(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 7.5 cm (left lens) and 35.5 cm (right lens) are separated by a distance 26 cm. Otheexpertta.com 13 Part (a) If an object is placed a distance do >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? R2 13% Part (b) If a 3.5 cm tall object is placed as indicated in part (a), and the image formed is 0.67 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? A 13% Part (e) What is the image distance, in centimeters, for the second lens? A 13% Part (f What is the magnification of the second lens? A 13% Part (g) What is the total magnification of this compound optical system? Grade Summary Deductions Potential 100% sino cos0 tan0 It 7 8 9 Submissions HOME Attempts remaining: 15 cotano asin acos0 4 5 6 (09 per attempt) detailed view atan() acotano si 1 2 3 cosh0 tanh0 cotanh0 Degrees Radians 0 BACKSPACE Feedback give up Submit HintExplanation / Answer
Part B)
ho = object height = 3.5 cm
hi = image heigt = 0.67 m
m = magnification = - hi/ho = - 0.67/3.5 = - 0.191
Part E)
for the first lens
m = - di/do
- 0.191 = - di/do
di = 0.191 do eq-1
using the lens equation
1/f = 1/di + 1/do
using eq-1
1/7.5 = 1/( 0.191 do ) + 1/do
do = 46.8 cm
using eq-1
di = 0.191 do = 0.191 x 46.8 = 8.94 cm
D = saperation between the two lenses = 26 cm
do' = object distance for the second lens = D - di = 26 - 8.94 = 17.1 cm
f' = focal length of second lens = 35.5 cm
di' = image distance for second lens
using the lens equation
1/f'= 1/d'i + 1/d'o
1/35.5 = 1/d'i + 1/17.1
d'i = - 33 cm
f)
for second lens , m' = - d'i/d'o = - (-33)/17.1 = 1.93
g)
Total magnification = m m' = (- 0.191) (1.93) = - 0.37
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