(13%) Problem 4: Consider the compound optical system shown in the diagram, wher

Question

(13%) Problem 4: Consider the compound optical system shown in the diagram, where two thin lenses of focal lengths 7.5 cm (left lens) and 37 cm (right lens) are separated by a distance 22 cm 0 Otheexpertta.com ? 13% Part (a) If an object is placed a distance d >f 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 3.5 cm tall object is placed as indicated in part (a), and the image formed is 0.39 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? S8 13% Part (e) what is the image distance, in centimeters, for the second lens? Grade Summary Deductions Potential i2 8% 92% Submissions Attempts remaining: 3 (4% per attempt) detailed view cos) tan in HOME cotanOa asin)acos) 12 3 0 atan)acotan) sinh) cosh0tanh0 cotanh0 Degrees Radians 4% 4% END 2 (0 BACKSPACE DEL CLEAR Submit Hint Hints: 1 % deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback 13% Part (f) What is the magnification of the second lens? 13% Part (g) What is the total magnification of this compound optical system? 13% Part (h) is the image created by the second lens real or virtual? Is it upright or inverted?

Explanation / Answer

for first lens:

f1 = 7.5 cm

m = - q1 / p1 = - 0.39 / 3.5

q1 = (0.39/3.5)p1

1/f1 = 1/p1 + 1/q1

1/7.5 = 1/p1 + 3.5/0.39p1

p1 = 74.8 cm

and q1 = (0.39/3.5)p1 = 8.34 cm  

object distance for 2nd lens, p2 = 22 - 8.34

p2 = 13.7 cm  

f2 = 37 cm

1/37 = 1/13.7 + 1/q2

q2 = - 21.7 cm

(E) di2 = q2 = -21.7 cm

(F) m2 = - q2 / p2 = - (-21.7)/(13.7)

m2 = 1.58  

(G) m = m1 m2 = (-0.39/3.5)(1.58)

m = - 0.18

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