees Radians Submit Hint Hints 5% deduction per hint. Hints remaining 1 Feedback 5% deduction per feedback.Explanation / Answer
From the given question,
for first lens
The magnification of first lens= 2.9/0-0.88= 0.3
(e)From the lens formula
1/v - 1/u= 1/f
1/3u - 1/(-u) = 1/6.5
4/3u=1/6.5
u=8.67cm
v=26cm
For second lens
Object distance= 29-26=3cm
image distance=v
focal length=42cm
1/v - 1/u = 1/f
1/v - 1/(-3) = 1/42
1/v = 1/42 - 1/3
1/v = -13/42
v=-3.23
Image distance for second lens is 3.23 cm towards left from second lens.
(f)magnification of second lens(m2)= 3.23/3=1.08
(g) Total magnification= m1 x m2=0.3 x 1.08 = 0.323
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