Consider the compound optical system shown in the diagram where two thin lenses

Question

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 distanc 29 cm. Please answer parts E, F, & G correctly. The answer to B = -0.3034, the answer to C = 8.5 and the answer to D = real.

(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 29 cm tall object is placed as indicated in part (a), and the image formed 0.88 cm tall, what is the magnification of the is 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(e) What is the image distance, in centimeters, for the second lens? 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? Final Correct Answer Student Answer Grade Virtual and inverted. Not available until end date 100% Grade Detail Student Feedback Correct see above 0% Final Answer Credit 100%

Explanation / Answer

part(a)

real , inverted

part(b)

magnification M = image height/object height

M1 = 0.88/2.9 = 0.303

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part(c)

M1 = -di1/d01

0.303 = -di1/d01

d01 = -di1/0.303

1/d01 + 1/di1 = 1/f1

0.303/-di1 + 1/di1 = 1/6.5

image distance di1 = 4.53 cm

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part(d)

real

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part(e)

for second lens

d02 = 29-4.53 = 24.47 cm

1/d02 = 1/di2 = 1/f2

1/24.47 + 1/di2 = 1/42

di2 = -58.6 cm

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part(f)

magnification M2 = -di2/d20 = 58.6/24.47 = 2.4

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part(g)

M = m1*m2 = -0.303*2.4 = -0.73

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part(h)

virtual inverted

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