Chapter 34, Problem 85 GO Your answer is partially correct. Try again. In the fi

Question

Chapter 34, Problem 85 GO Your answer is partially correct. Try again. In the figure below, stick figure 0 (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 (mounted in box 1) is a converging lens, 0 is at object distance p1 = 3.00 cm, and the distance between the lens and its focal point is 9.00 cm. Lens 2 (box 2) is a diverging lens at distance d = 17.0 cm, and the distance between the lens and its focal point is 9.00 cm. What is the overall magnification of the system? Number 1.10232 Units This answer has no units

Explanation / Answer

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    3   cm
do = object distance =    9   cm
di = image distance      
      
Thus, solving for di,      
      
di =    4.5   cm

Thus, this will be 17.0 - 4.5 cm = 12.5 cm from the second lens.
      
Thus, the magnification, m = -di/do,      
      
m1 =    -0.5  

For the second lens,

Note that      
      
1/f = 1/do + 1/di      
      
where      
      
f = focal length =    -9   cm
do = object distance =    21.5   cm
di = image distance      
      
Thus, solving for di,      
      
di =    -6.344262295   cm
      
Thus, the magnification, m = -di/do,      
      
m2 =    0.295081967  

Thus, the overall magnification is

mtot = m1m2 = -0.1475 [ANSWER]

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